Show that ($\lnot$q $\implies$ p) $\implies$ (p $\implies$ $\lnot$q) $\equiv$ ($\lnot$p $\lor$ $\lnot$q) using the conditional-disjunction equivalence

Question

Conditional-disjunction equivalence:

(p $\implies$ q) $\equiv$ ($\lnot$p $\lor$ q)

To show:

($\lnot$q $\implies$ p) $\implies$ (p $\implies$ $\lnot$q) $\equiv$ ($\lnot$p $\lor$ $\lnot$q)

Attempt:

$\lnot$($\lnot$q $\implies$ p) $\lor$ (p $\implies$ $\lnot$q) by conditional-disjuction equivalence

$\equiv$ $\lnot$($\lnot$($\lnot$q) $\lor$ p) $\lor$ ($\lnot$p $\lor$ $\lnot$q) by conditional-disjunction equivalence

$\equiv$ $\lnot$(q $\lor$ p) $\lor$ ($\lnot$p $\lor$ $\lnot$q) by double negation law

$\equiv$ ($\lnot$q $\land$ $\lnot$p) $\lor$ ($\lnot$p $\lor$ $\lnot$q) by DeMorgan law

I am stuck at the last step.

Answer 1

You are nearly there.

$(\neg q\wedge\neg p)\vee(\neg p\vee\neg q)\equiv[(\neg q\wedge\neg p)\vee\neg p]\vee\neg q$ by the associativity of $\vee$.

$\color{red}{[(\neg q\wedge\neg p)\vee\neg p]}\vee\neg q\equiv\neg p\vee \neg q$ by absorption on the red term.