Truth Tables and Boolean Algebra

Basic Engineering

F Hamer, R Horan & M Lavelle

The aim of this document is to provide a short, self assessment programme for students who wish to acquire a basic understanding of the fundamentals of Boolean Algebra through the use of truth tables.

1. Boolean Algebra (Introduction) 2. Conjunction (A ∧ B) 3. Disjunction (A ∨ B) 4. Rules of Boolean Algebra 5. Quiz on Boolean Algebra Solutions to Exercises Solutions to Quizzes

The full range of these packages and some instructions, should they be required, can be obtained from our web page Mathematics Support Materials. Section 1: Boolean Algebra (Introduction) 3 1. Boolean Algebra (Introduction) Boolean algebra is the algebra of propositions. Propositions will be denoted by upper case Roman letters, such as A or B, etc. Every proposition has two possible values: T when the proposition is true and F when the proposition is false.

The negation of A is written as ¬A and read as “not A”. If A is true then ¬A is A ¬A false. Conversely, if A is false then ¬A is T F true. This relationship is displayed in the F T adjacent truth table.

The second row of the table indicates that if A is true then ¬A is false. The third row indicates that if A is false then ¬A is true. Truth tables will be used throughout this package to verify that two propositions are logically equivalent. Two propositions are said to be logically equivalent if their truth tables have exactly the same values. Section 1: Boolean Algebra (Introduction) 4

Example 1 Show that the propositions A and ¬(¬A) are logically equivalent.

Solution From the deﬁnition of ¬ it follows that if ¬A is true then ¬(¬A) is false, whilst if A ¬A ¬(¬A) ¬A is false then ¬(¬A) is true. This is T F T encapsulated in the adjacent truth table. F T F From the table it can be seen that when A takes the value true, the proposition ¬(¬A) also takes the value true, and when A takes the value false, the proposition ¬(¬A) also takes the value false. This shows that A and ¬(¬A) are logically equivalent, since their logical values are identical. Logical equivalence may also be written as an equation which, in this case, is A = ¬(¬A) . Section 2: Conjunction (A ∧ B) 5 2. Conjunction (A ∧ B) If A and B are two propositions then the “conjunction” of A and B, written as A ∧ B, and read as “A and B ”, is the proposition which is true if and only if both of A and B are true. The truth table for this is shown.

There are two possible values for A B A ∧ B each of the propositions A, B, so there are 2 × 2 = 4 possible assign- T T T ments of values. This is seen in the T F F truth table. Whenever truth tables F T F are used, it is essential that every F F F possible value is included. Truth table for A ∧ B Example 2 Write out the truth table for the proposition (A∧B)∧C. Solution The truth table will now contain 2 × 2 × 2 = 8 rows, corresponding to the number of diﬀerent possible values of the three propositions. It is shown on the next page. Section 2: Conjunction (A ∧ B) 6

In the adjacent table, the A B C A ∧ B (A ∧ B) ∧ C ﬁrst three columns con- T T T T T tain all possible values for T T F T F A, B and C. The values T F T F F in the column for A ∧ B T F F F F depend only upon the ﬁrst F T T F F two columns. The val- F T F F F ues of (A ∧ B) ∧ C depend F F T F F upon the values in the F F F F F third and fourth columns. Truth table for (A ∧ B) ∧ C

Exercise 1. (Click on the green letters for the solutions.) (a) Write out the truth table for A ∧ (B ∧ C). (b) Use example 2 and part (a) to prove that (A ∧ B) ∧ C = A ∧ (B ∧ C) . Section 3: Disjunction (A ∨ B) 7 3. Disjunction (A ∨ B) If A and B are two propositions then the “disjunction” of A and B, written as A ∨ B, and read as “A or B ”, is the proposition which is true if either A or B, or both, are true. The truth table for this is shown.

