CSC 1300 – Problem Set 1 - Solutions

1. (20 points) For an integer 𝑥 consider the two sentences:

𝑃: 𝑥 < −3 𝑄: 𝑥 ≥ −3

(a) State 𝑃 ∧ 𝑄 𝑥 < −3 and 𝑥 ≥ −3

(b) State 𝑃 ∨ 𝑄 𝑥 < −3 or 𝑥 ≥ −3

(c) Use De Morgan’s Laws to state the negation of (a). Negation of (a): not the case that 𝑥 < −3 or 𝑥 ≥ −3 De Morgan’s Law : ¬(𝑃 ∧ 𝑄) ≡ ¬𝑃 ∨ ¬𝑄: 𝑥 ≱ −3 or 𝑥 ≮ −3 meaning 𝑥 ≥ −3 or 𝑥 < −3

(d) Use De Morgan’s Laws to state the negation of (b). Negation of (b): not the case that 𝑥 < −3 and 𝑥 ≥ −3 De Morgan’s Law : ¬(𝑃 ∨ 𝑄) ≡ ¬𝑃 ∧ ¬𝑄: 𝑥 ≮ −3 and 𝑥 ≱ −3 meaning 𝑥 ≥ −3 and 𝑥 < −3

2. (25 points) Let 𝑃, 𝑄, 𝑅, 𝑆, 𝑈, and 𝑊 be statements. Prove the following are true. Justify your answer using truth tables or using one or more of the laws governing logical equivalences.

(a) 𝑃 ∨ 𝑄 ∨ ¬𝑅 ≡ ¬𝑅 ∨ 𝑃 ∨ 𝑄

𝑃 ∨ 𝑄 ∨ ¬𝑅 ≡ 𝑃 ∨ 𝑄 ∨ ¬𝑅 [Association]

≡ ¬𝑅 ∨ 𝑃 ∨ 𝑄 [Communicative]

(b) 𝑃 ∨ 𝑄 ∧ ¬𝑅 ≡ 𝑃 ∨ 𝑄 ∧ (𝑃 ∨ ¬𝑅)

𝑃 ∨ 𝑄 ∧ ¬𝑅 ≡ 𝑃 ∨ 𝑄 ∧ 𝑃 ∨ ¬𝑅 [Distributive]

(c) 𝑃 → 𝑄 ∨ 𝑅 ≡ 𝑃 → 𝑄 ∨ 𝑃 → 𝑅

𝑃 → 𝑄 ∨ 𝑅 ≡ 𝑃 → 𝑄 ∨ (𝑃 → 𝑅) [Distributive]

(d) 𝑃 ∧ 𝑄 ⊕ 𝑅 ≡ (𝑃 ∧ 𝑄)⊕ (𝑃 ∧ 𝑅)

𝑃 ∧ 𝑄 ⊕ 𝑅 ≡ (𝑃 ∧ 𝑄)⊕ (𝑃 ∧ 𝑅) [Distributive]

(e) 𝑃 ∧ ¬𝑃 ∨ 𝑄 ∧ ¬𝑄 ∨ 𝑅 ∧ ¬𝑅 ∨ 𝑆 ∧ ¬𝑆 ∨ ¬𝑈 ∧ 𝑈 ∨𝑊 ≡ 𝑃 ∧ 𝑄 ∧ 𝑅 ∧ 𝑆 ∧ ¬𝑈 ∧𝑊

𝑷 ∧ ¬𝑷 ∨ 𝑸 ∧ ¬𝑄 ∨ 𝑅 ∧ ¬𝑅 ∨ 𝑆 ∧ ¬𝑆 ∨ ¬𝑈 ∧ 𝑈 ∨𝑊 ?≡ 𝑃 ∧ 𝑄 ∧ 𝑅 ∧ 𝑆 ∧ ¬𝑈 ∧𝑊

𝑃 ∧ ¬𝑃 ∨ 𝑄 ≡ 𝑃 ∧ ¬𝑃 ∨ (𝑃 ∧ 𝑄) [Distributive]

≡ 𝐹 ∨ (𝑃 ∧ 𝑄) [Contradiction]

≡ 𝑃 ∧ 𝑄 [Identity Law]

(𝑷 ∧ 𝑸) ∧ ¬𝑸 ∨ 𝑹 ∧ ¬𝑅 ∨ 𝑆 ∧ ¬𝑆 ∨ ¬𝑈 ∧ 𝑈 ∨𝑊 ?≡ 𝑃 ∧ 𝑄 ∧ 𝑅 ∧ 𝑆 ∧ ¬𝑈 ∧𝑊

(𝑃 ∧ 𝑄) ∧ ¬𝑄 ∨ 𝑅 ≡ ( 𝑃 ∧ 𝑄) ∧ ¬𝑄 ∨ ( 𝑃 ∧ 𝑄 ∧ 𝑅) [Distributive]

≡ ( 𝑃 ∧ ¬𝑄) ∧ (𝑄 ∧ ¬𝑄) ∨ ( 𝑃 ∧ 𝑅 ∧ (𝑄 ∧ 𝑅)) [Distributive]

