Chapter 01: Mathematical Logic

01 Mathematical Logic

Subtopics

1.1 Statement 1.2 Logical Connectives, Compound Statements and Truth Tables 1.3 Statement Pattern and Logical Equivalence Tautology, Contradiction and Contingency 1.4 Quantifiers and Quantified Statements 1.5 Duality 1.6 Negation of Compound Statement 1.7 Algebra of Statements (Some Standard equivalent Statements) 1.8 Application of Logic to Switching Circuits

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Std. XII : Perfect Maths - I

Type of Problems Exercise Q. Nos. Identify the statements and write 1.1 Q.1 down their Truth Value Miscellaneous Q.1 Express the statements in Symbolic 1.2 Q.1 Form/Write the statement in 1.4 Q.1, 2 Symbolic Form Miscellaneous Q.5 1.2 Q.2 Write the Truth values of 1.4 Q.3, 5 Statements 1.6 Q.1 Miscellaneous Q.2, 3, 9 Write the Negation of 1.3 Q.1 Statements/Using the Rules of 1.8 Q.1, 2, 4 Negation write the Negation of Miscellaneous Q.4, 11, 22 Statements Write the Verbal statement for the 1.4 Q.6 given Symbolic Statement Miscellaneous Q.6 Converse, Inverse and 1.4 Q.4 Contrapositive of the statement Miscellaneous Q.19, 21 Using Quantifiers Convert Open 1.6 Q.2 sentences into True statement 1.4 Q.7 Prepare the Truth Table/Find Truth 1.5 Q.1 Values of p and q for given cases Miscellaneous Q. 12, 15 Examine the statement Patterns 1.5 Q.3 (Tautology, Contradiction, Contingency) Miscellaneous Q.13, 14, 16 Using Truth Table, Verify Logical 1.5 Q.2 Equivalence Miscellaneous Q.7, 18 Write Dual of the statement 1.7 Q.1, 2, 3, 4 Algebra of statements (without 1.8 Q.3 using Truth Table verify the Logical Equivalence)/Rewrite the statement Miscellaneous Q.8, 17, 20 without using the conditional form Change the statements in the form if Miscellaneous Q.10 then Applications of logic to switching 1.9 Q.1 to 5 circuits Miscellaneous Q.23 to 29

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Chapter 01: Mathematical Logic

Introduction Note: The sentences like exclamatory, interrogative, Mathematics is an exact science. Every imperative etc., are not considered as statements as mathematical statement must be precise. Hence, the truth value for these statements cannot be there has to be proper reasoning in every determined. mathematical proof. Proper reasoning involves logic. The study of logic Open sentence helps in increasing one’s ability of systematic and logical reasoning. It also helps to develop the skills An open sentence is a sentence whose truth can vary of understanding various statements and their according to some conditions, which are not stated validity. in the sentence. Logic has a wide scale application in circuit Note: designing, computer programming etc. Hence, the Open sentence is not considered as statement in logic.

study of logic becomes essential. For example: Statement and its truth value i. x × 5 = 20 There are various means of communication viz., This is an open sentence as its truth depends verbal, written etc. Most of the communication on value of x (if x = 4, it is true and if x ≠ 4, it involves the use of language whereby, the ideas are is false). conveyed through sentences. ii. Chinese food is very tasty. There are various types of sentences such as: This is an open sentence as its truth varies i. Declarative (Assertive) from individual to individual. ii. Imperative (A command or a request) iii. Exclamatory (Emotions, excitement) Exercise 1.1 iv. Interrogative (Question) State which of the following sentences are Statement statements. Justify your answer. In case of the statements, write down the truth value. A statement is a declarative sentence which is either true or false but not both simultaneously. Statements i. The Sun is a star. are denoted by the letters p, q, r…. ii. May God bless you ! For example: iii. The sum of interior angles of a triangle is i. 3 is an odd number. 180 °. ii. 5 is a perfect square. iv. Every real number is a complex number. iii. Sun rises in the east. v. Why are you upset? iv. x + 3 = 6, when x = 3. vi. Every quadratic equation has two real roots. Truth Value vii. −9 is a rational number. A statement is either True or False. The Truth value 2 − − − of a ‘true’ statement is defined to be T (TRUE) and viii. x 3x + 2 = 0, implies that x = 1 or x = 2. that of a ‘false’ statement is defined to be F ix. The sum of cube roots of unity is one. x. Please get me a glass of water. (FALSE). xi. He is a good person. Note: 0 and 1 can also be used for T and F xii. Two is the only even prime number. respectively. xiii. sin 2 θ = 2sin θ cos θ for all θ ∈ R. Consider the following statements: xiv. What a horrible sight it was ! i. There is no prime number between 23 and 29. xv. Do not disturb. ii. The Sun rises in the west. 2 − − − iii. The square of a real number is negative. xvi. x 3x 4 = 0, x = 1. iv. The sum of the angles of a plane triangle is xvii. Can you speak in French? 180 °. xviii. The square of every real number is positive. Here, the truth value of statement i. and iv. is T and xix. It is red in colour. that of ii. and iii. is F. xx. Every parallelogram is a rhombus. 3

Std. XII : Perfect Maths - I Solution : Compound Statements: i. It is a statement which is true, hence its truth The new statement that is formed by combining two value is ‘T’. or more simple statements by using logical ii. It is an exclamatory sentence, hence, it is not a connectives are called compound statements. statement. iii. It is a statement which is true, hence its truth Component Statements: value is ‘T’. The simple statements that are joined using logical iv. It is a statement which is true, hence its truth connectives are called component statements. value is ‘T’. For example: v. It is an interrogative sentence, hence it is not a Consider the following simple statements, statement. i. e is a vowel vi. It is a statement which is false, hence its truth ii. b is a consonant value is ‘F’. vii. It is a statement which is false, hence its truth These two component statements can be joined by value is ‘F’. using the logical connective ‘or’ as shown below: viii. It is a statement which is false, hence its truth ‘e is a vowel or b is a consonant’ value is ‘F’. The above statement is called compound statement ix. It is a statement which is false, hence its truth formed by using logical connective ‘or’. value is ‘F’. x. It is an imperative sentence, hence it is not a Truth Table statement. A table that shows the relationship between truth xi. It is an open sentence, hence it is not a values of simple statements and the truth values of statement. compounds statements formed by using these simple xii. It is a statement which is true, hence its truth statements is called truth table. value is ‘T’. xiii. It is a statement which is true, hence its truth Note: value is ‘T’. The truth value of a compoud statement depends xiv. It is an exclamatory sentence, hence it is not a upon the truth values of its component statements. statement. Logical Connectives xv. It is an imperative sentence, hence it is not a ∧ statement. A. AND [ ] (Conjunction): xvi. It is a statement which is true, hence its truth If p and q are any two statements connected value is ‘T’. by the word ‘and’, then the resulting xvii. It is an interrogative sentence, hence, it is not compound statement ‘p and q’ is called a statement. conjunction of p and q which is written in the ∧ xviii. It is a statement which is false, hence its truth symbolic form as ‘p q’. value is ‘F’. (Since, 0 is a real number and For example: square of 0 is 0 which is neither positive nor p: Today is a pleasant day. negative). q: I want to go for shopping. xix. It is an open sentence, hence it is not a The conjunction of above two statements is statement. (The truth of this sentence depends ‘p ∧ q’ i.e. ‘Today is a pleasant day and I want upon the reference for the pronoun ‘It’.) to go for shopping’. xx. It is a statement which is false, hence its truth A conjunction is true if and only if both p and value is ‘F’. q are true. Logical Connectives, Compound Statements and Truth table for conjunction of p and q is as Truth Tables shown below:

∧ Logical Connectives: p q p q The words or group of words such as “and, or, if …. T T T then, if and only if, not” are used to join or connect T F F two or more simple sentences. These connecting F T F words are called logical connectives. F F F 4

Chapter 01: Mathematical Logic Note: Solution : The words such as but, yet, still, inspite, i. Let p : Mango is a fruit, q : Potato is a vegetable. though, moreover are also used to connect the ∴ The symbolic form of the given statement is simple statements. ∧ p q. These words are generally used by replacing ii. Let p : We play football, q : We go for cycling. ‘and’. ∴ The symbolic form of the given statement is ∨ ∨ B. OR [ ] (Disjunction): p q. If p and q are any two statements connected by iii. Let p : Milk is white, q : Grass is green. the word ‘or’, then the resulting compound ∴ The symbolic form of the given statement is statement ‘p or q’ is called disjunction of p and ∨ p q. q which is written in the symbolic form as iv. Let p : Rahul has physical disability, ∨ ‘p q’. q : Rahul stood first in the class. The word ‘or’ is used in English language in The given statement can be considered as ‘Rahul two distinct senses, exclusive and inclusive. has physical disability and he stood first in the For example: class.’ ∴ i. Rahul will pass or fail in the exam. The symbolic form of the given statement is p ∧ q. ii. Candidate must be graduate or post-graduate. v. Let p : Jagdish stays at home, In eg. (i), ‘or’ indicates that only one of the q : Shrijeet and Shalmali go for a movie. The given statement can be considered as two possibilities exists but not both which is ‘Jagdish stays at home and Shrijeet and Shalmali called exclusive sense of ‘or’. In eg. (ii), ‘or’ go for a movie.’ indicates that first or second or both the ∴ The symbolic form of the given statement is possibilities may exist which is called ∧ p q. inclusive sense of ‘or’. 2. Write the truth values of following A disjunction is false only when both p and q statements. are false. i. 3 is a rational number or 3 + i is a Truth table for disjunction of p and q is as complex number. shown below: ii. Jupiter is a planet and Mars is a star. iii. 2 + 3 ≠ 5 or 2 × 3 < 5 ∨ p q p q iv. 2 × 0 = 2 and 2 + 0 = 2 T T T v. 9 is a perfect square but 11 is a prime T F T number. F T T vi. Moscow is in Russia or London is in F F F France.

Solution : Exercise 1.2 3 i. Let p : is a rational number, 1. Express the following statements in q : 3 + i is a complex number. ∴ The symbolic form of the given statement is symbolic form: p ∨ q. i. Mango is a fruit but potato is a Since, truth value of p is F and that of q is T. vegetable. ∴ truth value of p ∨ q is T ii. Either we play football or go for cycling. ii. Let p : Jupiter is a planet, iii. Milk is white or grass is green. q : Mars is a star. iv. Inspite of physical disability, Rahul ∴ The symbolic form of the given statement is stood first in the class. p ∧ q. v. Jagdish stays at home while Shrijeet Since, truth value of p is T and that of q is F. and Shalmali go for a movie. ∴ truth value of p ∧ q is F 5

Std. XII : Perfect Maths - I iii. Let p : 2 + 3 ≠ 5, vii. Zero is not a complex number. q : 2 × 3 < 5. viii. Re (z) ≤ | z |. ∴ The symbolic form of the given statement is ix. The sun sets in the East. p ∨ q. x. It is not true that the mangoes are inexpensive. Since, truth value of both p and q is F. Solution : ∴ truth value of p ∨ q is F i. Rome is not in Italy. ≠ iv. Let p : 2 × 0 = 2, ii. 5 + 5 10 q : 2 + 0 = 2. iii. 3 is not greater than 4. ∴ The symbolic form of the given statement is iv. John is not good in river rafting. π p ∧ q. v. is not an irrational number. Since, truth value of p is F and that of q is T. vi. The square of a real number is not positive. ∴ truth value of p ∧ q is F vii. Zero is a complex number. viii. Re (z) > |z|. v. Let p : 9 is a perfect square, ix. The sun does not set in the East. q : 11 is a prime number. x. It is true that the mangoes are inexpensive. ∴ The symbolic form of the given statement is p ∧ q. D. If….then (Implication, →) (Conditional): Since, truth value of both p and q is T. If p and q are any two simple statements, then ∴ truth value of p ∧ q is T the compound statement, ‘if p then q’, vi. Let p : Moscow is in Russia, meaning “statement p implies statement q or q : London is in France. statement q is implied by statement p”, is ∴ The symbolic form of the given statement is called a conditional statement and is denoted p ∨ q. by p → q or p ⇒ q. Since, truth value of p is T and that of q is F. Here p is called the antecedent (hypothesis) ∴ ∨ truth value of p q is T and q is called the consequent (conclusion). C. Not [~] (Negation): For example: If p is any statement then negation of p Let p: I travel by train. i.e., ‘not p’ is denoted by ~p. Negation of any q: My journey will be cheaper. simple statement p can also be formed by Here the conditional statement is writing ‘It is not true that’ or ‘It is false that’, → before p. ‘p q: If I travel by train then my journey will be cheaper.’ For example: Conditional statement is false if and only if p : Mango is a fruit. antecedent is true and consequent is false. ~p : Mango is not a fruit. Truth table for conditional is as shown Truth table for negation is as shown below: below: p ~p → T F p q p q F T T T T T F F Note: If a statement is true its negation is false F T T and vice-versa. F F T

