Vtyjkyuk

$(pland q)rightarrow r$ and $(prightarrow r)lor (qrightarrow r)$

Have to try prove if they are logically equivalent or not using the laws listed below and also if need to use negation and implication laws. I was going to use associative law and then distributive but I wasn't sure how to get rid of the "implies"

Commutative laws: p ∧ q ≡ q ∧ p
p ∨ q ≡ q ∨ p

De Morgan’s laws: ∼(p ∧ q) ≡ ∼p ∨ ∼q
∼(p ∨ q) ≡ ∼p ∧ ∼q

Idempotent laws: p ∧ p ≡ p
p ∨ p ≡ p

Associative laws: (p ∧ q) ∧ r ≡ p ∧ (q ∧ r)
(p ∨ q) ∨ r ≡ p ∨ (q ∨ r)

Distributive laws: p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

With the laws that you provide you will not ba able to prove their equivalence. You need an equivalence involving implications. here is the one that is typically used:

Implication: $p rightarrow q equiv neg p lor q$

Use it as follows:

$(p land q) rightarrow r equiv$ (implication)

$neg (p land q) lor r equiv$ (deMorgan)

$(neg p lor neg q) lor r equiv$ (Idempotence)

$(neg p lor neg q) lor (r lor r) equiv$ (Association)

$neg p lor ( neg q lor (r lor r)) equiv$ (Association)

$neg p lor ((neg q lor r) lor r) equiv$ (commutation)

$neg p lor (r lor (neg q lor r)) equiv$ (Association)

$(neg p lor r) lor (neg q lor r) equiv$ (implication)

$(p rightarrow r) lor (q rightarrow r)$

The two statements are not equivalent.

Obviously you cannot use equivalence principles to demonstrate non-equivalence, so let's use a counterexample:

Let $p=True$, $q =False$, and $r=False$

then $(p land q) rightarrow r = (Tland F) rightarrow F = F rightarrow F = T$

But $(p rightarrow r) land (q rightarrow r) = (T rightarrow F) land (F rightarrow F) = Fland T =F$

You are missing a equivalence:

1. p → q ≡ ¬p v q

I find it easier to work on the right side:

 (p ∧ q)→r ≡ (p → r) ∨ (q → r) 

Taking only the right side:

(p → r) ∨ (q → r) (Using 1.)

(¬p v r) ∨ (¬q v r) (Commutative laws on the right parentheses)

(¬p v r) ∨ (r v ¬q) (Associative laws)

¬p v ((r ∨ r) v ¬q) (Idempotent laws)

¬p v ((r) v ¬q) (Commutative laws)

¬p v (¬q v r) (Associative laws)

((¬p v ¬q) v r) (De Morgan's laws)

¬(p ∧ q) v r (Using 1.)

(p ∧ q) → r