# Proving tautologies with $\iff$

I have a question regarding the process of proving that a statement form with $\!\iff\!$ is a tautology. For a simple example, let's say we have a statement:

$(A \iff (B\implies A))$

To attempt to prove that this is a tautology we assume that the statement form is false and try to obtain a contradiction. So if we do assume this then we have two choices, either the left side is true and the right side is false or vice-versa.

If the left side is true, then A is true, but to make the right side false A would have to be false and B true, therefore we have a contradiction since A cannot be both true and false.

If the left side is false, then A is false, but now for the right side to be true we just let B be false and the right side is true. Therefore in this case the statement form is false when A and B are both false.

Therefore the statement is not a tautology.

Am I correct in my assumption that to prove tautologies with statement forms that use $\iff\!\!$, we must always check both choices for left and right sides? Is there a quicker way to prove that statements such as these are tautologies?

• I mean, both directions have to hold. Of course, having a counterexample in one case means you don't have to check the other, so if you see it right away, you might as well not check the other. Commented Nov 6, 2015 at 20:52

$p\iff q$ is equivalent to $(p \land q) \lor (\neg p \land \neg q)$. The two disjuncts are the two things you checked: $(p \land q)$, and $(\neg p \land \neg q)$. The two formulas on either side of $\!\iff\!$ have to have the same truth value — either both true, or both false.
The other way to establish a biconditional is to note that $p\!\iff\! q$ is also equivalent to $(p \!\implies\! q) \land (q \!\implies\! p)$. So if you can prove that each of $p$ and $q$ implies the other, then you have proved $p\!\iff\! q$. Note that $p\!\implies\! q$ is true whenever the truth value of $p$ is $\le$ the truth value of $q$. Once again, you prove that the two truth values are equal, but here you do so by showing that they're less than or equal to each other. A variation on this approach is to prove $(p \!\implies\! q) \land (\neg p \!\implies\! \neg q)$, as the second conjunct of that is equivalent to $(q \!\implies\! p)$ (it's the contrapositive of the latter).