≡ ( 𝑃 ∧ ¬𝑄) ∧ 𝐹 ∨ ( 𝑃 ∧ 𝑅 ∧ (𝑄 ∧ 𝑅))

≡ 𝐹 ∨ ( 𝑃 ∧ 𝑅 ∧ (𝑄 ∧ 𝑅))

≡ 𝑃 ∧ 𝑅 ∧ (𝑄 ∧ 𝑅)

≡ 𝑃 ∧ 𝑄 ∧ (𝑅 ∧ 𝑅) [Associative]

≡ 𝑃 ∧ 𝑄 ∧ 𝑅

𝑃 ∧ 𝑄 ∧ 𝑹 ∧ ¬𝑹 ∨ 𝑺 ∧ ¬𝑆 ∨ ¬𝑈 ∧ 𝑈 ∨𝑊 ?≡ 𝑃 ∧ 𝑄 ∧ 𝑅 ∧ 𝑆 ∧ ¬𝑈 ∧𝑊

𝑹 ∧ ¬𝑹 ∨ 𝑺 ≡ (𝑅 ∧ ¬𝑅) ∨ (𝑅 ∧ 𝑆) [Distributive]

≡ 𝐹 ∨ (𝑅 ∧ 𝑆) [Contradiction]

≡ 𝑅 ∧ 𝑆

𝑃 ∧ 𝑄 ∧ 𝑅 ∧ 𝑺 ∧ ¬𝑺 ∨ ¬𝑼 ∧ 𝑈 ∨𝑊 ?≡ 𝑃 ∧ 𝑄 ∧ 𝑅 ∧ 𝑆 ∧ ¬𝑈 ∧𝑊

𝑺 ∧ ¬𝑺 ∨ ¬𝑼 ≡ (𝑆 ∧ ¬𝑆) ∨ (𝑆 ∧ ¬𝑈) [Distributive]

≡ 𝐹 ∨ (𝑆 ∧ ¬𝑈) [Contradiction]

≡ 𝑆 ∧ ¬𝑈 [Identity]

𝑃 ∧ 𝑄 ∧ 𝑅 ∧ 𝑆 ∧ ¬𝑼 ∧ 𝑼 ∨𝑾 ?≡ 𝑃 ∧ 𝑄 ∧ 𝑅 ∧ 𝑆 ∧ ¬𝑈 ∧𝑊

¬𝑼 ∧ 𝑼 ∨𝑾 ≡ (¬𝑼 ∧ 𝑈) ∨ (¬𝑼 ∧𝑊) [Distributive]

≡ 𝐹 ∨ (¬𝑼 ∧𝑊) [Contradiction]

≡ ¬𝑼 ∧𝑊 [Identity]

𝑃 ∧ 𝑄 ∧ 𝑅 ∧ 𝑆 ∧ ¬𝑈 ∧𝑊 ≡ 𝑃 ∧ 𝑄 ∧ 𝑅 ∧ 𝑆 ∧ ¬𝑈 ∧𝑊

3. (35 points) Consider the implication: If 𝑥 and 𝑦 are even, then 𝑥𝑦 is even.

(a) State the implication using the phrase “only if”

𝑥 and 𝑦 are even only if 𝑥𝑦 is even.

(b) State the implication using the word “sufficient” x being even, and y being even, is sufficient for xy to be even.

(c) State the converse of the implication If xy is even, then x and y are even.

(d) State the contrapositive of the implication If xy is odd, then x and y are odd.

(e) Is the implication true for some positive integers x and y? Is it true for all positive integers x and y? True for all.

(f) Is the converse of the implication true for some positive integers x and y? Is it true for all positive integers x and y? True for some.

(f) Is the contrapositive of the implication true for some positive integers x and y? Is it true for all positive integers x and y? True for all.

4. (20 points) Let 𝑃 and 𝑄 be statements. Determine whether the compound statements below are tautologies, contradictions, or neither. Justify your answer using truth tables or using one or more of the laws governing logical equivalences.

(a) (P ∧ ¬Q ) ∧ (P ∧ Q)

P Q ¬Q P∧ (¬Q) P∧Q (P∧ (¬Q))∧ (P∧Q)T T T F T FT F T T F FF T F F F FF F T F F F

Contradiction

(b) P ∧ Q → (P → Q)

P Q P∧Q P→Q (P∧Q)→(P→Q)T T T T TT F F F TF T F T TF F F T T

Tautology

(c) P ∧ (¬Q) → (P ∨ Q)

P Q ¬Q P∧ (¬Q) P∨Q (P∧ (¬Q))→(P∨Q)T T F F T TT F T T T TF T F F T TF F T F F T

Tautology

(d) P ∧ Q ↔ ((¬P) ∨ (¬Q))

P Q P∧Q ¬P ¬Q (¬P)∨ (¬Q) (P∧Q)↔((¬P)∨ (¬Q))T T T F F F FT F F F T T FF T F T F T FF F F T T T F

contradiction