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

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.

$$(\neg q\wedge\neg p)\vee(\neg p\vee\neg q)\equiv[(\neg q\wedge\neg p)\vee\neg p]\vee\neg q$$ 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.
• I Never saw ($\lnot$q $\lor$ $\lnot$p) as a single term to which the a law can be applied. Thank you. – Haslo Vardos Nov 7 '20 at 11:10