As before, there are two possi- A B A ∨ B ble values for each of the propo- T T T sitions A and B, so the number T F T of possible assignments of values F T T is 2 × 2 = 4 . This is seen in the F F F truth table. Truth table for A ∨ B Example 3 Write out the truth table for the proposition (A∨B)∨C. Solution The truth table will contain 2×2×2 = 8 rows, corresponding to the number of diﬀerent possible values of the three propositions. It is shown on the next page. Section 3: Disjunction (A ∨ B) 8

In the adjacent table, the A B C A ∨ B (A ∨ B) ∨ C ﬁrst three columns con- T T T T T tain all possible values for T T F T T A, B and C. The values T F T T T in the column for A ∨ B T F F T T depend only upon the ﬁrst F T T T T two columns. The val- F T F T T ues of (A ∨ B) ∨ C depend F F T F T upon the values in the F F F F F third and fourth columns. Truth table for (A ∨ B) ∨ C

Exercise 2. (Click on the green letters for the solutions.) (a) Write out the truth table for A ∨ (B ∨ C). (b) Use example 3 and part (a) to show that (A ∨ B) ∨ C = A ∨ (B ∨ C) . Section 3: Disjunction (A ∨ B) 9

Exercise 3. Write out the truth tables for the following propositions. (Click on the green letters for the solutions.) (a) ¬(A ∧ B), (b) (¬A) ∧ B, (c) (¬A) ∧ (¬B),(d) (¬A) ∨ (¬B),

Quiz Using the results of exercise 3, which of the following is true? (a) (¬A) ∧ B = (¬A) ∨ (¬B), (b) ¬(A ∧ B) = (¬A) ∨ (¬B), (c) ¬(A ∧ B) = (¬A) ∧ (¬B), (d) (¬A) ∧ B = (¬A) ∧ (¬B).

Exercise 4. Write out the truth tables for the following propositions. (Click on the green letters for the solutions.) (a) A ∧ (B ∨ C), (b) (A ∧ B) ∨ C, (c) (A ∧ B) ∨ (A ∧ C). Quiz Using the results of exercise 4, which of the following is true? (a) A ∧ (B ∨ C) = (A ∧ B) ∨ C, (b) A∧(B∨C) = (A∧B)∨(A∧C), (c) (A∧B)∨C = (A∧B)∨(A∧C),(d) None of these. The three diﬀerent operations ¬ , ∧ , and ∨ ,(“not”, “and” and “or”), form the algebraic structure of Boolean algebra. What remains to be determined is the set of rules governing their interactions. Section 4: Rules of Boolean Algebra 10 4. Rules of Boolean Algebra

(1a) A ∧ B = B ∧ A (1b) A ∨ B = B ∨ A (2a) A ∧ (B ∧ C) = (A ∧ B) ∧ C (2b) A ∨ (B ∨ C) = (A ∨ B) ∨ C (3a) A ∧ (B ∨ C) = (A ∧ B) ∨ (A ∧ C) (3b) A ∨ (B ∧ C) = (A ∨ B) ∧ (A ∨ C) (4a) A ∧ A = A (4b) A ∨ A = A (5a) A ∧ (A ∨ B) = A (5b) A ∨ (A ∧ B) = A (6a) A ∧ ¬A = F (6b) A ∨ ¬A = T (7) ¬(¬A) = A (8a) ¬(A ∧ B) = (¬A) ∨ (¬B) (8a) ¬(A ∨ B) = (¬A) ∧ (¬B) Section 4: Rules of Boolean Algebra 11

The following comments on these rules are useful:

All the rules can be veriﬁed by using truth tables.

Any rule labelled (a) can be obtained from the corresponding rule labelled (b) by replacing ∧ with ∨ and ∨ with ∧ , and vice versa.

Rules (1a) and (1b) can easily be shown to be true.

Rules (2a) and (2b) were proved in exercises 1 and 2 respectively.

Rule (3a) was proved in exercise 3 and the quiz immediately following. Rules (3a) and (3b) are called the distributive rules.

Rule (8a) was proved in exercise 4 and the quiz immediately following. Rules (8a) and (8b) are called De Morgan’s laws.

Rule (2a) means that A ∧ B ∧ C have the same meaning given by A ∧ (B ∧ C) or (A ∧ B) ∧ C. Similarly for A ∨ B ∨ C from (2b). Section 4: Rules of Boolean Algebra 12

Exercise 5. Use truth tables to prove the following propositions. (Click on the green letters for the solutions.) (a) A∨(B∧C) = (A∨B)∧(A∨C), (b) A ∧ (A ∨ B) = A, (c) ¬(A ∨ B) = ¬(A) ∧ ¬(B), (d) F ∨ X = X, (e) [A ∧ ((¬A) ∨ B)] ∨ B = B.

