Simplifying $ x \leftrightarrow y$

Question

In trying to simplify $ x \leftrightarrow y$, I'm getting that it's always false, which is nonsense, because even by common sense it should be true when $x=y$. How do I simplify it correctly? Here's my work:

$$ x \leftrightarrow y=(x\to y)\land(y\to x)=(\neg x \lor y)\land(\neg y\lor x)\\ =((\neg x \lor y)\land \neg y)\lor ((\neg x\lor y)\land x)\\ =((\neg x \land \neg y)\lor (y\land \neg y))\lor((\neg x\land x)\lor(y\land x))\\ =(\neg x\land \neg y)\lor (x\land y)\\ =(\neg x\land \neg y\land x)\lor (\neg x\land \neg y\land y)=F$$

Answer 1

I believe $(\lnot x \land \lnot y)\lor(x\land y)$ is simple enough (and until that line everything is fine). Even without the miscalculation, I believe that's where you should've stopped.

You're distributing incorrectly. We have $$ A\lor (B\land C) = (A\lor B)\land(A\lor C) $$ You've swapped the $\land$ and $\lor$ of the right-hand side in your last distribution. Instead, we have $$ (\lnot x \land \lnot y)\lor(x\land y) = ((\lnot x \land \lnot y)\lor x)\land ((\lnot x \land \lnot y)\lor y) $$ which may be further distributed and simplified to $$ ((\lnot x \lor x) \land (\lnot y\lor x))\land ((\lnot x\lor y) \land (\lnot y\lor y))\\ = (\lnot y\lor x)\land (\lnot x\lor y) $$ which is as simple as $(\lnot x \land \lnot y)\lor(x\land y)$, but not as immediately understandable (at least to me).

Answer 2

It's the penultimate equality. We do not have distributivity of $\wedge$ over $\vee$ as you seem to imply here.

Further, you can see that we you have before that is right, informally in terms of true and false statements. You've re-expressed "$x$ is true iff $y$ is true" as "either $x$ and $y$ are both false, or x and y are both true".