Exercise 1.3 Note: Equivalent forms of the conditional Write negations of the following statements: statement i. Rome is in Italy. p → q: ii. 5 + 5 = 10 a. p is sufficient for q. iii. 3 is greater than 4. b. q is necessary for p. iv. John is good in river rafting. c. p implies q. v. π is an irrational number. d. p only if q. vi. The square of a real number is positive. e. q follows from p. 6

Chapter 01: Mathematical Logic

E. Converse, Inverse and Contrapositive Exercise 1.4 statements: If p → q is given, then its 1. Express the following in symbolic form. converse is q → p i. I like playing but not singing. inverse is ~p → ~q ii. Anand neither likes cricket nor tennis. contrapositive is ~q → ~p iii. Rekha and Rama are twins. For example: iv. It is not true that ‘i’ is a real number. Let p : Smita is intelligent. v. Either 25 is a perfect square or 41 is q : Smita will join Medical. divisible by 7. i. q → p: If Smita joins Medical then she vi. Rani never works hard yet she gets is intelligent. good marks. ii. ~p → ~q: If Smita is not intelligent then vii. Eventhough it is not cloudy, it is still she will not join Medical. raining. iii. ~q → ~p: If Smita does not join Medical Solution : then she is not intelligent. i. Let p: I like playing, q: I like singing, ∴ The symbolic form of the given statement is Consider, the following truth table: p ∧ ~q. p q p→q ~p ~q q→p ~q →~p ~p →~q T T T F F T T T ii. Let p: Anand likes cricket, q: Anand likes tennis. T F F F T T F T ∴ The symbolic form of the given statement is ∧ F T T T F F T F ~p ~q. F F T T T T T T iii. In this statement ‘and’ is combining two nouns From the above table, we conclude that and not two simple statements. i. a conditional statement and its Hence, it is not used as a connective, so given contrapositive are always equivalent. statement is a simple statement which can be ii. converse and inverse of the conditional symbolically expressed as p itself. statement are always equivalent. iv. Let, p : ‘i’ is a real number. ↔ ∴ F. If and only if (Double Implication, ) The symbolic form of the given statement is ~p. (Biconditional): v. Let p : 25 is a perfect square, If p and q are any two statements, then q : 41 is divisible by 7. ‘p if and only if q’ or ‘p iff q’ is called the ∴ biconditional statement and is denoted by The symbolic form of the given statement is ∨ p ↔ q. Here, both p and q are called p q. implicants. vi. Let p : Rani works hard, q : Rani gets good marks. For example: ∴ Let p : price increases The symbolic form of the given statement is ∧ q : demand falls ~p q. Here the Biconditional statement is vii. Let p : It is cloudy, q : It is still raining. ‘p ↔ q : Price increases if and only if demand ∴ The symbolic form of the given statement is falls’. ∧ ~p q. A biconditional statement is true if and only if both the implicants have same truth value. 2. If p: girls are happy, q: girls are playing, Truth table for biconditional is as shown express the following sentences in symbolic below: form. i. Either the girls are happy or they are p q p ↔ q not playing. T T T ii. Girls are unhappy but they are T F F playing. F T F iii. It is not true that the girls are not F F T playing but they are happy. 7

Std. XII : Perfect Maths - I Solution : vi. Let p: 3 + 5 > 7, q: 4 + 6 < 10 ∨ ∧ ∴ i. p ~q ii. ~p q The symbolic form of the given statement is ∧ p ↔ q . iii. ~(~q p) Since, truth value of p is T and that of q is F. ∴ truth value of p ↔ q is F 3. Find the truth value of the following statements. 4. State the converse, inverse and i. 14 is a composite number or 15 is a contrapositive of the following conditional prime number. statements: ii. Neither 21 is a prime number nor it is i. If it rains then the match will be divisible by 3. cancelled. iii. It is not true that 4+3i is a real ii. If a function is differentiable then it is number. continuous. iv. 2 is the only even prime number and 5 iii. If surface area decreases then the divides 26. pressure increases. v. Either 64 is a perfect square or 46 is a iv. If a sequence is bounded then it is prime number. convergent. vi. 3 + 5 > 7 if and only if 4 + 6 < 10. Solution : Solution : i. Let p : It rains, q : the match will be cancelled. i. Let p : 14 is a composite number, ∴ The symbolic form of the given statement is q : 15 is a prime number. p → q. ∴ The symbolic form of the given statement is Converse: q → p p ∨ q. i.e., If the match is cancelled then it rains. Since, truth value of p is T and that of q is F. Inverse: ~p → ~q ∴ truth value of p ∨ q is T. i.e., If it does not rain then the match will not be cancelled. ii. Let p: 21 is a prime number, Contrapositive: ~q → ~p q: 21 is divisible by 3. i.e. If the match is not cancelled then it does ∴ The symbolic form of the given statement is not rain. ∧ ~p ~q. ii. Let p: A function is differentiable, Since, truth value of p is F and that of q is T q: It is continuous. ∴ truth value of ~p ∧ ~q is F. ∴ The symbolic form of the given statement is → iii. Let p: 4 + 3i is a real number. p q. Converse: q → p ∴ The symbolic form of the given statement is ~p. i.e. If a function is continuous then it is Since, truth value of p is F. differentiable. ∴ truth value of ~p is T. Inverse: ~p → ~q iv. Let p: 2 is the only even prime number, i.e. If a function is not differentiable then it q: 5 divides 26. is not continuous. ∴ The symbolic form of the given statement is Contrapositive: ~q → ~p p ∧ q. i.e. If a function is not continuous then it is Since, truth value of p is T and that of q is F not differentiable. ∴ ∧ truth value of p q is F. iii. Let p: Surface area decreases, v. Let p: 64 is a perfect square, q: The pressure increases. q: 46 is a prime number. ∴ The symbolic form of the given statement is ∴ The symbolic form of the given statement is p → q ∨ p q. Converse: q → p Since, truth value of p is T and that of q is F. i.e. If the pressure increases then the surface ∴ truth value of p ∨ q is T area decreases. 8

Chapter 01: Mathematical Logic Inverse: ~p → ~q iv. ~(p ∧ ∼r) ∨ (∼q ∨ s) i.e. If the surface area does not decrease ≡ ∼ (T ∧ T) ∨ (F ∨ F) then the pressure does not increase. ≡ ∼ (T) ∨ F Contrapositive: ~q → ~p ≡ F ∨ F i.e. If the pressure does not increase then the ≡ F surface area does not decrease. ∴ truth value of the given statement is false.

iv. Let p: A sequence is bounded, q: It is convergent. 6. If p: It is daytime, q: It is warm ∴ The symbolic form of the given statement is Give the compound statements in verbal p → q form denoted by Converse: q → p i. p ∧ ∼q [Oct 14] i.e. If a sequence is convergent then it is ii. p ∨ q bounded. iii. p → q Inverse: ~p → ~q iv. q ↔ p [Oct 14] i.e. If a sequence is not bounded then it is Solution : not convergent. i. It is daytime but it is not warm. [1 Mark] Contrapositive: ~q → ~p ii. It is daytime or it is warm. i.e. If a sequence is not convergent then it is iii. If it is daytime then it is warm. not bounded. iv. It is warm if and only if it is daytime.

[1 Mark] 5. If p and q are true and r and s are false

statements, find the truth value of the 7. Prepare the truth tables for the following: following statements: i. ∼p ∧ q ii. p → (p ∨ q) i. (p ∧ q) ∨ r iii. ∼p ↔ q ii. p ∧ (r → s) Solution : iii. (p ∨ s) ↔ (q ∧ r) i. ∼p ∧ q iv. ~ (p ∧ ∼r) ∨ (∼q ∨ s) Solution : p q ~p ~p ∧ q Given that p and q are T and r and s are F. T T F F i. (p ∧ q) ∨ r T F F F ≡ (T ∧ T) ∨ F F T T T ≡ T ∨ F F F T F

≡ T ii. p → (p ∨ q) ∴ truth value of the given statement is true. p q p ∨ q p → (p ∨ q) ii. p ∧ (r → s) T T T T ≡ ∧ → T (F F) T F T T ≡ T ∧ T F T T T ≡ T F F F T ∴ truth value of the given statement is true. ∼ ↔ iii. p q iii. (p ∨ s) ↔ (q ∧ r) p q ~p ~p ↔ q ≡ ∨ ↔ ∧ (T F) (T F) T T F F ≡ T ↔ F T F F T ≡ F F T T T ∴ truth value of the given statement is false. F F T F 9

Std. XII : Perfect Maths - I

∧ ≡ ∨ nd Statement Pattern and Logical Equivalence ii. ~(p q) ~p ~q (De-Morgan’s 2 Law) identica l A. Statement Pattern Let, p, q, r,… be simple statements. A p q ~p ~q p ∧ q ~(p ∧ q) ~p ∨ ~q compound statement obtained from these T T F F T F F simple statements and by using one or more T F F T F T T connectives ∧, ∨, ~, →, ↔ is called a F T T F F T T statement pattern. F F T T F T T Following points must be noted while preparing truth tables of the statement iii. p → q ≡ (~p) ∨ q patterns: identical i. Parentheses must be introduced wherever necessary. p q ~p p → q l ~p ∨ q For example: T T F T T ~ (p ∧ q) and ~ p ∧ q are not the same. T F F F F ii. If a statement pattern consists of ‘n’ F T T T T statements and ‘m’ connectives, then F F T T T n truth table consists of 2 rows and ↔ ≡ → ∧ → (m + n) columns. iv. p q (p q) (q p) [Mar 98, Oct 00, 01,04] B. Logical equivalence identical Two logical statements are said to be equivalent if and only if the truth values in p q p → q q → p p ↔ q (p → q) ∧ (q → p) their respective columns in the joint truth table T T T T T T are identical. T F F T F F If S1 and S2 are logically equivalent statement F T T F F F patterns, we write F F T T T T S1 ≡ S2. For example: Tautology, Contradiction and Contingency To prove: p ∧ q ≡ ~(p → ~q) Tautology [Mar 08] p q p ∧ q ~q (p → ~q) ~(p → ~q) A statement pattern having truth value always T, T T T F F T irrespective of the truth values of its component statement is called Tautology. T F F T T F F T F F T F For example, consider (p ↔ q) ↔ (q ↔ p) F F F T T F

p q ↔ ↔ ↔ ↔ ↔ In the above truth table, all the entries in the p q q p (p q) (q p) columns of p ∧ q and ~(p → ~q) are identical. T T T T T ∴ ∧ ≡ → T F F F T p q ~(p ~q). F T F F T Note: F F T T T i. ~(p ∨ q) ≡ ~p ∧ ~q (De-Morgan’s 1 st Law) [Mar 96] In the above truth table, all the entries in the last

identical column are T. ∴ The given statement pattern is a tautology.

p q ~p ~q p ∨ q ~(p ∨ q) ~p ∧ ~q Contradiction T T F F T F F T F F T T F F A statement pattern having truth value always F, F T T F T F F irrespective of the truth values of its component F F T T F T T statement is called a Contradiction. 10

Chapter 01: Mathematical Logic ∧ ∧ → For example, consider p ~ p ii. (p q) (~p) p ~ p p ∧ ~ p p q p ∧ q ~p (p ∧ q) → (~p) T F F T T T F F F T F T F F F T In the above truth table, all the entries in the last column F T F T T are F. F F F T T ∴ The given statement pattern is a contradiction. iii. (p → q) ↔ (~p ∨ q)

Contingency p q p → q ~p ~p ∨ q (p → q) ↔ A statement pattern which is neither a tautology nor (~p ∨ q) a contradiction is called Contingency. T T T F T T T F F F F T For example, consider (p ↔ q) ∧ ~(p → ~ q) F T T T T T (p ↔ q) ∧ F F T T T T p q p ↔ q ~q p →~q ~(p → ~q) → ~(p ~q) iv. (p ↔ r) ∧ (q ↔ p) T T T F F T T T F F T T F F p q r p ↔ r q ↔ p (p ↔ r) ∧ ↔ F T F F T F F (q p) F F T T T F F T T T T T T

T T F F T F In the above truth table, the entries in the last T F T T F F column are a combination of T and F. T F F F F F ∴ The given statement pattern is neither a F T T F F F tautology nor a contradiction, it is a F T F T F F contingency. F F T F T F Exercise 1.5 F F F T T T

1. Prepare the truth table of the following v. (p ∨ ~q) → (r ∧ p) statement patterns: → ∧ → (p ∨ ~q) → i. [(p q) q] p p q r ~q p ∨ ~ q r ∧ p ii. (p ∧ q) → (~p) (r ∧ p) iii. (p → q) ↔ (~p ∨ q) T T T F T T T iv. (p ↔ r) ∧ (q ↔ p) T T F F T F F v. (p ∨ ~q) → (r ∧ p) T F T T T T T Solution : T F F T T F F i. [(p → q) ∧ q] → p F T T F F F T p q p → q (p → q) ∧ q [(p → q) ∧ q] → p F T F F F F T T T T T T F F T T T F F T F F F T F F F T T F F

F T T T F F F T F T 2. Using truth tables, prove the following logical equivalences: Note: i. (p ∧ q) ≡ ~ (p → ~ q) Here, the statement pattern consists of 2 ii. p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q) statements and 3 connectives. Therefore, the truth table has, [Mar 13; Oct 15] 2 ∧ → ≡ → → Number of rows = 2 = 4 iii. (p q) r p (q r) [Oct 14] and number of columns = (2+3) = 5. iv. p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r) 11

Std. XII : Perfect Maths - I

Solution : iv. i. 1 2 3 4 5 6 7 8

p q r q ∧ r p ∨ (q ∧ r) p ∨ q p ∨ r (p ∨q) ∧ (p ∨r) 1 2 3 4 5 6 T T T T T T T T ∧ → → p q p q ~q p ~q ~ (p ~q) T T F F T T T T T T T F F T T F T F T T T T T F F T T F T F F F T T T T F T F F T F F T T T T T T T F F F T T F F T F F F T F F F F T F F F T F The entries in the columns 3 and 6 are identical. F F F F F F F F

∴ (p ∧ q) ≡ ~ (p → ~q) The entries in the columns 5 and 8 are identical ∴ p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r) ii.