Truth tables are useful for proving that two expressions are equivalent but, often, the same result is easier to obtain using Boolean algebra. Example 4 Simplify the expression [A ∧ (¬A ∨ B)] ∨ B. Solution Using (3a),

A ∧ (¬A ∨ B) = (A ∧ ¬A) ∨ (A ∧ B) = F ∨ (A ∧ B) , by (6a) , = A ∧ B , by ex 5(d) . Thus [A ∧ (¬A ∨ B)] ∨ B = (A ∧ B) ∨ B = B , by (5b) , conﬁrming the result of exercise 5(e), above. Section 5: Quiz on Boolean Algebra 13 5. Quiz on Boolean Algebra

Begin Quiz In each of the following, choose the simpliﬁed version of the given expression. (Use either truth tables, or Boolean algebra to simplify the expression.) 1. (A ∧ ¬C) ∨ (A ∧ B ∧ C) ∨ (A ∧ C). (a) T , (b) F , (c) A ∧ B, (d) A.

2. [A ∧ B] ∨ [A ∧ ¬B] ∨ [(¬A) ∧ B] ∨ [(¬A) ∧ (¬B)]. (a) A ∧ B, (b) A ∨ B, (c) T , (d) F .

3. (A ∧ B ∧ C) ∨ (¬A) ∨ (¬B) ∨ (¬C). (a) T , (b) F , (c) A ∧ B, (d) A ∧ C.

End Quiz Solutions to Exercises 14 Solutions to Exercises Exercise 1(a) The truth table for A ∧ (B ∧ C) is constructed as follows:

In the first three columns, A B C B ∧ C A ∧ (B ∧ C) write all possible values T T T T T for the propositions A,B T T F F F and C, and use these to T F T F F calculate the values in the T F F F F fourth column. The final F T T T F column is found by taking F T F F F the “conjunction” of the F F T F F first and fourth columns. F F F F F

Click on the green square to return Solutions to Exercises 15

Exercise 1(b) The logical equivalence of the propositions (A∧B)∧C and A∧(B∧C) can be seen by comparing their truth tables. The former is given in example 2 and the latter in exercise 1(a). These truth tables are identical, so

(A ∧ B) ∧ C = A ∧ (B ∧ C) .

Click on the green square to return Solutions to Exercises 16

Exercise 2(a) The truth table for A ∨ (B ∨ C) is shown below:

The ﬁrst three columns A B C (B ∨ C) A ∨ (B ∨ C) contain all possible values T T T T T for the propositions A, B T T F T T and C. The fourth col- T F T T T umn represents the “dis- T F F F T junction” (B ∨ C) and F T T T T the last column shows the F T F T T “disjunction” of elements F F T T T from the ﬁrst and fourth F F F F F columns.

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Exercise 2(b)

To prove the logical equivalence (A ∨ B) ∨ C = A ∨ (B ∨ C) compare the truth tables for both of these propositions. The former is given in example 3 and the latter in exercise 2(a). These truth tables are identical and therefore (A ∨ B) ∨ C = A ∨ (B ∨ C).

Click on the green square to return Solutions to Exercises 18

Exercise 3(a)

The truth table for the proposition ¬(A ∧ B) is given below.

The adjacent truth table A B A ∧ B ¬(A ∧ B) contains 2×2 = 4 rows cor- T T T F responding to all possible T F F T values for the propositions F T F T A and B. F F F T

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Exercise 3(b)

The truth table for the proposition (¬A) ∧ B is given below.

In the adjacent truth table the third column contains A B ¬A (¬A) ∧ B the “negation” of a given T T F F proposition A and the last T F F F column represents its “con- F T T T junction” with the proposi- F F T F tion B.

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Exercise 3(c)

The truth table for the proposition (¬A) ∧ (¬B) is given below.

In the truth table the third and fourth columns con- A B ¬A ¬B (¬A) ∧ (¬B) tain the “negation” of the T T F F F propositions A and B re- T F F T F spectively, the last column F T T F F is the “conjunction” of ¬A F F T T T and ¬B.

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Exercise 3(d)

The truth table for the proposition (¬A) ∨ (¬B) is below.

