Prove that not every boolean function is equal to a boolean function constructed by only using $∧$ and $∨$.
If p,q = (0,1)
(p$∧$q)$∨$q = (0$∧$1)$∨$1 = 1
(p$∧$q)$∨$~q = (0$∧$1)$∨$~1 = 0
Therefore (p$∧$q)$∨$q $≠$ (p$∧$q)$∨$~q.
Is my proof good enough? I realize that this question has been asked more than once here but I didn't quite understand the proofs that they gave. Any feedback will be greatly appreciated thanks!
I don't think it's good enough because you are just showing that two particular boolean functions are not always equal. Hint: use a single variable, two is unnecessarily complicated.