Can $\neg p \implies p$ be true? Can $\neg p \implies p$ be true?
How about $p \implies \neg p$?
I was told yes, but it doesn't make sense to me. Any help would be appreciated!
 A: Absolutely.  In fact, the general principle $$(\lnot p\implies p)\implies p$$ is often used in mathematical proofs; one shows $\lnot p\implies p$ and from this one concludes $p$.
For example, suppose there are three beans, each either red or blue; we want to show there are at least two beans of the same color.  ($p$ is “there are two beans of the same color”.)
Assume the contrary ($\lnot p$), that no two beans have the same color.  Then of the first two beans, one  must be red and one must be blue; say the red one is bean $B_r$ and the blue one is $B_b$.
Now if the third bean $B_3$ is red then $B_r$ and $B_3$ are the same color, and if it is blue then $B_b$ and $B_3$ are the same color, so there are two beans of the same color.  Thus $\lnot p \implies p$, and from this we can conclude $p$, that there must be two beans of the same color.
A: Use the truth table. From the truth table you can see that an implication statement is true if the hypothesis is false. An implication statement can be false only when the hypothesis holds and the conclusion doesn't. So $p\implies \neg p$ holds when $p$ is false and $\neg p\implies p $ holds when $\neg p$ is false.
A: If $p$ is true, then $q\to p$ is true, for ANY $q$.  Now just take $q=\neg p$.
A: Yes. Because you can imply everything from false, so you need to make sure the first term becomes false.
For:  $¬p \rightarrow p$
$$p = true$$
$$(false \rightarrow true)$$
$$\Rightarrow true$$
And for: $p \rightarrow ¬p$
$$p = false$$
$$(false \rightarrow true)$$
$$\Rightarrow true$$