In this truth table the third and fourth columns A B ¬A ¬B (¬A) ∨ (¬B) contain the “negation” of T T F F F propositions A and B re- T F F T T spectively, while the last F T T F T column is the “disjunction” F F T T T of ¬A and ¬B.

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Exercise 4(a) The truth table for the proposition A ∧ (B ∨ C) is below.

In the adjacent table, the A B C B ∨ C A ∧ (B ∨ C) ﬁrst three columns contain T T T T T all 8 possible values for T T F T T propositions A, B and C. T F T T T The fourth column shows T F F F F the “disjunction” of B and F T T T F C, while the last one is F T F T F A ∧ (B ∨ C). F F T T F F F F F F

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Exercise 4(b) The truth table for the proposition (A ∧ B) ∨ C is below.

A B C A ∧ B (A ∧ B) ∨ C The ﬁrst three columns T T T T T contain all possible values T T F T T for propositions A, B and T F T F T C. The fourth column T F F F F shows the “conjunction” F T T F T of A and B, while the last F T F F F one is (A ∧ B) ∨ C. F F T F T F F F F F

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Exercise 4(c) The truth table for the proposition (A ∧ B) ∨ (A ∧ C) is below.

The fourth and A B C A ∧ B A ∧ C (A ∧ B) ∨ (A ∧ B) ﬁfth columns show the “con- T T T T T T junction” of A T T F T F T and B, and A T F T F T T and C, respec- T F F F F F tively. The ﬁnal F T T F F F column is the F T F F F F “disjunction” of F F T F F F these. F F F F F F

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Exercise 5(a) The truth tables for A∨(B ∧C) and (A∨B)∧(A∨C) are given below.

A B C A ∨ (B ∧ C) A B C (A ∨ B) ∧ (A ∨ C) T T T T T T T T T T F T T T F T T F T T T F T T T F F T T F F T F T T T F T T T F T F F F T F F F F T F F F T F F F F F F F F F Comparing them shows that A ∨ (B ∧ C) = (A ∨ B) ∧ (A ∨ C) .

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Exercise 5(b) The truth table for (A ∨ B) is given below.

A B A ∨ B A ∧ (A ∨ B) T T T T T F T T F T T F F F F F

Comparing the ﬁrst and the last columns above shows that A ∧ (A ∨ B) = A.

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Exercise 5(c) The truth table for ¬(A ∨ B) and ¬(A) ∧ ¬(B) are given below.

A B ¬(A ∨ B) A B (¬A) ∧ (¬B) T T F T T F T F F T F F F T F F T F F F T F F T

The truth tables are identical so these propositions are logically equivalent, i.e. ¬(A ∨ B) = ¬(A) ∧ ¬(B) .

Click on the green square to return Solutions to Exercises 28

Exercise 5(d) For an arbitrary proposition X, the “disjunction” of X and “false” (i.e. the proposition F ) has the following truth table.

F X F ∨ X F T T F F F

Therefore the relation F ∨ X = X is correct.

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Exercise 5(e) For any two propositions A and B, construct the following truth table:

A B (¬A) ∨ B A ∧ ((¬A) ∨ B) [A ∧ ((¬A) ∨ B)] ∨ B T T T T T T F F F F F T T F T F F T F F

Comparing the last column with the second one we ﬁnd that [A ∧ ((¬A) ∨ B)] ∨ B = B.

Click on the green square to return Solutions to Quizzes 30 Solutions to Quizzes Solution to Quiz:

The truth tables from exercise 3(a) and exercise 3(d) are, respectively,

A B ¬(A ∧ B) A B (¬A) ∨ (¬B) T T F T T F T F T T F T F T T F T T F F T F F T

Since these are identical, it follows that ¬(A ∧ B) = (¬A) ∨ (¬B) .

End Quiz Solutions to Quizzes 31

Solution to Quiz: The truth tables below are from exercise 4(a) and exercise 4(c), respectively.

A B C A ∧ (B ∨ C) A B C (A ∧ B) ∨ (A ∧ B) T T T T T T T T T T F T T T F T T F T T T F T T T F F F T F F F F T T F F T T F F T F F F T F F F F T F F F T F F F F F F F F F

From these, it can be seen that A ∧ (B ∨ C) = (A ∧ B) ∨ (A ∧ C) . End Quiz