# providing bi-conditional without truth table

I am being stuck in this question. I was asked to prove that without using truth tables. I was taught how to prove that using truth table.

$(p \iff q) \equiv (p \Rightarrow q)\land(q \Rightarrow p)$

So, my first step is to use conditional identities: $$(p→q)∧(q→p) \equiv (\neg p \lor q) ∧(\neg q \lor p),$$ but I think I can't get any clue to prove that. How should I approach this?

• What you wrote down as what you want to prove is usually how the bi-conditional is defined. How did you define the bi-conditional, if not like that? The only other way to define it that I have in mind would be via its truth table values for p and q but you said that isn't how you are supposed to think of it.
– MM8
Jan 26 '17 at 17:38
• Maybe you defined it as $(p \land q) \lor (\neg p \land \neg q)$, in which case you are just a few more steps of simplification and parenthesis-rules away from the proof.
– MM8
Jan 26 '17 at 17:42

Recall that biconditionnals $Q\iff P$ and $P \iff Q$ both mean the same thing.
Thus, $P\iff Q$ is true if and only if $Q\iff P$ is true so $P$ and $Q$ are interchangeable in $\left(P\rightarrow Q\right)\wedge\left(Q\rightarrow P\right)$ .
Careful though, $P \rightarrow Q$ does not mean the same thing as $Q\rightarrow P$. The converse of $P \rightarrow Q$, namely $Q\rightarrow P$, can have different truth value.