1 2 3 4 5 6 7 8 3. Using truth tables examine whether the (p ∧ q) ∨ following statement patterns are tautology, p q p ↔ q ~p ~q p ∧ q ~p ∧ ~q contradiction or contingency. (~p ∧ ~q) i. (p ∧ ~ q) ↔ (p → q) [Mar 13] T T T F F T F T ii. (~p ∧ q) ∧ (q → p) T F F F T F F F iii. (p ∧ q) ∨ (p ∧ r) F T F T F F F F iv. [(p ∨ q) ∨ r] ↔ [p ∨ (q ∨ r)] F F T T T F T T ∨ ∧ ∨ v. (p q) (p r) Solution : The entries in columns 3 and 8 are identical. i. ∴ ↔ ≡ ∧ ∨ ∧ p q (p q) (~p ~q) [1 Mark] 1 2 3 4 5 6 [1 mark each for column 3 and column 8] (p ∧ ~ q) ↔ p q ~q p ∧ ~q p → q (p → q) iii. T T F F T F T F T T F F 1 2 3 4 5 6 7 F T F F T F p → F F T F T F p q r p ∧ q q → r (p ∧ q) → r (q → r) In the above truth table, all the entries in the last T T T T T T T column are F. T T F T F F F ∴ (p ∧ ~q) ↔ (p → q) is a contradiction. [1 Mark] [1 mark each for column 5 and column 6] T F T F T T T T F F F T T T ii. F T T F T T T (~p ∧ q) ∧ p q ~p ~p ∧ q q → p F T F F F T T (q → p) T T F F T F F F T F T T T T F F F T F F F F F T T T F T T T F F

F F T F T F The entries in the columns 6 and 7 are identical. In the above truth table, all the entries in the last ∴ (p ∧ q) → r ≡ p → (q → r) [1 Mark] column are F. [1 mark each for column 6 and column 7] ∴ (~p ∧ q) ∧ (q → p) is a contradiction. 12

Chapter 01: Mathematical Logic iii. i. Universal Quantifier: p q r p ∧ q p ∧ r (p ∧ q) ∨ (p ∧ r) The symbol ‘ ∀’ stands for “all values of” and T T T T T T is known as universal quantifier. T T F T F T For example, Consider A = {1, 2, 3} T F T F T T ∀ ∈ T F F F F F Let p: x A, x < 4 F T T F F F Here, the statement p uses the quantifier ‘for ∀ F T F F F F all’( ). F F T F F F This statement is true if and only if each and F F F F F F every element of set A satisfies the condition ‘x < 4’ and is false otherwise. In the above truth table, the entries in the last Here, the given statement is true for all the column are a combination of T and F. elements of set A, as 1, 2, 3 satisfy the ∴ (p ∧ q) ∨ (p ∧ r) is a contingency. ∈ condition, ‘ x A, x < 4’. iv. ii. Existential Quantifier: p q r p ∨ q q ∨ r (p ∨ q) p ∨ (q ∨ r) [(p ∨q) ∨r] The symbol ‘ ∃’ stands for ‘there exists’ and is ∨ r ↔ known as existential quantifier. ∨ ∨ [p (q r)] For example, Consider A = {4, 14, 66, 70}. T T T T T T T T Let p: ∃ x ∈ A such that x is an odd number. T T F T T T T T Here, the statement p uses the quantifier ‘there T F T T T T T T ∃ T F F T F T T T exists’ ( ). F T T T T T T T This statement is true if atleast one element of F T F T T T T T set A satisfies the condition ‘ x is an odd number’ and is false otherwise. F F T F T T T T F F F F F F F T Here, the given statement is false as none of the elements of set A satisfy the condition, In the above truth table, all the entries in the last ‘x ∈ A such that x is an odd number’. column are T. ∴ ∨ ∨ ↔ ∨ ∨ Quantified statement [(p q) r] [p (q r)] is a tautology. v. The statement containing quantifiers is known as p q r p ∨ q p ∨ r (p ∨ q) ∧ (p ∨ r) quantified statement. Generally, an open sentence T T T T T T with a quantifier becomes a statement and is called quantified statement. T T F T T T T F T T T T For example: T F F T T T Use quantifiers to convert open sentence x + 2 < 4 F T T T T T into a statement. F T F T F F Solution : F F T F T F ∃ x ∈ N such that x + 2 < 4, is a true F F F F F F ∈ statement, since x = 1 N satisfies x + 2 < 4. In the above truth table, the entries in the last Exercise 1.6 column are a combination of T and F. ∴ (p ∨ q) ∧ (p ∨ r) is a contingency. 1. If A = {3, 4, 6, 8}, determine the truth value

of each of the following: Quantifiers and Quantified Statements i. ∃ x ∈ A, such that x + 4 = 7 ∀ ∈ Quantifiers ii. x A, x + 4 < 10. ∀ ∈ ≥ Quantifiers are the symbols used to denote a group iii. x A, x + 5 13. of words or a phrase. Generally, two types of iv. ∃ x ∈ A, such that x is odd. quantifiers are used. They are as follows: v. ∃ x ∈ A, such that ( x − 3) ∈ N 13

Std. XII : Perfect Maths - I Solution : Remarks: i. Since x = 3 ∈ A, satisfies x + 4 = 7 i. The symbol ‘ ∼’ is not changed while finding ∴ the given statement is true. the dual. ∴ Its truth value is ‘T’. ii. Dual of a dual is the statement itself. ii. Since x = 6, 8 ∈ A, do not satisfy x + 4 < 10, iii. The special statements ‘t’ (tautology) and ‘c’ ∴ the given statement is false. (contradiction) are duals of each other. ∴ Its truth value is ‘F’ iv. T is changed to F and vice-versa. iii. Since x = 3, 4, 6 ∈ A, do not satisfy x + 5 ≥ 13, Principle of Duality: ∴ the given statement is false. If a compound statement S 1 contains only ~, ∧ and ∨ ∴ Its truth value is ‘F’. and statement S 2 arises from S 1 by replacing ∧ by ∨ ∨ ∧ iv. Since x = 3 ∈ A, satisfies the given statement, and by , then S 1 is a tautology if and only if S 2 is

∴ the given statement is true. a contradiction. ∴ Its truth value is ‘T’. Exercise 1.7

∈ − ∈ v. Since x = 4, 6, 8 A, satisfy ( x 3) N, 1. Write the duals of the following statements: ∴ the given statement is true. i. (p ∧ q) ∨ r ∴ Its truth value is ‘T’. ii. T ∨ (p ∨ q) 2. Use quantifiers to convert each of the iii. p ∧ [∼q ∨ (p ∧ q) ∨ ∼r] following open sentences defined on N, into Solution : a true statement: i. (p ∨ q) ∧ r 2 i. x = 25 ii. 2x + 3 < 15 ii. F ∧ (p ∧ q) 2 iii. x − 3 = 11 iv. x + 1 ≤ 5 iii. p ∨ [~q ∧ (p ∨ q) ∧ ~r] 2 v. x − 3x + 2 = 0 Solution : 2. Write the dual statement of each of the ∃ ∈ 2 i. x N, such that x = 25. following compound statements: This is a true statement since x = 5 ∈ N 2 i. Vijay and Vinay cannot speak Hindi. satisfies x = 25. ii. ∃ x ∈ N, such that 2 x + 3 < 15. ii. Ravi or Avinash went to Chennai. This is a true statement since x = 1, 2, 3, 4, 5 iii. Madhuri has curly hair and brown eyes. [Mar 14] ∈ N satisfy 2 x + 3 < 15. ∃ ∈ − Solution : iii. x N such that x 3 = 11. i. Vijay or Vinay cannot speak Hindi. ∈ This is a true statement since x = 14 N ii. Ravi and Avinash went to Chennai. − satisfies x 3 = 11. iii. Madhuri has curly hair or brown eyes. [1 Mark] 2 iv. ∃ x ∈ N, such that x + 1 ≤ 5. This is a true statement since x = 1, 2 ∈ N 3. Write the duals of the following statements: 2 ≤ satisfy x + 1 5. i. (p ∨ q) ∨ r ≡ p ∨ (q ∨ r) v. ∃ x ∈ N such that x2 − 3x + 2 = 0. ii. p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r) This is a true statement since x = 1, 2 ∈ N Solution : satisfy x2 − 3x + 2 = 0. i. (p ∧ q) ∧ r ≡ p ∧ (q ∧ r) ∧ ∨ ∧ ∨ ∧ ii. p (q r) ≡ (p q) (p r) Duality

4. Write duals of each of the following statements Two compound statements S1 and S 2 are said to be where t is a tautology and c is a contradiction. duals of each other, if one can be obtained from the ∧ ∧ ∧ ∨ other by interchanging ‘∧’ and ‘∨’ and vice-versa. i. p q c ii. ~ p (q c) ∧ ∨ ∧ The connectives ‘ ∧’ and ‘ ∨’ are duals of each other. iii. (p t) (c ~ q) Also, a dual is obtained by replacing t by c and c by Solution : t, where ‘t’ denotes tautology and ‘c’ denotes i. p ∨ q ∨ t ii. ~ p ∨ (q ∧ t) contradiction. iii. (p ∨ c) ∧ (t ∨ ~ q) 14

Chapter 01: Mathematical Logic

c. Associative Law: Negation of compound statement i. (p ∨ q) ∨ r ≡ p ∨ (q ∨ r) ≡ p ∨ q ∨ r i. Negation of conjunction: ii. (p ∧ q) ∧ r ≡ p ∧ (q ∧ r) ≡ p ∧ q ∧ r Negation of the conjunction of two simple statements is the disjunction of their negations. d. Distributive Law: ∧ ≡ ∨ ∨ ∧ ≡ ∨ ∧ ∨ i.e. ~(p q) ~p ~q i. p (q r) (p q) (p r) ∧ ∨ ≡ ∧ ∨ ∧ ii. Negation of disjunction: ii. p (q r) (p q) (p r) Negation of the disjunction of two simple e. Identity Law: statements is the conjunction of their i. p ∨ F ≡ p negations. ii. p ∧ F ≡ F i.e. ~(p ∨ q) ≡ ~p ∧ ~q iii. p ∨ T ≡ T iii. Negation of negation: ∧ ≡ iv. p T p The negation of negation of a simple statement is the statement itself. f. Complement Law: ∨ ≡ i.e. If p is a simple statement then ∼(∼p) ≡ p i. p ~p T ii. p ∧ ~p ≡ F iv. Negation of conditional (implication) statement: g. Involution Law: The negation of a conditional statement p → q is i. ~T ≡ F p but not q. ii. ~F ≡ T → ≡ ∧ i.e. ~(p q) p ~q iii. ~(~p) ≡ p v. Negation of biconditional (double h. DeMorgan’s Law: implication) statement: i. ~(p ∨ q) ≡ ~p ∧ ~q ∧ ≡ ∨ The negation of a biconditional statement ii. ~(p q) ~p ~q p ↔ q is the negation of p → q or q → p. i. Absorption Law: ↔ ≡ ∧ ∨ ∧ ∨ ∧ ≡ i.e. ~(p q) (p ~q) (q ~p) i. p (p q) p ∧ ∨ ≡ Note: ii. p (p q) p Negation of statement pattern involving one or j. Conditional Law: more simple statements p, q, r, …. and one or i. p → q ≡ ~p ∨ q more connectives ∼, ∧ or ∨ is obtained by ii. p ↔ q ≡ (~p ∨ q) ∧ (~q ∨ p) replacing ∨ by ∧, ∧ by ∨. Statements p, q, r ∼ ∼ ∼ Exercise 1.8 are replaced by their negations p, q, r etc. and vice-versa. 1. Write the negations of following statements. vi. Negation of a quantified statement: i. All equilateral triangles are isosceles. While finding the negations of quantified ii. Some real numbers are not complex statements, the word ‘all’ is replaced by numbers. ‘some’ and ‘for every’ is replaced by ‘there iii. Every student has paid the fees. exists’ and vice-versa. iv. ∀ n ∈ N, n + 1 > 2. 2 v. ∀ x ∈ N, x + x is even number. Algebra of Statements vi. ∃ n ∈ N, such that n 2 = n. Some standard equivalent statements: vii. ∃ x∈ R, such that x2 < x. a. Idempotent Law: viii. All students of this college live in the i. p ∨ p ≡ p hostel. ii. p ∧ p ≡ p ix. Some continuous functions are b. Commutative Law: differentiable. i. p ∨ q ≡ q ∨ p x. Democracy survives if the leaders are ii. p ∧ q ≡ q ∧ p not corrupt. 15

