# Show whether the following statement in a tautology or a contradiction?

What I am given:

[(p∧r)∧(p→ q)]→q

What I did:

⇔ [(p∧r)∧( ¬p V q)] → q: Implication

⇔ ¬ [(p∧r)∧( ¬p V q)] V q: Implication

⇔ ¬ (p∧r) V ¬ ( ¬p V q) V q: De Morgan

⇔ (¬p V ¬r) V (¬ ¬p ∧ ¬q) V q: De Morgan

⇔(¬p V ¬r) V (p ∧ ¬q) V q: Double Negation

⇔(¬p V ¬r) V (q V p) ∧ (q V ¬q): Distributive

⇔(¬p V ¬r) V (q V p) ∧ T: Tautology

⇔(¬p V ¬r) V (q V p): Identity

⇔(¬p V p) V (q V r): Associative

⇔T V (q V r): Tautology

⇔ T Domination law

I do not think this is right. Can someone provide insight?

Thanks

• Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax. May 28, 2019 at 18:27
• @dantopa It's a good question already. Don't template the newbies. May 28, 2019 at 18:33

The outer operator is an implication, so if this is false then $$q$$ must be false and $$(p\land r)\land(p\to q)$$ must be true, which in turn means that $$p,r,p\to q$$ must all be true. But then assigning $$p,r$$ as true means $$p\to q$$ is false, a contradiction. So the whole expression is a tautology.
As for the bonus question, $$p\mid p$$ (NAND) is equal to $$\neg p$$. This can be seen by truth table.