verify logical equivalence without truth table

Question

$(p\land q)\rightarrow r$ and $(p\rightarrow r)\lor (q\rightarrow r)$

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)

Answer 1

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

Implication: $p \to q \equiv \neg p \lor q$

Use it as follows:

$$ \begin{align*} (p \land q) \to r &\equiv \neg (p \land q) \lor r &(\text{Implication}) \\ &\equiv (\neg p \lor \neg q) \lor r &(\text{De Morgan}) \\ &\equiv (\neg p \lor \neg q) \lor (r \lor r) &(\text{Idempotence}) \\ &\equiv \neg p \lor (\neg q \lor (r \lor r)) &(\text{Association}) \\ &\equiv \neg p \lor ((\neg q \lor r) \lor r) &(\text{Association}) \\ &\equiv \neg p \lor (r \lor (\neg q \lor r)) &(\text{Commutation}) \\ &\equiv (\neg p \lor r) \lor (\neg q \lor r) &(\text{Association}) \\ &\equiv (p \to r) \lor (q \to r) &(\text{Implication}) \end{align*} $$

Answer 2

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 $(p \land q) \rightarrow r = (T\land F) \rightarrow F = F \rightarrow F = T$

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