Question

Prove

a)p→q, r→s⊢p∨r→q∨s

b)(p ∨ (q → p)) ∧ q ⊢ p

Answer #1

Prove or disprove that [(p → q) ∧ (p → r)] and [p→ (q ∧ r)] are
logically equivalent.

What is the correct meaning of the logical expression p→q∨r∧s
?
((p→q)∨r)∧s
p→((q∨r)∧s)
(p→(q∨r))∧s
p→(q∨(r∧s))

Prove ((P ∨ ¬Q) ∧ (¬P ∨ R)) → (Q → R)
Hint: this starts with the usual setup for an implication, then
repeatedly uses disjunctive syllogism.

Prove: (p ∧ ¬r → q) and p → (q ∨ r) are biconditional using
natural deduction NOT TRUTH TABLE

1. Prove p∧q=q∧p
2. Prove[((∀x)P(x))∧((∀x)Q(x))]→[(∀x)(P(x)∧Q(x))]. Remember to
be strict in your treatment of quantifiers
.3. Prove R∪(S∩T) = (R∪S)∩(R∪T).
4.Consider the relation R={(x,y)∈R×R||x−y|≤1} on Z. Show that
this relation is reflexive and symmetric but not transitive.

Using rules of inference prove.
(P -> R) -> ( (Q -> R) -> ((P v Q) -> R) )
Justify each step using rules of inference.

Give direct and indirect proofs of:
a. p → (q → r), ¬s ∨ p, q ⇒ s → r.
b. p → q, q → r, ¬(p ∧ r), p ∨ r ⇒ r

p → q, r → s ⊢ p ∨ r → q ∨ s
Solve using natural deduction rules.

Give the indirect proofs of:
p→q,¬r→¬q,¬r⇒¬p.p→q,¬r→¬q,¬r⇒¬p.
p→¬q,¬r→q,p⇒r.p→¬q,¬r→q,p⇒r.
a∨b,c∧d,a→¬c⇒b.

Use the laws of propositional logic to prove the following:
1) (p ∧ q ∧ ¬r) ∨ (p ∧ ¬q ∧ ¬r) ≡ p ∧ ¬r
2) (p ∧ q) → r ≡ (p ∧ ¬r) → ¬q

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