# Logic Text Chapter 3 Solutions

## Chapter 3: Truth Tables

### Basic

Question {3.1}

The asterisk is under the column of the main operator.

1:

 y ~ y 0 1 0 1 0 1 *

2 and 12:

 y p ~ y ∨ p y ≡ p 0 0 1 0 1 0 0 1 0 0 1 1 0 1 1 0 0 1 1 0 0 1 0 0 1 0 0 1 1 0 1 1 1 1 1 1 * *

3 and 8:

 y c ~ y ≡ c ~ (y ≡ c) 0 0 1 0 0 0 0 0 1 0 0 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 1 1 0 0 1 1 0 1 0 1 0 1 1 1 * *

4 and 14:

 y c p y ⊃ (c ⊃ p) (y ⊃ c) ⊃ p 0 0 0 0 1 0 1 0 0 1 0 0 0 0 0 1 0 1 0 1 1 0 1 0 1 1 0 1 0 0 1 1 0 0 0 1 1 0 0 0 1 1 0 1 1 1 1 0 1 1 1 1 1 0 0 1 1 0 1 0 1 0 0 1 0 1 0 1 1 1 0 1 1 1 0 0 1 1 1 1 0 1 0 1 0 0 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 * *

5 and 10:

 y c p (y & ~ p) ⊃ c y ≡ (c ∨ ~ p) 0 0 0 0 0 1 0 1 0 0 0 0 1 1 0 0 0 1 0 0 0 1 1 0 0 1 0 0 0 1 0 1 0 0 0 1 0 1 1 0 0 1 1 1 0 0 1 1 0 0 0 1 1 1 0 0 1 1 0 1 1 0 0 1 1 1 0 0 0 1 1 0 1 1 0 1 0 1 1 0 0 1 1 0 1 0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 1 1 1 1 0 1 1 1 1 0 0 1 1 1 1 1 1 1 0 1 * *

6, 11 and 13:

 y c y & c y ∨ c y ⊃ c 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 0 1 1 0 1 1 1 0 1 0 0 1 1 0 1 0 0 1 1 1 1 1 1 1 1 1 1 1 * * *

7:

 y p ~ (y & p) 0 0 1 0 0 0 0 1 1 0 0 1 1 0 1 1 0 0 1 1 0 1 1 1 *

9:

 c ~ ~ c 0 0 1 0 1 1 0 1 *

15:

 y c (y ⊃ c) ≡ (c ⊃ y) 0 0 0 1 0 1 0 1 0 0 1 0 1 1 0 1 0 0 1 0 1 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 *

Question {3.2}

They are all tautologies except for the fourth one.

 ((p & q) ⊃ r) ⊃ (p ⊃ r) 1 0 0 1 0 0 1 0 0

Let p = 1, q = 0 and r = 0. That makes the formula false.

Question {3.3}

1:

 p ~ ~ p 0 0 1 0 1 1 0 1

2:

 p q p & q q & p 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 1 0 1 0 0 0 0 1 1 1 1 1 1 1 1 1

3:

 p q p ⊃ q ~ p ∨ q 0 0 0 1 0 1 0 1 0 0 1 0 1 1 1 0 1 1 1 0 1 0 0 0 1 0 0 1 1 1 1 1 0 1 1 1

4:

 p q r p ⊃ (q ⊃ r) (p ⊃ q) ⊃ r ~ p ∨ ~ (q & ~ r) 0 0 0 0 1 0 1 0 0 1 0 0 0 1 0 1 1 0 0 1 0 0 0 1 0 1 0 1 1 0 1 0 1 1 1 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 0 1 1 0 0 1 0 1 0 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 0 1 0 0 1 1 1 0 0 1 0 1 0 1 1 1 0 1 1 1 0 0 1 1 0 1 1 1 0 0 0 1 1 1 0 1 0 1 0 0 1 1 1 0 0 0 1 0 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 0 1

The first and third propositions are equivalent to each other, and both are not equivalent to the second.

5. p & ~ p is equivalent to ~(qq), as both are contradictions. r ∨ ~ r and ss are equivalent as both are tautologies.

Question {3.4}

1:

 p q p p ⊃ q q 0 02 0 0 1 0 0 0 1 0 0 1 1 1 1 0 1 1 0 0 0 1 1 1 1 1 1 1

The argument form is valid, as there is no row where the premises are true & the conclusion false.

2:

 p q p q p ≡ q 0 0 0 0 0 1 0 0 1 0 1 0 0 1 1 0 1 0 1 0 0 1 1 1 1 1 1 1

The argument form is valid as there is no row where the premises are both true & the conclusion false.