Std. XII : Perfect Maths - I xi. The necessary and sufficient condition iii. ~[(~p ∧ ~q) ∨ (p ∧ ~q)] for a person to be successful is to be ≡ ~(~p ∧ ~q) ∧ ~(p ∧ ~q) honest. …(Negation of disjunction) xii. Some quadratic equations have ≡ [~(~p) ∨~ (~q)] ∧ [~p ∨ ~(~q)] unequal roots. ....(Negation of conjunction) Solution : ≡ (p ∨ q) ∧ (~p ∨ q) ….(Negation of negation) i. Some equilateral triangles are not isosceles. 3. Without using truth tables, show that ii. All real numbers are complex numbers. i. p ∧ (q ∨ ∼p) ≡ p ∧ q iii. Some students have not paid the fees. ii. (p ∧ q) ∨ (∼p ∧ q) ∨ (∼q ∧ r) ≡ q ∨ r iv. ∃ n ∈ N, such that n + 1 ≤ 2. Solution : ∧ ∨ 2 i. L.H.S. = p (q ~p) v. ∃ x ∈ N, such that x + x is not an even ≡ (p ∧ q) ∨ (p ∧ ~p) number. …(Distributive law) vi. ∀ n ∈ N, n 2 ≠ n. ≡ (p ∧ q) ∨ F ….(Complement law) ≡ p ∧ q ….(Identity law) ∀ ∈ 2 ≥ vii. x R, x x. ∴ L.H.S. = R.H.S. viii. Some students of this college do not live in the ∴ p ∧ (q ∨ ∼p) ≡ p ∧ q hostel. ii. L.H.S . = (p ∧ q) ∨ (~p ∧ q) ∨ (~q ∧ r) ≡ [(p ∨ ~p) ∧ q] ∨ (~q ∧ r) ix. All continuous functions are not differentiable. … (Associative and distributive law) x. Let, p: The leaders are not corrupt. ≡ (T ∧ q) ∨ (~q ∧ r) q: Democracy survives. ....(Complement law) ∴ The given statement is of the form, p → q ≡ q ∨ (~q ∧ r) Its negation is of the form p ∧ ∼ q. ....(Identity law) i.e. ‘the leaders are not corrupt, but democracy ≡ (q ∨ ~q) ∧ (q ∨ r) does not survive.’ ....(Distributive law) xi. Let, p: A person is successful. ≡ T ∧ (q ∨ r) ....(Complement law) q: A person is honest. ≡ q ∨ r ....(Identity law) ∴ The given statement is of the form, p ↔ q ∴ L.H.S. = R.H.S. Its negation is of the form, ∴ (p ∧ q) ∨ (∼p ∧ q) ∨ (∼q ∧ r) ≡ q ∨ r

(p ∧ ∼ q) ∨ (q ∧ ∼ p) 4. Form the negations of the following statements i.e. ‘a person is successful, but he is not honest by giving justification. or a person is honest, but he is not successful.’ i. (p ∧ q) → (~p ∨ r) xii. All quadratic equations have equal roots. ii. (q ∨ ~r) ∧ (p ∨ q) Solution: 2. Using the rules of negation, write the i. ~[(p ∧ q) → (~p ∨ r)] negations of the following. ≡ (p ∧ q) ∧ ~(~p ∨ r) ∧ → ∨ ∧ i. p (q r) ii. (~p q) r …. (Negation of implication) iii. (∼p ∧ ∼q) ∨ (p ∧ ∼q) ≡ ∧ ∧ ∧ Solution : (p q) [~(~p) ~r] i. ~[p ∧ (q → r)] …. (Negation of disjunction) ≡ ~p ∨ ~(q → r) ….(Negation of conjunction) ≡ (p ∧ q) ∧ (p ∧ ~r) ∨ ∧ …. (Negation of negation) ≡ ~p (q ~r) .…(Negation of implication) ii. ~[(q ∨ ~r) ∧ (p ∨ q)] ii. ~[(~p ∨ q) ∧ r] ≡ ~(q ∨ ~r) ∨ ~(p ∨ q) ≡ ~(~p ∨ q) ∨ ~r ....(Negation of conjunction) ….(Negation of conjunction) ≡ [~(~p) ∧ ~q] ∨ ~r .... (Negation of disjunction) ≡ [~q ∧ ~(~r)] ∨ (~p ∧ ~q) ≡ (p ∧ ~q) ∨ ~r ….(Negation of negation) …. (Negation of disjunction) 16

Chapter 01: Mathematical Logic ≡ ∧ ∨ ∧ (~q r) (~p ~q) i. Switches S1 and S 2 connected in series: …. (Negation of negation)

≡ (~q ∧ r) ∨ (~q ∧ ~p) ….(Commutative law) S1 S2 ≡ ∧ ∨ ~q (r ~p) ….(Distributive law) L

Application of Logic to Switching Circuits Let p: the switch S 1 is closed The working of an electric switch is similar to a logical statement which has exactly two outcomes, q: the switch S 2 is closed namely, T or F. A switch also has two outcomes or l: the lamp glows results (current flows and current does not flow) In this case, the lamp glows, if and only if depending upon the status of the switch i.e. (ON or both the switches are closed. OFF). This analogy is very useful in solving Thus we have, p ∧ q ≡ l. problems of circuit design with the help of logic.

Input – Output (switching) table: Switch p q p ∧ q An electric switch is a two state device used for T (1) T (1) T (1) turning a current ‘on’ or ‘off’. T (1) F (0) F (0) S S F (0) T (1) F (0) closed (On) switch open (Off ) switch F (0) F (0) F (0)

As shown in the above figures, if the switch is on ii. Switches S1 and S 2 are in parallel i.e., circuit is closed, current passes through the circuit and vice-versa. S1

Consider a simple circuit having a switch ‘S’, a S2 battery and a lamp ‘L’. When the switch ‘S’ is L closed (i.e., ON, current is flowing through the circuit), the lamp glows (is on). Similarly, when the swtich is open (i.e., OFF, current is not flowing Let p : the switch S 1 is closed through the circuit), the lamp does not glow (is off). q : the switch S 2 is closed l : the lamp glows In this case, the lamp glows, if at least one of S the switches is closed. Battery Lamp L ∨ ≡ Thus, we have p q l, Input – output (switching) table:

Thus, if p is a statement ‘the switch is closed’ and if p q p ∨ q l is a statement ‘the lamp glows’ then p is equivalent T (1) T (1) T (1) to l i.e p ≡ l. T (1) F (0) T (1) F (0) T (1) T (1) Note: i. ~p means ‘the switch is open’. In this case, F (0) F (0) F (0) the lamp will not glow and thus ∼ p ≡ ∼ l. The above two networks can be combined to form a complicated network as shown ii. If a switch is ‘ON’ then its truth value is T or below: 1 and if the switch is ‘OFF’, its truth value is

F or 0. S1 S2

L If there are two switches, then they can be connected in the following ways: S3 17

Std. XII : Perfect Maths - I

Let p : The switch S 1 is closed ∴ Symbolic form of the given circuit is, q : The switch S 2 is closed [(p ∧ q) ∨ (∼ p ∧ ∼ r)] ∧ (p ∨ q ∨ r) ≡ l r : The switch S 3 is closed l : The lamp glows Generally l is not written and therefore the In this case, the lamp glows, if S1 and S 2 both are symbolic form is closed or if S 3 is closed. [(p ∧ q) ∨ (∼ p ∧ ∼ r)] ∧ (p ∨q ∨ r) Thus we have, (p ∧ q) ∨ r ≡ l. Note: i. If two or more switches in a circuit are Exercise 1.9 open or closed simultaneously, then they are denoted by same letter and are called 1. Represent the following circuits ‘equivalent switches.’ symbolically and write the input-output or ii. Any two switches in a circuit having switching table. opposite states are called complementary switches. S1 For example, if S 1 and S 2 are the two i. switches such that when S 1 is closed, S 2 is open and vice-versa, then the switches S2 S1 and S 2 are called complementary ′ S′ S′ switches and S 2 is denoted as S1 . In 1 2 such a situation, one of them is ′ considered as p and the other as ~p or p . L iii. Two circuits are called equivalent if output of the two circuits is always same.

iv. A circuit is called simpler if it contains S S lesser number of switches. ii. 1 2

Example : S3 Express the following circuit in the symbolic form: ′ ′ S1 S2

S1 S2 S1 L S2

S′ S′ S3 1 3 iii. S 1 S3

S2 L Solution: S ′ 2 S1 Let p : The switch S 1 is closed. q : The switch S 2 is closed. r : The switch S 3 is closed. ∼ S′ L p : The switch 1 is closed. ∼ S′ r: The switch 3 is closed. Solution : l : Lamp L is ‘on’. i. Let p: The switch S 1 is closed. − The lamp L is ‘on’ if and only if q: The switch S 2 is closed. i. Switch S 1 and Switch S 2 are closed. ~p: The switch S′ is closed or the switch S 1 is Q 1 ( S1 and S 2 are in series.) open. S′ S′ or ii. Switch 1 and 3 are closed. ′ ~q: The switch S2 is closed or the switch S 2 is Q S′ S′ ( 1 and 3 are in series) open.

and iii. Switch S 1 or S 2 or S 3 are closed. ∴ The symbolic form of the given circuit is (Q S1, S 2, S 3 are in parallel) (p ∨ q) ∨ (~p ∧ ~q) 18

Chapter 01: Mathematical Logic

Input-output Table: r: The switch S 3 is closed ~p: The switch S′ is closed or the switch S 1 (p ∨ q) ∨ 1 p q ~p ~q p ∨ q ~p ∧ ~q (~p ∧ ~q) is open. 1 1 0 0 1 0 1 ∴ The symbolic form of the given circuit is ∨ ∧ ∧ ∨ 1 0 0 1 1 0 1 (p q) q (r ~p) 0 1 1 0 1 0 1 Input- output Table: 0 0 1 1 0 1 1 (p ∨ q) Switching Table: (p ∨ q) p q r ~p p ∨ q (r ∨ ~p) ∧ q ∧ p q ~p ~q p ∨ q ~p ∧ ~q (p ∨ q) ∨ (~p ∧ ~q) ∧ q (r ∨ ~p) T T F F T F T 1 1 1 0 1 1 1 1 T F F T T F T F T T F T F T 1 1 0 0 1 1 0 0 F F T T F T T 1 0 1 0 1 0 1 0 1 0 0 0 1 0 0 0 ii. Let p: The switch S 1 is closed. 0 1 1 1 1 1 1 1 q: The switch S 2 is closed. 0 1 0 1 1 1 1 1 r: The switch S 3 is closed. 0 0 1 1 0 0 1 0 ~p: The switch S′ is closed or the switch S 1 is 1 0 0 0 1 0 0 1 0 open. S′ 2 Switching table: ~q: The switch 2 is closed or the switch S is open. (p ∨ q) (p ∨ q) ∧ q ∧ ∴ The symbolic form of the given circuit is p q r ~p p ∨ q r ∨ ~p ∧ q (r ∨ ~p) [(p ∧ q) ∨ (~p ∧ ~q)] ∧ r. Let a ≡ (p ∧ q), b ≡ (∼p ∧ ∼q) T T T F T T T T T T F F T T F F Input-output Table: T F T F T F T F p q r ~p ~q a b a ∨ b (a ∨ b) ∧ r T F F F T F F F 1 1 1 0 0 1 0 1 1 F T T T T T T T F T F T T T T T 1 1 0 0 0 1 0 1 0 F F T T F F T F 1 0 1 0 1 0 0 0 0 F F F T F F T F 1 0 0 0 1 0 0 0 0

0 1 1 1 0 0 0 0 0 2. Construct the switching circuits of the 0 1 0 1 0 0 0 0 0 following statements. 0 0 1 1 1 0 1 1 1 i. [p ∨ (~p ∧ q)] ∨ [( ∼q ∧ r) ∨ ∼p] 0 0 0 1 1 0 1 1 0 [Mar 15] ii. (p ∧ q ∧ r) ∨ [∼p ∨ (q ∧ ∼r)] Switching table: iii. [(p ∧ r) ∨ (~q ∧ ∼r)] ∨ (~p ∧ ~r) p q r ~p ~q a b a ∨ b (a ∨ b) ∧ r Solution : T T T F F T F T T i. T T F F F T F T F Let p: The switch S 1 is closed T F T F T F F F F q: The switch S 2 is closed. T F F F T F F F F r: The switch S 3 is closed. ′ F T T T F F F F F ~p: The switch S1 is closed or the switch S 1 is F T F T F F F F F open. F F T T T F T T T ′ ~q: The switch S2 is closed or the switch S 2 is F F F T T F T T F open. ′ iii. Let p: The switch S 1 is closed ~r : The switch S3 is closed or the switch S 3 is q: The switch S 2 is closed open. [1 Mark] 19

Std. XII : Perfect Maths - I

Consider the given statement,

S1 S2 S3 [p ∨ (∼ p ∧ q )] ∨ [( ∼q ∧ r) ∨ ∼ p].

