# Prove $p \lor q \Leftrightarrow (\neg p) \rightarrow q$, but with caveats.

In this question, the professor asks us in parts i through iii to prove using truth tables that:

i. $$\neg (p \lor q) \Leftrightarrow (\neg p) \land (\neg q)$$

ii. $$\neg (p \land q) \Leftrightarrow (\neg p) \lor (\neg q)$$

iii. $$\neg (\neg p) \Leftrightarrow p$$

Then in part iv. he changes the game and asks us to "use (i-iii) to derive (by the use of a succession of $$\Leftrightarrow$$'s)" that $$p \lor q \Leftrightarrow (\neg p) \rightarrow q$$. That is, a truth table is now specifically not allowed.

I am just stuck. I get that statements i. through iii. are now "tools" that we can use to prove that $$p \lor q \Leftrightarrow (\neg p) \rightarrow q$$. I understand that $$p \lor q$$ means one or both must be true, so if one is false (i.e. $$\neg p$$), then the other must be true. But I don't see how I am supposed to prove that a statement is equivalent to an implication using i. through iii. if none of these statements i. through iii. involve an implication!

I tried submitting this combination of $$\Leftrightarrow$$ statements and exposition:

"$$p \lor q \Leftrightarrow \neg (\neg (p \land q)) \Leftrightarrow \neg((\neg p) \land (\neg q))$$ where this last statement means that $$p$$ and $$q$$ cannot be simultaneously false, so if $$p$$ is false, then $$q$$ must be true,"

However, the professor said there's an easier way, and I should be able to prove the statement using ONLY $$\Leftrightarrow$$ statements, and no exposition. I feel like I'm just missing something obvious. Or maybe he is assuming we're going to use outside knowledge and not JUST statements i. through iii.?

EDIT: In case anyone ever comes across this, the answer is that you are to use a combination of the rules i. through iii. as described, and also common, outside facts. So you have $$(p \lor q) \equiv \neg(\neg p) \lor q \equiv \neg p \rightarrow q$$.

• How do you define $p\to q$? – John Douma Jan 20 at 18:49
• You are right. With only (i)-(iii) you cannot, because there is nowhere the symbol $\to$ in them... thus, it cannot "pop up" from nowhere. – Mauro ALLEGRANZA Jan 20 at 18:50
• What is "exposition" ? – Mauro ALLEGRANZA Jan 20 at 18:51
• Good question, "exposition" meaning "explanation in words, instead of just symbols." – 1Teaches2Learn Jan 20 at 18:52
• Ok... $p \lor q$ is "either $p$ or $q$ (inclusive or)" that means that at least one of them must be TRUE. If so, if $p$ is FALSE, then $q$ must necessarily be TRUE. And this is : from Left to Right... – Mauro ALLEGRANZA Jan 20 at 18:55

You're absolutely right. With no rule involving $$\rightarrow$$ in (i)-(iii) this is impossible.
But, typically it is given that $$p \rightarrow q \Leftrightarrow \neg p \lor q$$