Does this prove that [p ∧ (p → q)] → q is a tautology?

Question

Do I make any mistakes in this proof?

[p ∧ (p → q)] → q ≡

¬[p ∧ (p → q)] ∨ q ≡

¬[p ∧ (¬p ∨ q)] ∨ q ≡

[¬p ∨ ¬(¬p ∨ q)] ∨ q ≡

[¬p ∨ (¬¬p ∧ ¬q)] ∨ q ≡

[¬p ∨ (p ∧ ¬q)] ∨ q ≡

[(¬p ∨ p) ∧ (¬p ∨ ¬q)] ∨ q ≡

[ T ∧ (¬p ∨ ¬q)] ∨ q ≡

(¬p ∨ ¬q) ∨ q ≡

¬p ∨ (¬q ∨ q) ≡

¬p ∨ T ≡

T

Is there a better way to do this? I am struggling to do the propositional calculus equivalent of multiplying by 1?

For example, on a different question, a person did a solution where they did

is there a way to apply a similar tool/technique in this situation?

Thank you so much. Any help is greatly appreciated.

Answer 1

Your proof seems correct. Here's an alternative method:

$a\to b$ is $F$ iff $a=T$ and $b=F$

So $p \land (p\to q)$ must be $T$, which means $p=T$ and so $q=T$. But then we have $T\to T\equiv T$ and not $F$. So there is no case where the expression is $F$.