S′ 1 p ∨ (∼p ∧ q): represents that switch S 1 is S′ connected in parallel with the series combination S2 3 ′ of S1 and S 2.

(∼q ∧ r ) ∨ ∼p : represents that switch S′ is 1 connected in parallel with the series combination L S′ of 2 and S 3. iii. Let p: The switch S 1 is closed Therefore, [p ∨ (∼p ∧ q )] ∨ [( ∼q ∧ r) ∨ ∼p] q: The switch S 2 is closed. represents that the circuits corresponding to [p ∨ r: The switch S 3 is closed. S′ (∼ p ∧ q )] and [( ∼q ∧ r) ∨ ∼ p] are connected in ~p: The switch 1 is closed or the switch S 1 is parallel with each other. [1 Mark] open. ~q: The switch S′ is closed or the switch S 2 is ∴ Switching Circuit corresponding to the given 2 statement is: open. S′ ~r : The switch 3 is closed or the switch S 3 is S1 open. Consider the given statement, [(p ∧ r) ∨ (∼q ∧ ∼r)] ∨ (∼p ∧ ∼r) S′ S2 1 ∧ ∨ ∼ ∧ ∼ (p r) ( q r): represents that the series combination of S 1 and S 3 and series ′ S 3 S2 combination of S′ and S′ are connected in 2 3 parallel with each other. [1 Mark] S′ ∼p ∧ ∼r : represents that switches S′ and S′ 1 1 3 are connected in series. L Therefore, [(p ∧ r) ∨ (∼q ∧ ∼r)] ∨ (∼ p ∧ ∼r) represents that the circuits corresponding to ii. Let p : The switch S 1 is closed. [(p ∧ r) ∨ (∼q ∧ ∼r)] and ( ∼p ∧ ∼r) are q : The switch S 2 is closed. connected in parallel with each other. r : The switch S 3 is closed. ∴ Switching Circuit corresponding to the given ′ statement is: ~p : The switch S1 is closed or the switch S 1

is open. S3 S1 S′ ~r : The switch 3 is closed or the switch S′ 0 S′ 2 3 S3 is open. Consider the given statement, S′ S′ 1 3 (p ∧ q ∧ r) ∨ [∼ p ∨ (q ∧ ∼ r)]

p ∧ q ∧ r: represents that switches S 1, S 2, S 3 L

are connected in series. 3. Give an alternative arrangement for the ∼ p ∨ (q ∧ ∼r): represents that the series following circuit, so that the new circuit has ′ combination of S 2 and S3 is connected in two switches only. Also write the switching ′ parallel with S1 . table. [July 16] S S′ Therefore, (p ∧ q ∧ r) ∨ [∼ p ∨ (q ∧ ∼ r) 1 2 represents that the circuits corresponding to S′ S (p ∧ q ∧ r) and [ ∼ p ∨ (q ∧ ∼ r)] are connected 1 2 in parallel with each other. S′ S′ ∴ Switching circuit corresponding to the given 1 2 statement is: L 20

Chapter 01: Mathematical Logic Solution : Solution : 1 Let p: The switch S is closed. i. Let p: The switch S 1 is closed. q: The switch S 2 is closed. ′ q: The switch S 2 is closed. ~p: The switch S1 is closed or the switch S 1 is open. ′ r: The switch S 3 is closed. ~q: The switch S2 is closed or the switch S 2 is open. ′ ∴ symbolic form of the given circuit is ~p: The switch S1 is closed or the switch ∧ ∨ ∧ ∨ ∧ (p ~q) (~p q) (~p ~q) [1 Mark] S1 is open. ≡ ∧∼ ∨ ∼ ∧ ∨∼ (p q) [ p (q q)] S′ …(Associative and Distributive law) ~q: The switch 2 is closed or the switch ≡ (p ∧∼q) ∨ (∼p∧T) ….(Complement law) S2 is open. ≡ (p ∧∼q) ∨∼p ....(Identity law) ∴ The symbolic form of the given circuit is ≡ (p ∨ ~p) ∧ (~q ∨∼ p) ....(Distributive law) (p ∧ q) ∨ (~p ∧ q) ∨ (r ∧ ~q) [1 Mark] ≡ ∨ ∧ ∧∼ (p ~p) (~p q) ….(Commutative law) ii. Switching table: ≡ T∧(~p ∨ ~q) ….( Complement law) ≡ ∧ ∨ ∼ ∧ ∨ ∧ ∼ ≡ ∼p∨∼q ….(Identity law) Let a (p q) ( p q) (r q)

[1 Mark] The alternative arrangement for given circuit is: p q r p∧q ∼p ∼q ∼p∧q r∧∼q a

T T T T F F F F T

SS' 1 1 T T F T F F F F T T F T F F T F T T S'2 [1 Mark] S2 T F F F F T F F F

F T T F T F T F T L F T F F T F T F T Switching Table: F F T F T T F T T (p ∧ ∼q) ∨ F F F F T T F F F

p q ~p ~q p ∧ ~q ~p ∧ q ~p ∧ ~q (∼p ∧ q) ∨ (∼p ∧∼q) iii. For simplified switching circuit, T T F F F F F F Consider (p ∧ q) ∨ (~p ∧ q) ∨ (r ∧ ~q) T F F T T F F T ≡ [(p ∨ ~p) ∧ q] ∨ (r ∧ ~q) F T T F F T F T ....(Associative and Distributive law) F F T T F F T T ≡ (T ∧ q) ∨ (r ∧ ~q)....(Complement law) 4. Find ≡ q ∨ (r ∧ ~q) ....(Identity law) i. symbolic form ii. switching table and ≡ (q ∨ r) ∧ (q ∨ ~q) ....(Distributive law) iii. draw simplified switching circuit for ≡ (q ∨ r) ∧ T ....(Complement law) the following switching circuit. ∨ ≡ q r ....(Identity law) ∴ Simplified switching circuit is: S1 S2

S′ S2S1 1 S2

S ′ 3 S2 S3S2

L L

21

Std. XII : Perfect Maths - I ′ 5. Find the symbolic form of the following ∼q : The switch S2 is closed or the switch S 2 is switching circuit, construct its switching open. table and interpret your result. ∴ The symbolic form of the given circuit is

[p ∧ (~p ∨ ~q)] ∧ q

S1 ∧ ∨ ∧ ∧ ≡ [(p ~p) (p ~q)] q ′ ′ ….[Associative and Distributive law] S1 S2 ≡ [F ∨ (p ∧ ~q)] ∧ q ….[Complement law] S2 ≡ (p ∧ ~q) ∧ q ….[Identity law] ≡ (p ∧ q) ∧ (~q ∧ q) ….[Distributive law] ≡ (p ∧ q) ∧ F ….[Complement law] L ≡ F ….[Identity law] Solution : ∴ Irrespective of the status of the switches, the Let p : The Switch S 1 is closed. current will not flow in the circuit, that is, the q : The Switch S 2 is closed. circuit will always be open. ∼ S′ p : The Switch 1 is closed or the switch S1 is Miscellaneous Exercise - 1 open. ′ 1. Which of the following sentences are ∼q : The switch S2 is closed or the switch S 2 is statements in logic? Justify your answer. open. π ∴ The symbolic form of the given circuit is: i is a real number. ! (p ∨q) ∧ (∼p) ∧ (∼q) ii. 5 = 120 iii. Himalaya is an ocean and Ganga is a Switching table: river. p q ∼p ∼q p∨q (p ∨q) ∧ (∼p) ∧ (∼q) iv. Please get me a cup of tea. T T F F T F v. Bring me a notebook. T F F T T F vi. Alas ! We lost the match 2 2 F T T F T F vii. cos 2 θ = cos θ − sin θ, for all θ ∈ R. 2 ≥ F F T T F F viii. If x is a real number then x 0. Solution: In the above truth table, all the entries in the last column are ‘F’, i. It is a statement. ii. It is a statement. ∴ the given circuit represents a contradiction. iii. It is a statement. ∴ Irrespective of whether the switches S 1 and S 2 are open or closed, the given circuit will iv. It is an imperative sentence, hence it is not a always be open (i.e. off). statement. v. It is an imperative sentence, hence it is not a 6. Simplify the given circuit by writing its logical statement. expression. Also write your conclusion. vi. It is an exclamatory sentence, hence it is not a statement. ′ S1 vii. It is a statement. viii. It is a statement. S2 S1

S′ 2. Write the truth values of the following 2 statements: i. The square of any odd number is even

L or the cube of any even number is even. Solution : 5 5 Let, p : The switch S 1 is closed ii. is irrational but 3 + is a q : The switch S 2 is closed complex number. [Oct 14] ′ ∼p : The switch S1 is closed or the switch S 1 is iii. ∃ n ∈ N, such that n + 5 > 10. [Oct 14] open. iv. ∀ n ∈ N, n + 3 > 5. 22

Chapter 01: Mathematical Logic v. If ABC is a triangle and all its sides Solution : are equal then each angle has i. Since x = 5 ∈ A, satisfies x + 2 = 7. measure 30 °. ∴ the given statement is true. 2 vi. ∀ n ∈ N, n + n is an even number ∴ Its truth value is ‘T’. 2− while n n is an odd number. ii. Since, x = 7, 9 ∈ A, do not satisfy x + 3 < 10. Solution : ∴ the given statement is false. i. Let p: The square of any odd number is even. ∴ Its truth value is ‘F’. q: The cube of any even number is even. ∈ ≥ ∴ The symbolic form of the given statement is iii. Since, x = 4, 5, 7, 9 A, satisfy x + 5 9. ∴ p ∨ q. the given statement is true. ∴ Since the truth value of p is F and that of q is Its truth value is ‘T’. T, iv. Since, x = 4 ∈ A, satisfies ‘ x is even’. ∴ truth value of p ∨ q is T ∴ the given statement is true. ii. Let p: 5 is irrational. Its truth value is ‘T’. v. Since x = 9 ∈ A does not satisfy 2 x ≤ 17. 5 q: 3 + is a complex number. ∴ the given statement is false. ∴ The symbolic form of the given statement is ∴ Its truth value is ‘F’. p ∧ q Since the truth value of p is T and that of q is F, 4. Write negations of the following statements i. Some buildings in this area are ∴ truth value of p ∧ q is F. [1 Mark] multistoried. iii. Consider the statement, ∃ n ∈ N, n + 5 > 10 ii. All parents care for their children. Clearly n ≥ 6, n ∈N satisfy n + 5 > 10. iii. ∀n ∈ N, n + 7 > 6. ∴ iv. ∃ x ∈ A, such that x + 5 > 8. its truth value is T. [1 Mark] Solution : ∀ ∈ iv. Consider the statement, n N, n + 3 > 5 i. All buildings in this area are not multistoried. Q n = 1 and n = 2 ∈ N do not satisfy n + 3 > 5 ii. Some parents do not care for their children. ∴ truth value of p is F. iii. ∃ n ∈ N, such that n + 7 ≤ 6. iv. ∀ x ∈ A, x + 5 ≤ 8. v. Let p: ABC is a triangle and all its sides are equal. 5. Write the following statements in symbolic q: Each angle has measure 30 °. form: ∴ The symbolic form of the given statement is i. Ramesh is cruel or strict. p → q. ii. I am brave is necessary and sufficient Since the truth value of p is T and that of q is F, condition to climb the Mount Everest. ∴ → iii. I can travel by train provided I get truth value of p q is F my ticket reserved. vi. Let p: ∀ n ∈ N, n 2 + n is an even number. 2 iv. Sandeep neither likes tea nor coffee q: ∀n∈N, n − n is an odd number. but enjoys a soft-drink. ∴ The symbolic form of the given statement is v. ABC is a triangle only if p ∧ q. AB + BC > AC. Since, the truth value of p is T and q is F, vi. Rajesh is studious but does not get ∴ ∧ truth value of p q is F good marks. Solution : 3. If A = {4, 5, 7, 9}, determine the truth i. Let p: Ramesh is cruel, value of each of the following quantified q: Ramesh is strict. statements. ∴ The symbolic form of the given statement is ∨ i. ∃ x ∈ A, such that x + 2 = 7. p q. ii. ∀ x ∈ A, x + 3 < 10. ii. Let p: I am brave. iii. ∃ x ∈ A, such that x + 5 ≥ 9. q: I can climb the Mount Everest. iv. ∃ x ∈ A, such that x is even. ∴ The symbolic form of the given statement is v. ∀ x ∈ A, 2 x ≤ 17. p ↔ q. 23