3:

 p q p & q p ≡ q 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 1 1 0 1 0 0 1 0 0 1 1 1 1 1 1 1 1

The argument form is valid as there is no row where the premise is true & the conclusion false.

4:

 p q p q ⊃ p q 0 0 0 0 1 0 0 0 1 0 1 0 0 1 1 0 1 0 1 1 0 * 1 1 1 1 1 1 1

The argument form is invalid as there is a row where the premises are true & the conclusion false, marked with the star.

5:

 p q p q p & q 0 0 0 0 0 0 0 0 1 0 1 0 0 1 1 0 1 0 1 0 0 1 1 1 1 1 1 1

The argument is valid as there is no row where the premises are both true & the conclusion false.

6:

 p q p p ∨ q 0 0 0 0 0 0 0 1 0 0 1 1 1 0 1 1 1 0 1 1 1 1 1 1

The argument is valid as there is no row where the premise is true & the conclusion false.

7:

 p q p ≡ q p ≡ ~ q ~ p 0 0 0 1 0 0 0 1 0 1 0 0 1 0 0 1 0 1 0 1 1 0 1 0 1 0 0 1 1 1 0 0 1 1 1 1 1 1 1 0 0 1 0 1

The argument is valid as there is no row where the premises are true & the conclusion false. (There is no row where premises are true.)

8:

 p q r p ⊃ (q ⊃ r) q ⊃ (p ⊃ r) 0 0 0 0 1 0 1 0 0 1 0 1 0 0 0 1 0 1 0 1 1 0 1 0 1 1 0 1 0 0 1 1 0 0 1 1 0 1 0 0 1 1 0 1 1 1 1 1 1 0 1 1 1 0 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 1 0 1 1 0 1 1 1 1 1 1 0 1 0 1 0 0 1 0 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1

The argument is valid – there is no row where the premise is true and the conclusion false. They are true in exactly the same rows, so they are equivalent.

9:

 p p ⊃ ~ p ~ p 0 0 1 1 0 1 0 1 1 0 0 1 0 1

The argument is valid, as there’s no row where the premise is true and the conclusion false. (Again, they’re equivalent.)

10:

 p ~ ~ p p 0 0 1 0 0 1 1 0 1 1

The argument is valid. ~~p and p are equivalent, what’s more.

11:

 p q r p ⊃ q (r ⊃ p) ⊃ (r ⊃ q) 0 0 0 0 1 0 0 1 0 1 0 1 0 0 0 1 0 1 0 1 0 0 1 1 0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 0 1 1 0 1 1 1 0 0 1 1 1 1 1 0 0 1 0 0 0 1 1 1 0 1 0 1 0 1 1 0 0 1 1 1 0 1 0 0 1 1 0 1 1 1 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

The argument is valid – there is no row where the premise is true and the conclusion false. They are not true in exactly the same rows, so they are not equivalent.

12:

 p q p ⊃ q ~ q ⊃ ~ p 0 0 0 1 0 1 0 1 1 0 0 1 0 1 1 0 1 1 1 0 1 0 1 0 0 1 0 0 0 1 1 1 1 1 1 0 1 1 0 1

The argument is valid. There’s no row with the premise true & conclusion false.

13:

 p q p ~ p ⊃ q 0 0 0 1 0 0 0 0 1 0 1 0 1 1 1 0 1 0 1 1 0 1 1 1 0 1 1 1

The argument is valid. There’s no row with the premise true & conclusion false.

14:

 p q p ⊃ (p ⊃ q) p ⊃ q 0 0 0 1 0 1 0 0 1 0 0 1 0 1 0 1 1 0 1 1 1 0 1 0 1 0 0 1 0 0 1 1 1 1 1 1 1 1 1 1

The argument is valid. There’s no row with the premise true & conclusion false.

15:

 p q p q ⊃ q 0 0 0 0 1 0 0 1 0 1 1 1 1 0 1 0 1 0 1 1 1 1 1 1

The argument is valid. There’s no row with the premise true & conclusion. In fact, the conclusion is a tautology, and any argument like this is valid.

16:

 p q r p ∨ q ~ q ∨ r p ∨ r 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 1 0 1 1 0 1 1 0 1 0 0 1 1 0 1 0 0 0 0 0 0 1 1 0 1 1 0 1 1 1 0 1 1 1 0 0 1 1 0 1 0 1 0 1 1 0 1 0 1 1 1 0 1 0 1 1 1 1 1 1 1 0 1 1 1 0 1 0 0 1 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1

The argument is valid – there is no row where the premise is true and the conclusion false.

