Is $(p \lor q) \to (\neg p \to (p \lor q))$ a tautology?

Question

This question came up in a recent assignment. I was asked to find a logically equivalent alternative to the statement in the title from the options below.

1. (p ∨ q) ∨ ¬p
2. (¬p & ¬q) ∨ (p ∨ q)
3. (p & q) ∨ (p ∨ q)
4. (¬p & ¬q) ∨ (¬p & q)

No matter how many times I redo this question I always come up with both options 1 and 2 being a logical equivalent. Have I missed something somewhere or is there an error in the question?

Answer 1

Hint

Rewrite $(¬p \to (p \lor q))$ as $(p \lor (p \lor q))$ using Material Implication.

Then use Associativity and Idempotency to simplify it to : $p \lor q$.

What you get is :

$(p \lor q) \to (p \lor q)$.

Answer 2

rewrite (p ∨ q) -> [¬p -> (p ∨ q)] as (¬(p ∨ q))∨[¬p -> (p ∨ q)] as p-> q ac be rewritten as (¬p) ∨q

so (¬(p ∨ q))∨[¬p -> (p ∨ q)]=(¬(p ∨ q))∨((¬(¬p))∨(p ∨ q) can be written as (¬(p ∨ q))∨(p∨(p∨q))

or (¬(p ∨ q))∨(p∨q) which is a tautology

Answer 3

Let $r=p\lor q$. Suppose the negation of your proposition; that is, consider $$\lnot(r\to(\lnot p\to r)).$$ The only way an implication can be false is if the first part is true while its conclusion is false, so we have $r$ and $\lnot(\lnot p\to r)$, but by the same token, we have $\lnot p$ and $\lnot r$, which contradicts $r$. Hence your proposition must be a tautology.