Std. XII : Perfect Maths - I iii. Let p: I can travel by train, Solution : q: I get my ticket reserved. The symbolic forms of the given statements are: ∴ The symbolic form of the given statement is i. ~p → ~q ii. p ↔ q → ∨ ∧ q p. iii. ~(~p q) iv. p ~q v. q → p iv. Let p: Sandeep likes tea, q: Sandeep likes coffee. Truth table for all the above statements: r: Sandeep enjoys a soft-drink. p q ∼p ∼q (∼p∨q) (i) (ii) (iii) (iv) (v) ∴ The symbolic form of the given statement is T T F F T T T F F T ∧ ∧ (~p ~q) r. T F F T F T F T T T v. Let p: ABC is a triangle, F T T F T F F F F F F F T T T T T F F T q: AB + BC > AC. ∴ The symbolic form of the given statement is Note: In the above table, the numbers (i), (ii), (iii), (iv) p → q. and (v) represent corresponding statements. vi. Let p: Rajesh is studious, In the above table, the columns of statements (i) and (v) q: Rajesh gets good marks. are same. ∴ They are logically equivalent. ∴ The symbolic form of the given statement is p ∧ ~q. Similarly, the columns of statements (iii) and (iv) are same. ∴ They are also logically equivalent.

6. If p : The examinations are approaching, 8. Rewrite the following statements without q : Students study hard, give a verbal using the conditional form: statement for each of the following: i. If prices increase then the wages rise. i. p ∧ ∼q ii. p ↔ q ii. If it is cold, we wear woolen clothes. iii. ∼p → q iv. p ∨ q iii. I can catch cold if I take cold water bath. Solution : v. ∼q → ∼p All these statements are of the form p → q ≡ ∼p ∨ q. Solution : ∴ The statements without using conditional form i. The examinations are approaching but the will be, students do not study hard. i. Prices do not increase or the wages rise. ii. The examinations are approaching if and only ii. It is not cold or we wear woolen clothes. if the students study hard. iii. I do not take cold water bath or I catch cold. iii. If the examinations are not approaching, then the students study hard. 9. If p, q, r are statements with truth values T, iv. The examinations are approaching or the F, T respectively, determine the truth students study hard. values of the following: v. If the students do not study hard then the i. q → (p ∨ ∼r) ∼ ∧ ∨ ∼ examinations are not approaching. ii. ( r p) q iii. (p → q) ∨ r 7. If p: It is raining, q: The weather is humid, iv. (r ∧ q) ↔ ∼p [July 16] which of the following statements are v. (p ∨ q) → (q ∨ r) logically equivalent? Justify ! Solution : i. If it is not raining then the weather is Truth value of p, q and r are T, F and T respectively. not humid. i. q → (p ∨ ~r) ii. It is raining if and only if the weather is ≡ F → (T ∨ ~T) humid. ≡ F → (T ∨ F) iii. It is not true that it is not raining or ≡ F → T the weather is humid. ≡ T iv. It is raining but the weather is not Hence, the truth value is ‘T’. humid. ii. (~r ∧ p) ∨ ~q v. The weather is humid only if it is ≡ (~T ∧ T) ∨ ~F raining. ≡ (F ∧ T) ∨ T 24

Chapter 01: Mathematical Logic ≡ F ∨ T Solution : ≡ T i. Let, p: 6 is an even number. Hence, the truth value is ‘T’. q: 36 is a perfect square . ∴ ∨ iii. (p → q) ∨ r The given statement is of the form p q ∼ ∧ ∼ ≡ (T → F) ∨ T Its Negation is p q i.e. 6 is not an even number and 36 is not a perfect ≡ F ∨ T square. ≡ T Hence, the truth value is ‘T’. ii. Let, p: Diagonals of a parallelogram are iv. (r ∧ q) ↔ ∼p perpendicular. q : It is a rhombus. ≡ (T ∧ F) ↔ ∼T ∴ The given statement is of the form p → q. ≡ (T ∧ F) ↔ F [1 Mark] Its negation is p ∧ ∼q. ≡ F ↔ F ≡ T i.e. diagonals of a parallelogram are perpendicular Hence, the truth value is ‘T’ [1 Mark] but it is not a rhombus. v. (p ∨ q) → (q ∨ r) iii. Let, p : 10 > 5 and 5 < 8 ≡ (T ∨ F) → (F ∨ T) q : 8 < 7 ∴ → ≡ T → T The given statement is of the form p q ∧ ∼ ≡ T Its negation is p q i.e, 10 > 5 and 5 < 8 but 8 ≥ 7. Hence, the truth value is ‘T’. iv. Let, p: A person is rich. 10. Change each of the following statement in q: He is a software engineer. the form if… then… ∴ The given statement is of the form p ↔ q. i. I shall come provided I finish my Its negation is (p ∧ ∼ q) ∨ (q ∧ ∼ p) work. i.e., a person is rich and he is not a software ii. Rights follow from performing the engineer or a person is software engineer and duties sincerely. he is not rich. iii. x = 1 only if x2 = x. iv. The sufficient condition for being rich v. Let, p: Mangoes are delicious. is to be rational. q: Mangoes are expensive. v. Getting bonus is necessary condition ∴ The given statement is of the form p ∧ q ∼ ∨ ∼ for me to purchase a car. Its negation is, p q. Solution : i.e., mangoes are not delicious or they are not i. If I finish my work then I shall come. expensive. ii. If the duties are performed sincerely then the rights follow. vi. Let, p: Sky is not blue. iii. If x = 1 then x2 = x. ∴ The given statement is of the form ( ∼p) iv. If a man is rational, then he is rich. Its negation will be, ∼(∼p) ≡ p v. If I purchase a car then I get bonus. i.e., sky is not blue.

11. Write negations of the following statements: vii. Let, p: The weather is fine. i. 6 is an even number or 36 is a perfect q: My friends are coming. square. r: We go for a picnic. ∴ → ∧ ii. If diagonals of a parallelogram are The given statement is of the form p (q r) perpendicular then it is a rhombus. Its negation is, p ∧ ∼(q ∧ r) ≡ p ∧ (∼q ∨ ∼ r) iii. If 10 > 5 and 5 < 8 then 8 < 7. i.e., the weather is fine but my friends are not iv. A person is rich if and only if he is a coming or we are not going for a picnic.

software engineer. 12. Construct the truth table for each of the v. Mangoes are delicious but expensive. vi. It is false that the sky is not blue. following statement patterns: vii. If the weather is fine then my friends i. p → (q → p) will come and we go for a picnic. ii. (~ p ∨ ~ q) ↔ [~ (p ∧ q)] 25

Std. XII : Perfect Maths - I iii. ~(~p ∧ ~q) ∨ q T F F F F T F F T iv. [(p ∧ q) ∨ r] ∧ [~r ∨ (p ∧ q)] F T T T T T T T T F T F T T F F T T v. [( ∼p ∨ q) ∧ (q → r)] → (p → r) F F T T T T T T T Solution : F F F T T T T T T i. p → (q → p)

13. Using truth tables show that following p q q → p p → (q → p) statement patterns are tautologies. T T T T i. [(p → q) ∧ ∼q] → (∼p) T F T T ii. (p → q) ∨ (q → p) F T F T iii. [p → (q → r)] ↔ [(p ∧ q) → r] F F T T Solution : i. ii. (~p ∨ ~ q) ↔ [~ (p ∧ q)] [(p → q) [(p → q) ∧ p q ~p ~q (p → q) (~p ∨ ~q ) ↔ ∧ ~q ] ~q] → (~p) p q ~p ~q ~p ∨ ~q p ∧ q ~(p ∧ q) [~(p ∧ q)] T T F F T F T T T F F F T F T T F F T F F T T F F T T F T T F T T F T F T F T T F T F T T F F T T T T T

F F T T T F T T In the above truth table, all the entries in the iii. ~(~p ∧ ~q) ∨ q last column are T. ∴ The given statement pattern is a tautology. p q ~p ~q ~p ∧ ~q ~(~p ∧~q) ~(~p ∧ ~q) ∨ q ii. T T F F F T T p q p → q q → p (p → q) ∨ (q → p) T F F T F T T T T T T T F T T F F T T T F F T T F F T T T F F F T T F T iv. [(p ∧ q) ∨ r] ∧ [~r ∨ (p ∧ q)] F F T T T

[(p ∧ q) ∨ [~r ∨ [(p ∧ q) ∨ r] ∧ In the above truth table, all the entries in the p q r p ∧ q ~r r] (p ∧ q)] [~r ∨ (p ∧ q)] last column are T. ∴ The given statement pattern is a tautology. T T T T T F T T T T F T T T T T iii. q [p → (q → r)] T F T F T F F F [p → [(p ∧ q) p q r → (p ∧ q) ↔[(p ∧ q) T F F F F T T F (q → r)] → r] F T T F T F F F r → r] F T F F F T T F T T T T T T T T F F T F T F F F T T F F F T F T T F T T T F T T F F F F F T T F T F F T T F T T v. [( ∼p ∨ q) ∧ (q → r)] → (p → r) F T T T T F T T [(~p ∨ q) ∧ F T F F T F T T (~p ∨ q) ∧ p q r ~p ~p ∨q q → r p → r (q → r)] → (q → r) F F T T T F T T ( p → r) F F F T T F T T T T T F T T T T T In the above truth table, all the entries in the T T F F T F F F T last column are T. T F T F F T F T T ∴ The given statement pattern is a tautology. 26

Chapter 01: Mathematical Logic 14. Using truth tables show that following ii. (p ∨ q) is T and (p ∨ q) → q is F. statement patterns are contradictions. Consider the following truth table: i. [(p ∨ q) ∧ ∼p] ∧ (∼q) p q p ∨ q (p ∨ q) → q ii. (p ∧ q) ∧ (∼p ∨ ∼q) Solution : T T T T i.

T F T F [(p ∨ q) ∧ ~p] p q ~p ~q p ∨ q [(p ∨ q) ∧ ~p] ∧ (~q) F T T T T T F F T F F F F F T T F F T T F F F T T F T T F ∴ If (p ∨ q) is T and (p ∨ q) → q is F, then p is F F T T F F F true and q is false i.e., truth value of p is T and In the above truth table, all the entries in the that of q is F. last column are F. iii. (p ∧ q) is F and (p ∧ q) → q is T ∴ The given statement pattern is a contradiction. Consider the following truth table: ii. p q p ∧ q (p ∧ q) → q (p ∧ q) ∧ p q ~p ~q p ∧ q (~p ∨ ~q) (~p ∨ ~q) T T T T

T T F F T F F T F F T T F F T F T F F T T F F T F F T F T F F T T F T F F F F T

In the above truth table, all the entries in the last column are F. ∴ If (p ∧ q) is F and (p ∧ q) → q is T, then there ∴ The given statement pattern is a contradiction. are three possibilities for the truth value of

p and q. 15. Find truth values of p and q in the Either, p is T and q is F or following cases: p is F and q is T or i. (p ∨ q) is T and (p ∧ q) is T. both p and q are F. ii. (p ∨ q) is T and (p ∨ q) → q is F.

iii. (p ∧ q) is F and (p ∧ q) → q is T. 16. Determine whether the following statement Solution : patterns are tautologies, contradictions or ∨ ∧ i. (p q) is T and (p q) is T. contingencies. Consider the following truth table: i. (p → q) ∧ (p ∧ ~q) ∧ ∨ ∧ ∨ ∧ ∨ ∧ ∨ ∧ ii. (p q) (~p q) (p ~q) (~p ~q) p q p q p q ∧ → → iii. [p (p q)] q ∨ ∨ ∧ ∧ T T T T iv. [(p ~q) (~p q)] r Solution:

T F T F i.

