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Question 1

• A set of premises is given below

(𝑝 ⟶ 𝑞) ∨ (~𝑟)
(~𝑝) ∧ (𝑞 ∨ 𝑟)
𝑟 ⟶ (~𝑝)

Determine which of the following statements is a valid conclusion from the above set of premises using truth tables or by providing a logical explanation.

(~ 𝑝) ∨ (~𝑟)
𝑞 ⟶ (𝑝 ∧ (~𝑞))
𝑝 ⟶ (𝑞 ∧ ~𝑟)

• Construct a chain of logical equivalences to show that

(~𝑞 ∧ 𝑟) ⟶ (𝑝 ⟶ 𝑞) ≡ (~𝑞) ⟶ (𝑝 ⟶ ~𝑟).

Do not use truth tables in this part of the question.

• Use the Rules of Inference to prove that the following argument form is valid.

𝑞 ∨ 𝑟
(𝑝 ∧ 𝑞) ⟶ s
~ s
∴ 𝑝 ⟶ 𝑟

Do not use truth tables in this part of the question.

Question 2

• Give a counter-example to show that the following statement is false.

∀𝑥 ∈ ℕ ∀𝑦 ∈ ℝ ∀𝑧 ∈ ℝ ((𝑥2 < 𝑦2) ∨ (𝑦2 < 𝑧2)) ⟶ ((𝑥 < 𝑦) ∨ (𝑦 < 𝑧))

• Provide the negation of the statement, giving your answer without using any logical negation symbol. Equality and inequality symbols such as =, ≠, <, > are allowed.

∃𝑥 ∈ ℤ ∀𝑦 ∈ ℕ ∀𝑧 ∈ ℕ ((𝑥 ≠ 0) ∧ (𝑥𝑦)𝑧 = 1) ⟶ ((𝑧 = 0) ∨ (𝑥𝑦 = 1))

• Let 𝐷 be the set

𝐷 = {−10, −9, −7, −6, −4, −3, −2,0,1,2,3,4,5,6,9,10,12,13,14}.

Suppose that the domain of the variable 𝑥 is 𝐷. Write down the truth set of the predicate.

((𝑥 > 1) ⟶ (𝑥 is even)) ⟶ (𝑥 is divisible by 4).

• Let 𝑃,𝑄, 𝑅, 𝑆 denote predicates. Use the Rules of Inference to prove that the following argument form is valid.

∀𝑥 (𝑃(𝑥) ⟶ (∀𝑦 𝑄(𝑦)))
∀𝑥 (𝑅(𝑥) ⟶ (∃𝑦 ~𝑄(𝑦)))
∃𝑥 (𝑅(𝑥) ∧ 𝑆(𝑥))
∴ ∀𝑥 ~𝑃(𝑥