17:

 p q p ⊃ q q ⊃ p 0 0 0 1 0 0 1 0 0 1 0 1 1 1 0 0 * 1 0 1 0 0 0 1 1 1 1 1 1 1 1 1 1

The argument is invalid. There’s a row with the premise true & conclusion false (with the star).

18:

 p q r p ⊃ (q ⊃ r) (p ⊃ q) ⊃ r 0 0 0 0 1 0 1 0 0 1 0 0 0 0 0 1 0 1 0 1 1 0 1 0 1 1 0 1 0 0 1 1 0 0 0 1 1 0 0 * 0 1 1 0 1 1 1 1 0 1 1 1 1 1 0 0 1 1 0 1 0 1 0 0 1 0 1 0 1 1 1 0 1 1 1 0 0 1 1 1 1 0 1 0 1 0 0 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1

The argument is invalid – there is a row where the premise is true and the conclusion false. (See the asterisk.)

19:

 p q r (p & q) ⊃ r p ⊃ (~ q ∨ r) 0 0 0 0 0 0 1 0 0 1 1 0 1 0 0 0 1 0 0 0 1 1 0 1 1 0 1 1 0 1 0 0 0 1 1 0 0 1 0 1 0 0 0 1 1 0 0 1 1 1 0 1 0 1 1 1 1 0 0 1 0 0 1 0 1 1 1 0 1 0 1 0 1 1 0 0 1 1 1 1 1 0 1 1 1 1 0 1 1 1 0 0 1 0 0 1 0 0 1 1 1 1 1 1 1 1 1 1 0 1 1 1

The argument is valid – there is no row where the premise is true and the conclusion false.

20:

 p q r s p ⊃ q r ⊃ s (q ⊃ r) ⊃ (p ⊃ s) 0 0 0 0 0 1 0 0 1 0 0 1 0 1 0 1 0 0 0 0 1 0 1 0 0 1 1 0 1 0 1 0 1 1 0 0 1 0 0 1 0 1 0 0 0 1 1 1 0 1 0 0 0 1 1 0 1 0 1 1 1 0 1 1 1 0 1 1 0 1 0 0 0 1 1 0 1 0 1 0 0 1 0 1 0 0 1 0 1 0 1 1 0 1 1 1 0 0 1 0 1 1 0 1 1 0 0 1 1 1 0 0 1 1 1 1 0 1 0 0 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 0 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 1 0 0 0 1 1 0 1 0 1 1 1 1 1 0 1 0 1 0 0 1 0 0 0 1 1 0 1 0 0 1 0 1 1 1 0 0 1 1 1 0 1 1 1 1 1 1 1 1 0 0 1 1 1 0 1 0 1 0 0 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 0 0 1 1 1 1 1 1 1 0 1 1 1 1 0 0 1 1 1 0 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

The argument form is valid. There’s no row where the premises are both true and the conclusion is false.

Question {3.5}

The argument form goes like this: (j & ~ b) ⊃ ~ j therefore jb

This is a valid argument form:

 j b (j & ~ b) ⊃ ~ j j ⊃ b 0 0 0 0 1 0 1 1 0 0 1 0 0 1 0 0 0 1 1 1 0 0 1 1 1 0 1 1 1 0 0 0 1 1 0 0 1 1 1 0 0 1 1 0 1 1 1 1

Question {3.6}

The argument form goes like this: ~(b & ~ m) therefore mb It is invalid.

 ~ (b & ~ m) m ⊃ b 1 0 0 0 1 1 0 0

Question {3.7}

The argument form is: ec, cp, p ⊃ ~ e therefore ~ e It is a valid form.

Question {3.8}

The argument form is: pc, pm, cs therefore ~(~ m &~ s) This is valid.

Question {3.9}

The argument form is: (r ⊃ ~ w) ⊃ r, therefore r. Surprisingly, this is valid. If the conclusion, r, is false, then (r ⊃ ~ w) is true, and r is false, so the premise is also false.

Question {3.10}

(jc) & (ed) therefore (jd) ∨ (ec) Surprisingly, this is valid.

Question {3.11}

Here’s a truth table for exclusive disjunction (I’ll use the symbol x for it).

 p q p x q 0 0 0 0 0 0 1 0 1 1 1 0 1 1 0 1 1 1 0 1

The exclusive disjunction of p and q is true when exactly one of p and q is true.

Question {3.14}

Each formula here is a tautology. I think that this shows that ’ ⊃ ’ is not a good translation of ‘if’ in English.