→ ∧ → ∧ F T T F p q p q ~q p ~q (p q) (p ∧ ~q)

F F F F T T T F F F ∴ If (p ∨ q) is T and (p ∧ q) is T, then both p and T F F T T F q have to be true i.e. their truth value must be F T T F F F T. F F T T F F 27

Std. XII : Perfect Maths - I In the above truth table, all the entries in the Solution: last column are F. i. Consider, RHS = ∼(p ∧ ∼q) ∧ ∼(q ∧ ∼p) ∴ (p → q) ∧ (p ∧ ~q) is a contradiction. ≡ ∼ ∧ ∼ ∨ ∧ ∼ [(p q) (q p)] ii. ….[Negation of disjunction] ≡ ∼[∼(p ↔ q)] (p ∧ q) ∨ ….[Negation of double implication] (~p ∧ q) ∨ p q ~p ~q p ∧ q ~p ∧ q p ∧ ~q ~p ∧ ~q ≡ p ↔ q (p ∧ ~q) ∨ ….[Negation of negation] (~p ∧ ~q) = LHS. T T F F T F F F T ∴ p ↔ q ≡ ∼(p ∧ ∼q) ∧ ∼(q ∧ ∼p) T F F T F F T F T F T T F F T F F T ii. Consider, F F T T F F F T T LHS = ∼(p ∨ q) ∨ (∼p ∧ q) ≡ (∼p ∧ ∼q) ∨ (∼p ∧ q) In the above truth table, all the entries in the …. (Negation of disjunction) [1 Mark] last column are T. ≡ ∼p ∧ (∼q ∨ q) ∴ ∧ ∨ ∧ ∨ ∧ ∨ ∧ (p q) (~p q) (p ~q) (~p ~q) is a ….(Distributive law) [1 Mark] tautology. ≡ ∼p ∧ T ....(Complement law) iii. ≡ ∼ p ....(Identity law) [1 Mark] = RHS. p q p → q p ∧ (p → q) [p ∧ (p → q)] → q ∴ ∼(p ∨ q) ∨ (∼p ∧ q) ≡ ∼p T T T T T T F F F T iii. Consider, RHS = (p ∨ q) ∧ ∼p F T T F T ≡ (p ∧ ∼ p) ∨ (q ∧ ∼p) F F T F T …. [Distributive law] In the above truth table, all the entries in the ≡ (p ∧ ∼ p) ∨ (∼p ∧ q) last column are T. ….[Commutative law] ∴ ∧ → → ≡ ∨ ∼ ∧ [p (p q)] q is a tautology. F ( p q) …. [Complement law] iv. ≡ ∼ p ∧ q …. [Identity law] (p ∨ ~q) ∨ [(p ∨ ~q) ∨ p q r ~p ~q p ∨ ~q ~p ∧ q = LHS (~p ∧ q) (~p ∧ q)] ∧ r ∴ ∼ ∧ ≡ ∨ ∧ ∼ p q (p q) p T T T F F T F T T T T F F F T F T F 18. Using truth tables prove the following T F T F T T F T T logical equivalences: T F F F T T F T F i. p ↔ q ≡ (p ∧ q) ∨ (∼p ∧ ∼q) F T T T F F T T T ii. (p ∧ q) → r ≡ p → (q → r) F T F T F F T T F Solution : F F T T T T F T T i. Ref Exercise 1.5 Q2 (ii) F F F T T T F T F ii. Ref Exercise 1.5 Q2 (iii)

In the above truth table, the entries in the last column are a combination of T and F. ∴ ∨ ∨ ∧ ∧ 19. Write converse, inverse and contrapositive [(p ~q) (~p q)] r is a contingency. of the following conditional statements:

i. If an angle is a right angle then its 17. Using the rules of logic, prove the following measure is 90 °. logical equivalences. ↔ ≡ ∧ ∧ ∧ ii. If two triangles are congruent then their i. p q ~(p ~q) ~(q ~p) areas are equal. [Mar 15] ∨ ∨ ∧ ≡ ii. ~(p q) (~p q) ~p [Mar 16] iii. If f(2) = 0 then f( x) is divisible by iii. ~p ∧ q ≡ (p ∨ q) ∧ ~p (x − 2). 28

Chapter 01: Mathematical Logic Solution : ≡ (~q ∨ p) ∨ q ….[Negation of negation] i. Let p : An angle is a right angle, ≡ (p ∨ ~q) ∨ q ....[Commutative law] ° q : Its measure is 90 . ≡ p ∨ (~q ∨ q) ....[Associative law] ∴ The symbolic form of the given statement is ≡ p ∨ T ....[Complement law] → p q. ≡ T ....[Identity law] Converse: q → p Since, the truth value of the given statement pattern i.e. If the measure of an angle is 90 °, then it is a is T, therefore, it is a tautology. right angle. Inverse: ~p → ~q 21. Consider following statements. i.e. If an angle is not a right angle, then its i. If a person is social then he is happy. measure is not 90 °. ii. If a person is not social then he is not happy. Contrapositive: ~q → ~p iii. If a person is unhappy then he is not i.e. If the measure of an angle is not 90 ° then it social. is not a right angle. iv. If a person is happy then he is social. ii. Let p : Two triangles are congruent, Identify the pairs of statements having q : Their areas are equal. same meaning. ∴ The symbolic form of the given statement is Solution : p → q. Let p: A person is social, q: He is happy. Converse: q → p The symbolic forms of the given statements are: i.e. If areas of two triangles are equal then they i. p → q are congruent. [1 Mark] ii. ~p → ~q Inverse: ~p → ~q iii. ~q → ~p i.e. If two triangles are not congruent then their iv. q → p areas are not equal. Contrapositive: ~q → ~p Statements (i) and (iii) have same meaning. i.e. If areas of two triangles are not equal then they Since, a statement and its contrapositive are are not congruent. [1 Mark] equivalent. Also, statements (ii) and (iv) have the same iii. Let p: f(2) = 0, meaning. Since, converse and inverse of a − q: f( x) is divisible by ( x 2) compound statement are also equivalent. ∴ The symbolic form of the given statement is p → q. 22. Using the rules of logic, write the negations Converse: q → p of the following statements: i.e. If f( x) is divisible by ( x − 2), then f(2) = 0. i. (p ∨ q) ∧ (q ∨ ∼r) ∼ ∧ ∨ ∧ ∼ Inverse: ~p → ~q ii. ( p q) (p q) ∧ ∨ i.e. If f(2) ≠ 0, then f( x) is not divisible by iii. p (q r) → ∧ (x − 2). iv. (p q) r Contrapositive: ~q → ~p Solution : i.e. If f( x) is not divisible by ( x−2) then f(2) ≠ 0. i. ~[(p ∨ q) ∧ (q ∨ ~r)] ≡ ∨ ∨ ∨ ~(p q) ~(q ~r) 20. Without using truth table, prove that ….[Negation of conjunction] ≡ ∧ ∨ ∧ [(p ∨ q) ∧ ∼p] → q is a tautology. (~p ~q) (~q ~ (~r)] Solution : ….[Negation of disjunction] [(p ∨ q) ∧ ∼p] → q ≡ (~p ∧ ~q) ∨ (~q ∧ r) ≡ [(p ∧ ~p) ∨ (q ∧ ~p)] → q ….[Distributive law] ….[Negation of negation] ≡ [F ∨ (q ∧ ~p)] → q ….[Complement law] ≡ (∼ p ∧ ∼q) ∨ (r ∧ ∼q) ≡ (q ∧ ~p) → q ….[Identity law] ....[Commutative law] ≡ ~(q ∧ ~p) ∨ q …[Conditional law] ≡ (~p ∨ r) ∧ ~q ≡ [(~q ∨ ~(~p)] ∨ q ....[Negation of conjunction] ….[Distributive law] 29

Std. XII : Perfect Maths - I ii. ∼[( ∼ p ∧ q) ∨ (p ∧ ∼ q)] 24. Construct the switching circuits of the ≡ ∼(∼ p ∧ q) ∧ ∼ (p ∧ ∼q) following statements: ….[Negation of disjunction] i. (p ∧ ~q ∧ r) ∨ [p ∧ (~q ∨ ~r)] ≡ [∼(∼ p) ∨ ∼q] ∧ [∼ p ∨ ∼(∼ q)] ii. [(p ∧ r) ∨ (~q ∧ ~r)] ∧ (~p ∧ ~r) ….[Negation of conjunction] Solution : ≡ (p ∨ ∼q) ∧ [∼ p ∨ q) i. Let p: the switch S 1 is closed. ….[Negation of negation] q: the switch S 2 is closed. iii. ∼[p ∧ (q ∨ r)] ≡ ∼p ∨ ∼(q ∨ r) r: the switch S 3 is closed. ′ ….[Negation of conjunction] ~q: the switch S2 is closed or the switch S 2 ≡ ∼p ∨ (∼ q ∧ ∼ r) is open. S′ ….[Negation of disjunction] ~r: the switch 3 is closed or the switch S 3 iv. ~[(p → q) ∧ r] is open. ≡ ~(p → q) ∨ ~r Consider the given statement, ....[Negation of conjunction] (p ∧ ~q ∧ r) ∨ [p ∧ (~q ∨ ~r)] ≡ ∧ ∨ ′ (p ~q) ~r p ∧ ~q ∨ r: represents that the switches S 1, S2 ....[Negation of implication] S and 3 are connected in series. 23. Express the given circuits in symbolic form. p ∧ (~q ∨ ~r): represents that parallel ′ ′ combination of S2 and S3 is connected in i. S1 S2 series with S 1. Therefore, (p ∧ ~q ∧ r) ∨ [p ∧ (~q ∨ ~ r)] represents that the circuits corresponding to S′ 1 [(p ∧ ~q ∧ r) and [p ∧ (~ q ∨ ~r)] are connected in parallel with each other. Hence, the switching circuit of given statement is S S′ 1 2 S′ S1 2 S3 L

SS′ ' 22 ii. S1 L S S1 SS1 S′ 1 1 3

S2 S3 L ii. Let p: the switch S 1 is closed q: the switch S 2 is closed r: the switch S 3 is closed ′ Solution : ~p: the switch S1 is closed or the switch S 1 i. Let p: the switch S 1 is closed. is open. q: the switch S 2 is closed. ′ ~q: the switch S2 is closed or the switch S 2 ~p: the switch S′ is closed or the switch S 1 1 is open. is open. ′ ~r: the switch S3 is closed or the switch S 3 is ~q: the switch S′ is closed or the switch S 2 2 open. is open. Consider the given statement, ∴ The symbolic form of the given circuit is [(p ∧ r) ∨ (~q ∧ ~r)] ∧ (~p ∧ ~r) (p ∧ q) ∨ (~p) ∨ (p ∧ ~q) (p ∧ r) ∨ (~q ∧ ~r): represents that series ii. Let p: the switch S 1 is closed combination of S 1 and S 3 and series combination q: the switch S 2 is closed ′ ′ of S2 and S3 are connected in parallel. r: the switch S 3 is closed ∧ S′ S′ ∴ The symbolic form of the given circuit is ~p ~r : represents that 1 and 3 are connected (p ∨ q) ∧ (p ∨ r) in series. 30

Chapter 01: Mathematical Logic Therefore, [(p ∧ r) ∨ (~q ∧ ~ r)] ∧ (~p ∧ ~ r) The symbolic form of the given circuit is, represents that the circuits corresponding to [(p ∧ q) ∨ (~r ∨ ~s ∨ ~t) ] ∧ [(p ∧ q) ∨ (r ∧ s ∧ t)] [(p ∧ r) ∨ (~q ∧ ~ r)] and (~p ∧ ~r) are ≡ [(p ∧ q) ∨ ~(r ∧ s ∧ t)] ∧ [(p ∧ q) ∨ (r ∧ s ∧ t)] connected in series. ….[De-Morgan’s law] Hence, switching circuit of the given statement ≡ (p ∧ q) ∨ [~(r ∧ s ∧ t) ∧ (r ∧ s ∧ t)] is ….[Distributive law] ′ S1 S1 S3 SS11 S1 S′ 3 ≡ (p ∧ q) ∨ F ….[Complement law]

S′ S′ 2 3 ≡ p ∧ q ….[Identity law]

L Hence, the simplified circuit is

25. Simplify the following circuits so that the S1 S2 new circuit has minimum number of switches. Also draw the simplified circuit. L i. ′ S1 S2 26. Check whether the following switching circuits are logically equivalent. Justify ! S′ S 1 2 A. i.

