Propositional Logic Formal Proof

Question

I'm trying to formally prove the following equation, as practice ahead of my logic exam. However, I'm having a little difficulty working out the steps. Here are the rules that I'm using;

A ∧ A ≡ A, A ∨ A ≡ A idempotence
A ∧ B ≡ B ∧ A, A ∨ B ≡ B ∨ A commutativity
A ∧ (B ∧ C ) ≡ (A ∧ B) ∧ C , A ∨ (B ∨ C ) ≡ (A ∨ B) ∨ C associativity
A ∧ (A ∨ B) ≡ A, A ∨ (A ∧ B) ≡ A absorption
A ∧ (B ∨ C ) ≡ (A ∧ B) ∨ (A ∧ C ) distributivity
A ∨ (B ∧ C ) ≡ (A ∨ B) ∧ (A ∨ C ) distributivity
A ∧ (¬A) ≡ false, A ∨ (¬A) ≡ true negation
¬(¬A) ≡ A double negation
¬(A ∧ B) ≡ (¬A) ∨ (¬B), ¬(A ∨ B) ≡ (¬A) ∧ (¬B) de Morgan
A ⇒ B ≡ (¬A) ∨ B implication
A ⇔ B ≡ (A ⇒ B) ∧ (B ⇒ A) bi-implication

And this is the equation;

 (p⇒r) ∧ (q⇒r) ≣ (p∨q) ⇒ r

I've figured that I use, Implication, Commuatability and Distributivity but I'm stuck at this point. Appreciate any help!

I'm voting to close this question as off-topic because it is about logic and [math.se] instead of programming or software development. — Pang, Apr 30 '16 at 09:01

Leandro Caniglia · Answer 1 · 2016-10-19T20:12:25.383

here is a formal proof

(p∨q) ⇒ r ≣ ¬(p∨q) ∨ r          implication
          ≣ (¬p∧¬q) ∨ r         de Morgan
          ≣ (¬p∨r) ∧ (¬q∨r)     distributivity and commutativity
          ≣ (p⇒r) ∧ (q⇒r)       implication

Note however that nobody thinks this way and that the actual exercise should consist in giving an explanation of why both expressions are equivalent.

EXPLANATION

Given that p ⇒ (p∨q), from (p∨q) ⇒ r, we get p ⇒ (p∨q) ⇒ r and therefore p ⇒ r. Since the same argument is valid for q we also get q ⇒ r. So, (p ⇒ r) ∧ (q ⇒ r).
Conversely, if (p ⇒ r) ∧ (q ⇒ r) then r must be true whenever any of p and q happens to be true. In other words, whenever (p∨q) is true. Hence (p∨q) ⇒ r.

Propositional Logic Formal Proof

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