Question

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

Answer #1

[16pt] Which of the following formulas are semantically
equivalent to p → (q ∨ r): For each
formula from the following (denoted by
X) that is equivalent to p → (q
∨ r), prove the validity of X
« p → (q ∨
r) using natural deduction. For
each formula that is not
equivalent to p → (q ∨ r), draw its truth table
and clearly mark the entries that result in the
inequivalence. Assume the binding priority used in...

Prove: ~p v q |- p -> q by natural deduction

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

1) Show that ¬p → (q → r) and q → (p ∨ r) are logically
equivalent. No truth table and please state what law you're using.
Also, please write neat and clear. Thanks
2) .Show that (p ∨ q) ∧ (¬p ∨ r) → (q ∨ r) is a tautology. No
truth table and please state what law you're using. Also, please
write neat and clear.

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

Prove
a)p→q, r→s⊢p∨r→q∨s
b)(p ∨ (q → p)) ∧ q ⊢ p

Use a truth table to determine whether the following argument is
valid.
p
→q ∨ ∼r
q →
p ∧ r
∴ p →r

Given: (P & ~ R) > (~R & Q), Q> ~P Derive: P >
R. use propositional logic and natural derivation rules.

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

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.

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