S S 1 2 S′ S′ 1 2

S 3 L ii. S S S S1 2 1 2 S1 S ii. 2 S′ 3 ′ S1 S3 S4 ′ S5 B. i. S S3 S4 5

S1 S2

L S S3 2 Solution : i. Refer to Exercise 1.9 Q.3. ii. S1 ii. Let, p: The switch S 1 is closed. q: The switch S 2 closed. r: The switch S 3 is closed. S2 S3 s: The switch S 4 is closed. Solution : t: The switch S 5 is closed. Let, ′ ~r : The switch S3 is closed or the switch S 3 is p : The switch S 1 is closed. open. q : The switch S 2 is closed. S′ r : The switch S 3 is closed. ~s: The switch 4 is closed or the switch S 4 is open. A. i. The symbolic form of given circuit is, ′ ~t: The switch S5 is closed or the switch S 5 is p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r) ….(i) open. ….[distributive law] 31

Std. XII : Perfect Maths - I ii. The symbolic form of given circuit is, ≡ [(q ∨ p) ∧ T] ∧ r ….[Complement law] (p ∧ q) ∨ (p ∧ r) …. (ii) ≡ (q ∨ p) ∧ r ….[Identity law] ∴ The given circuits are logically equivalent ≡ (p ∨ q) ∧ r ….[Commutative law] ....[From (i) and (ii)] ∴ The switching circuit corresponding to the B. i. The symbolic form of the given circuit is given statement is: (p ∨ q) ∧ (q ∨ r) ....(i) ii. The symbolic form of the given circuit is, S1 S3 p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r) ….[distributive law] ....(ii) S2 ∴ The given circuits are not logically

equivalent ....[From (i) and (ii)] L

27. Give alternative arrangement of the 28. Draw the simplified circuit of the following following circuit, so that the new circuit has switching circuit. minimum switches only. ′ S1

S S′ 1 S2 1 S2 ′ S S S1 2 3 S1

S1 S2 S3 S′ S ′ 2 1 S2 S3

L S1 Solution : Let, p : The switch S 1 is closed. S2 q : The switch S 2 is closed. r : The switch S 3 is closed. L ∼ S′ p: The switch 1 is closed or the switch S 1 is Solution : open. Let p : The switch S 1 is closed. ′ ~q: The switch S2 is closed or the switch S 2 is q : The switch S 2 is closed. ′ open. ~p : The switch S1 is closed or the switch S 1 is open. ′ The symbolic form of given circuit is, ~q : The switch S2 is closed or the switch S 2 is open. ∧ ∧ ∨ ∧ ∧ ∨ ∧ ∧ (p q ~p) (~p q r) (p q r) The symbolic form of the given circuit is ∨ ∧ ∧ (p ~q r) (~p ∨ q) ∨ (p ∨ ~q) ∨ (p ∨ q) ≡ ∧ ∧ ∨ ∧ ∧ ∨ ∧ ∧ (p ~p q) (~p q r) (p q r) ≡ (~p ∨ q) ∨ [p ∨ (q ∨ ~q)] ∨ (p ∧ ~q ∧ r) ….[Commutative law] ….[Commutative and Distributive law] ≡ (F ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) ≡ (~p ∨ q) ∨ (p ∨ T) ….[Complement law] ….[Complement law] ≡ (~p ∨ q) ∨ T ….[Identity law] ≡ ∨ ∧ ∧ ∨ ∧ ∧ ∨ ∧ ∧ F (~p q r) (p q r) (p ~q r) ≡∼ p ∨ (q ∨ T) ….[Commutative law] ….[Identity law] ≡ ~p ∨ T ….[Identity law] ≡ ∧ ∧ ∨ ∧ ∧ ∨ ∧ ∧ (~p q r) (p q r) (p ~q r) ≡ T ….[Identity law] ….[Identity Law] ∴ The symbolic form of the given circuit is a ≡ ∨ ∧ ∧ ∨ ∧ ∧ [(~p p) (q r)] (p ~q r) tautology. Hence, the current will always flow ….[Distributive law] through the circuit irrespective of whether the ≡ ∧ ∧ ∨ ∧ ∧ [T (q r)] (p ~q r) switches are open or closed. ….[Complement law] ≡ ∧ ∨ ∧ ∧ (q r) (p ~q r) ….[Identity law] ≡ [q ∨ (p ∧ ~q)] ∧ r ….[Distributive law] ≡ ∨ ∧ ∨ ∧ [(q p) (q ~q)] r ….[Distributive law] L 32

Chapter 01: Mathematical Logic

29. Represent the following switching circuit in Multiple Choice Questions symbolic form and construct its switching table. Write your conclusion from the 1. Which of the following is a statement? switching table. (A) Stand up! (B) Will you help me?

S1 S1 (C) Do you like social studies?

S′ (D) 27 is a perfect cube. 2 2. Which of the following is not a statement? S′ S2 S3 3 (A) Please do me a favour. (B) 2 is an even integer. (C) 2 + 1 = 3.

L (D) The number 17 is prime.

3. Which of the following is an open statement? Solution : (A) x is a natural number. Let p : The switch S 1 is closed. (B) Give me a glass of water. q : The switch S 2 is closed. (C) Wish you best of luck. r : The switch S 3 is closed. (D) Good morning to all. ~p : The switch S′ is closed or the switch S 1 is open 1 4. Which of the following is not a proposition in S′ ~q : The switch 2 is closed or the switch S 2 is open logic. S′ ~r : The switch 3 is closed or the switch S 3 is (A) 3 is a prime. open. (B) 2 is a irrational. The symbolic form of the given circuit is (C) Mathematics is interesting. ∨ ∨ ∧ ∨ ∧ (p ~q ~r) [p (q r)] (D) 5 is an even integer. ∴ The switching table corresponding to the given statements is : 5. If p: The sun has set q: The moon has risen, [p ∨ (~q) ∨ then the statement ‘The sun has not set or the p ∨ p ∨ p q r ~q ~r q ∧ r (~r)] ∧ moon has not risen’ in symbolic form is (~q) ∨ (~r) (q ∧ r) [p ∨(q ∧r)] written as (A) ~p ∨ ~q (B) ~p ∧ q T T T F F T T T T (C) p ∧ ~q (D) p ∨ ~q T T F F T T F T T T F T T F T F T T 6. Assuming p: She is beautiful, q: She is clever, ∧ T F F T T T F T T the verbal form of p (~q) is (A) She is beautiful but not clever. F T T F F F T T F (B) She is beautiful and clever. F T F F T T F F F (C) She is not beautiful and not clever. F F T T F T F F F (D) She is beautiful or not clever. F F F T T T F F F 7. Let p: ‘It is hot’ and q: ‘It is raining’. The final column of the above table is equivalent The verbal statement for (p ∧ ~q) → p is to the column of ‘p’ i.e. column corresponding to (A) If it is hot and not raining, then it is hot. (B) If it is hot and raining, then it is hot. switch S 1. Hence, the given circuit is equivalent to (C) If it is hot or raining, then it is not hot. the circuit where only switch S 1 is present. (D) If it is hot and raining, then it is not hot. Hence, switching circuit is as follows: 8. Using the statements p: Kiran passed the examination, S1 S1 s : Kiran is sad. the statement ‘It is not true that Kiran passes therefore he is sad’ in symbolic form is L (A) ~p → s (B) ~ (p → ~ s) (C) ~p → ~ s (D) ~ (p → s) 33

Std. XII : Perfect Maths - I 9. Assuming p: She is beautiful, q: She is clever, 18. If p and q be two statements then the the verbal form of ~p ∧ (~q) is conjunction of the statements, p ∧ q is false (A) She is beautiful but not clever. when (B) She is beautiful and clever. (A) both p and q are true. (C) She is not beautiful and not clever. (B) either p or q are true (D) She is beautiful or not clever. (C) either p or q or both are false. (D) both p and q are false. 10. The converse of the statement ‘If it is raining then it is cool’ is 19. The negation of the statement, “The question (A) If it is cool then it is raining. paper is not easy and we shall not pass” is (B) If it is not cool then it is raining. (A) The question paper is not easy or we (C) If it is not cool then it is not raining. shall not pass. (D) If it is not raining then it is not cool. (B) The question paper is not easy implies we shall not pass. 11. If p and q are simple propositions, then p ∧ q (C) The question paper is easy or we shall is true when pass. (A) p is true and q is false. (D) We shall pass implies the question paper (B) p is false and q is true. is not easy. (C) p is true and q is true. 20. The statement (p ∧ q) ∧ (~ p ∨ ~ q) is (D) p is false q is false. (A) a contradiction. 12. Which of the following is logically equivalent (B) a tautology. to ~[~p → q] (C) neither a contradiction nor a tautology. ∨ (A) p ∨ ~q (B) ~p ∧ q (D) equivalent to p q. ∧ ∧ (C) ~p q (D) ~p ~q 21. The proposition p ∧ ~p is a 13. The logically equivalent statement of p → q (A) tautology and contradiction. is (B) contingency. (C) tautology. (A) ~p ∨ q (B) q → ~p ∨ ∨ (D) contradiction. (C) ~q p (D) ~q ~p 22. The proposition p → ~ (p ∧ q) is a ∨ 14. The logically equivalent statement of ~p ~q (A) tautology is (B) contradiction ∧ ∧ (A) ~p ~q (B) ~(p q) (C) contingency (C) ~(p ∨ q) (D) p ∧ q (D) either (A) or (B) 15. The contrapositive of (p ∨ q) → r is 23. The false statement in the following is (A) ~r → ~p ∧ ~q (B) ~r →(p ∨ q) (A) p ∧ (~p) is a contradiction. → ∨ → ∨ → → → (C) r (p q) (D) p (q r) (B) (p q) (~q ~p) is a contradiction. (C) ~(~p) → p is a tautology. 16. Which of the following propositions is true? (D) p ∨ (~p) is a tautology. (A) p → q ≡ ~p → ~q (B) ~(p → ~q) ≡ ~p ∧ q 24. Negation of ~(p ∨ q) is (C) ~(p ↔ q) ≡[~ (p → q) ∧ ~ (q →p)] (A) ~p ∨ ~q → ≡ ∧ (B) ~p ∧ ~q (D) ~(~p ~q) ~p q (C) p ∧ ~q 17. When two statements are connected by the (D) p ∨ ~q connective ‘if and only if’ then the compound statement is called 25. The dual of ~(p ∨q) ∨[p ∨(q ∧~r)]is, (A) conjunction of the statements. (A) ~(p ∧q) ∧ [p ∨ (q ∧~r)] (B) disjunction of the statements. (B) (p ∧q) ∧ [p ∧ (q ∨ ~ r)] (C) biconditional statement. (C) ~(p ∧q) ∧ [p ∧ (q ∧ r)] (D) conditional statement. (D) ~(p ∧q) ∧ [p ∧ (q ∨ ~ r)] 34

Chapter 01: Mathematical Logic 26. The symbolic form of the following circuit, where p: switch S 1 is closed.and q: switch S 2 is closed, is- (A) (p ∨q) ∧ [∼p∨(p ∧~q)] (B) (~p ∧q) ∨ [~p ∨ (p ∧~q)] (C) (p ∨q) ∨ [~p ∧(p ∨~q)] (D) (p ∧q) ∨ [~p ∧(p ∧~q)]

S1 S11

S 2 S1S1S1

S′ S 1 S′ 2 L

27. If A = {2, 3, 4, 5, 6}, then which of the following is not true? [Oct 13] (A) ∃ x ∈ A such that x + 3 = 8 (B) ∃ x ∈ A such that x + 2 < 5 (C) ∃ x ∈ A such that x + 2 < 9 ∀ ∈ ≥ (D) x A such that x + 6 9 28. If p ˄ q = F, p → q = F, then the truth value of p and q is : [Oct 15] (A) T, T (B) T, F (C) F, T (D) F, F

29. The negation of p ∧ (q → r) is [Mar 16] (A) p ∨ (~ q ∨ r) (B) ~ p ∧ (q → r) (C) ~ p ∧ (~ q → ~ r) ∨ ∧ (D) ~ p (q ~ r) 30. Inverse of the statement pattern (p ∨ q) → (p ∧ q) is [July 16] (A) (p ∧ q) → (p ∨ q) (B) ∼ (p ∨ q) → (p ∧ q) (C) (∼ p ∨ ∼ q) → (∼ p ∧ ∼ q) (D) (∼ p ∧ ∼ q) → (∼ p ∨ ∼ q)

Answers to Multiple Choice Questions

1. (D) 2. (A) 3. (A) 4. (C) 5. (A) 6. (A) 7. (A) 8. (D) 9. (C) 10. (A) 11. (C) 12. (D) 13. (A) 14. (B) 15. (A) 16. (D) 17. (C) 18. (C) 19. (C) 20. (A) 21. (D) 22. (C) 23. (B) 24. (B) 25. (D) 26. (C) 27. (D) 28. (B) 29. (D) 30. (D)

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