How to prove this logical equivalence using different laws?

Question

Prove that $﹁p → (q→r)$ and $q → (p∨r)$ are logically equivalent using different laws.

this is my answer:

$﹁p → (q→r) = q → (p∨r)$

$(q→r) = ﹁q∨r$ implication equivalence

$﹁p → (q→r) = p∨(﹁q∨r)$ material implication

$(p∨﹁q) ∨ (p∨r)$ distribution

$(q → p) ∨ (p∨r)$ implication

$q → (p ∨ p) ∨ r = q → (p∨r)$ tautology

Hi guys can you please check my answer if it is correct...Thanks in advance

Answer 1

There's no need to use distribution since $p \lor(\lnot q \lor r) \equiv p \lor \lnot q \lor r$ due to associativity of $\lor$, and by commutativity of $\lor$, that gives us $$\lnot q \lor p \lor r \equiv \lnot q \lor (p \lor r) \equiv q\rightarrow (p \lor r)$$

What is highly questionable in your proof is the second line of the following:

$(q → p) ∨ (p∨r)$

$q → (p ∨ p) ∨ r = q → (p∨r)$

You need to say more to justify this move. In the end, though, as I point out above, it is unnecessary to have distributed $\lor $ over $\lor$.

Answer 2

1. q→(p ∨ r) Given
2. ¬q ∨ (p ∨ r) Switcheroo
3. p ∨( ¬q ∨ r) Associative
4. ¬p →(q→r) Switcheroo

Answer 3

1) $\lnot p\to(q\to r)$

2) $\lnot[\lnot p]\lor [q\to r]$ using implication law

3) $p\lor [q\to r]$ using double negation law

4) $p\lor [\lnot q\lor r]$ using implication law

5) $\lnot q\lor [p\lor r]$ using associative law

6) $\lnot q\to [p\lor r]$ using implication law