Use laws of logic to prove ¬ ( p → ¬ q ) ∨ ( q ∧ ( p ∨ ¬ q)) is equivalent to ( p and q)

Question

Please write a detailed answer using the laws used to get to (p and q)

Answer 1

Hi, here's how I did it:

~(p → ~q) ∨ ( q ∧ ( p ∨ ¬ q))

This is the statement you've started with. We can build a truth table to show that ~(p → ~q) ≡ (p ∧ q) (try this yourself!). This gives us

(p ∧ q) ∨ ( q ∧ ( p ∨ ¬ q))

Now, by the associative law, we can regroup the parentheses to get

p ∧ (q ∨ ( q ∧ ( p ∨ ¬ q)))

Notice here that in the parentheses we have q ∨ (something). It doesn't really matter what that something evaluates to, since logically q or something = q. We thus have

p ∧ (q)

which is what we needed to show.

Hope that helps! Let me know if you have any questions.