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Home , Negation, Reference, Truth value

PHIL12A Section answers, 16 February 2011

Julian Jonker

1 How much do you know?

1. Show that the following sentences are equivalent.

(a) (Ex 4.16) A ∨ ∨ A and A ∨ B AB (A ∨ B) ∨ A A ∨ B TT TTT T T T T T TF TTF T T T T F FT FTT T F F T T FF FFF F F F F F

(b) (Ex 4.18) A ∨ (B ∧ ) and (A ∨ B) ∧ (A ∨ C) ABC A ∨ (B ∧ C) (A ∨ B) ∧ (A ∨ C) TTT T T TTT TTT T TTT TTF T T TFF TTT T TTF TFT T T FFT TTF T TTT TFF T T FFF TTF T TTF TFF T T FFF TTF T TTF FTT F T TTT FTT T FTT FTF F F TFF FTT F FFF FFT F F FFT FFF F FTT FFF F F FFF FFF F FFF

1 2. For each of the following , show that the conclusion is a tautological consequence of the using tables.

(I’ll use the shortcut method in each case, starting by placing a T under the main connective of each and an F under the main connective of the conclusion, and the working backwards until I hit a , at which point I will leave an x under the sentence to which I cannot assign a consistent .

(a) (Ex 4.20)

1 (A ∧ B) ∨ C

2 (B ∨ C)

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

(b) (Ex 4.22)

1 A

2 B ∨ C

3 ((B ∧ A) ∨ (C ∧ A)

AB ∨ C ((B ∧ A) ∨ (C ∧ A) TT F TT FTFFT T Fx T Fx FFTFFFT

2 (c)

1 A ∨ B

2 B ∨ C

3 ¬B

4 A ∧ C

A ∨ BB ∨ C ¬ B A ∧ C TTT F TFFTTF F TTFFTTTF Tx F Tx

(d) The following conclusion is in fact not a tautological consequence of the premises, as the truth assignment below illustrates.

1 A ∨ B

2 C ∨ D

3 ¬A ∨ ¬C

4 B ∧ D

A ∨ BC ∨ D ¬ A ∨ ¬ C B ∧ D FTTTTFTFTFT TFF

3. Recall the following equivalences:

Associativity of ∧: P ∧ (Q ∧ R) ⇔ (P ∧ Q) ∧ R ⇔ P ∧ Q ∧ R

Associativity of ∨: P ∨ (Q ∨ R) ⇔ (P ∨ Q) ∨ R ⇔ P ∨ Q ∨ R

Commutativity of ∧: P ∧ Q ⇔ Q ∧ P

Commutativity of ∨: P ∨ Q ⇔ Q ∨ P

Idempotence of ∧: P ∧ P ⇔ P

Idempotence of ∨: P ∨ P ⇔ P

Double : ¬¬P ⇔ P

3 De Morgan’s laws: ¬(P ∨ Q) ⇔ ¬P ∧ Q) ¬(P ∧ Q) ⇔ ¬P ∨ ¬Q

Using these equivalences to put the following sentences into , using a chain of equiv- alences. At each step of the chain, indicate which principle you are using:

(a) (Ex 4.36) (¬A ∨ B) ∨ (B ∨ C)

(¬A ∨ B) ∨ (B ∨ C) ⇔ ¬A ∨ B ∨ B ∨ C (Associativity of ∨)

⇔ ¬A ∨ B ∨ C (Idempotence of ∨)

(b) (A ∨ B) ∧ C ∧ (¬(¬B ∧ ¬A) ∨ B)

(A ∨ B) ∧ C ∧ (¬(¬B ∧ ¬A) ∨ B) ⇔ (A ∨ B) ∧ C ∧ (¬¬B ∨ ¬¬A ∨ B) (De Morgan’s laws)

⇔ (A ∨ B) ∧ C ∧ (B ∨ A ∨ B) ()

⇔ (A ∨ B) ∧ C ∧ (A ∨ B) (Idempotency of ∨)

⇔ (A ∨ B) ∧ C (Idempotency of ∧)

(c) ¬(¬¬A ∧ ¬¬B) ∨ C ∨ (¬B ∨ ¬A) ∨ ¬A

¬(¬¬A ∧ ¬¬B) ∨ C ∨ (¬B ∨ ¬A) ∨ ¬A ⇔ ¬(A ∧ B) ∨ C ∨ (¬B ∨ ¬A) ∨ ¬A (Double negation)

⇔ (¬A ∨ ¬B) ∨ C ∨ (¬B ∨ ¬A) ∨ ¬A (De Morgan’s laws)

⇔ (¬A ∨ ¬B) ∨ C ∨ (¬A ∨ ¬B) ∨ ¬A (Commutativity of ∨)

⇔ (¬A ∨ ¬B) ∨ C ∨ ¬A (Commutativity of ∨)

⇔ ¬A ∨ ¬B ∨ C ∨ ¬A (Associativity of ∨)

⇔ ¬A ∨ ¬B ∨ C (Idempotency of ∨)

2 Something slightly harder, if there’s time.

1. Think of two sentences P and Q such that P is a of Q, but not a tautological conse- quence.

An example: P:‘x = 2’, Q:‘x2 = 4’

4 2. (Ex 4.25) Is A ∧ B a tautological consequence of A ∨ B? Give an example of two different sentences A and B in the blocks language such that A ∧ B is a logical consequence of A ∨ B.

A ∧ B is not a tautological consequence of A ∨ B: consider for example the row of the in which A is true and B is .

However, if A is interpreted as LeftOf(a,b) and B is interpreted as RightOf(b,a), then we have that A ∧ B is a logical consequence of A ∨ B.

3. When using truth tables to determine whether one sentence is a tautological consequence of a of sentences, you normally write out all possible truth values for each atomic sentence. Once you have more than three atomic sentences, this becomes laborious. Is there a quicker way to do it? Show how to do it with one of the examples above.

See the examples above!

3 Challenge question

Consider the arbitrary sentence φ consisting of parentheses, ∧ , ∨ symbols, ¬ symbols and n atomic sentences. (Make up an example of such a complex sentence and draw its truth table in order to help with the questions that follow.) What are the rows of its truth table? How many rows in its truth table? Can a truth value be assigned to each row of the truth table of φ? How would you justify this? Consider for example φ = ¬(¬(A ∨ B) ∧ C). The truth table for this sentence has three reference columns and eight rows. Since the sentence consists of just atomic sentences, parentheses and truth-functional connectives, we can draw up a truth table and assign a truth value in each row on the basis of the truth values of the atomic sentences and the definitions of the truth-functional connectives. Now imagine that you have such a truth table. Consider the rows which make φ true. For each such row, there is a way to compose a sentence Li using the atomic sentences in φ, ∧ symbols and ¬ symbols. Can you work out how to do it? (Think about what this row of the truth table means, and how you could express this using just the atomic sentences of φ, ∧ symbols and ¬ symbols.) Here is the truth table for φ:

5 ABC ¬ (¬ (A ∨ B) ∧ C) TTT TFTTTFT TTF TFTTTFF TFT TFTTFFT TFF TFTTFFF FTT TFFTTFT FTF TFFTTFF FFT FTFFFTT FFF TTFFFFF

Each row of the truth table produces a truth value under the main connective of the sentence that is a function of the truth values of the atomic sentences. A truth table is just an intuitive way of illustrating this. However, we could

talk about the function F¬(¬(A∨B)∧C) as a function from all the possible combinations of truth values of the atomic sentences to the two truth values T and F. So each row represents the value that this function takes on when given the truth values under the atomic sentences as its arguments. In other , we could represent the second row of

the truth table in the following way: F¬(¬(A∨B)∧C)(T, T, F) = T. For each arbitrary sentence we could produce a function in the same way mapping the combinations of the atomic sentences to T and F. Now consider that for each row there are a finite number of true atomic sentences upon which the truth value of the sentence depends. If the sentence is true in this row, then its truth value is equivalent to the conjunction of those atomic sentences which are true and the of those atomic sentences which are false. For example,

F¬(¬(A∨B)∧C)(T, T, F) = T = FA∧B∧¬C (T, T, F). So we let L2 = A ∧ B ∧ ¬C. Similarly, for each row in which

φ is true, there is a sentence Li consisting of the conjunction of the atomic sentences that are true in that row and the negations of those atomic sentences which are false. If any such sentence Li is true, it is because the atomic sentences of which it consists take on the truth values that match the ith row of the truth table, and by construction φ is true in that row. So whenever any Li is true, φ is also true.

Now convince yourself that φ is equivalent to the disjunction of the Ls – that is, L1 ∨ L2 ∨ ...Lj, where j is the number of true rows of the truth table. If you have got this far, well done. You have demonstrated why it is that every sentence we can so far, no matter how complex the sentence, can be written in negation normal form. (Actually, you have shown something more: that every sentence can be written in .)

Suppose there are j true rows of the truth table of φ. Since φ is true when L1 is true, or when L2 is true, ... , or when Lj is true, it follows that φ is true when at least one of the disjuncts of L1 ∨ L2 ∨ ... ∨ Lj is true. Also, since the truth table of φ exhausts the possible combinations of truth values of the atomic sentences of φ, we know that when φ is true it is true in some ith row of its truth table. But in that case Li is true, and so the disjunction

L1 ∨ L2 ∨ ... ∨ Lj is true. This means that φ is equivalent to L1 ∨ L2 ∨ ... ∨ Lj, a sentence in disjunctive normal form. Since φ was arbitrary, we conclude that we can represent any sentence consisting of atomic sentences, parentheses and truth-functional connectives in disjunctive normal form.

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