Non-Equivalence

Context
Natural deduction

Theorem

 * $$\lnot \left ({p \iff q}\right) \dashv \vdash \left({\lnot p \land q}\right) \lor \left({p \land \lnot q}\right)$$
 * $$\lnot \left ({p \iff q}\right) \dashv \vdash \lnot \left({p \implies q}\right) \lor \lnot \left({q \implies p}\right)$$
 * $$\lnot \left ({p \iff q}\right) \dashv \vdash \left({p \lor q} \right) \land \lnot \left({p \land q}\right)$$

Thus we see that negation of equivalence means the same thing as "either-or but not both".

Proof
$$\lnot \left ({p \iff q}\right) \vdash \left({\lnot p \land q}\right) \lor \left({p \land \lnot q}\right)$$:

$$\left({\lnot p \land q}\right) \lor \left({p \land \lnot q}\right) \vdash \lnot \left ({p \iff q}\right)$$:

The above reasoning is completely reversible.

$$\lnot \left ({p \iff q}\right) \vdash \lnot \left({p \implies q}\right) \lor \lnot \left({q \implies p}\right)$$:

$$\lnot \left({p \implies q}\right) \lor \lnot \left({q \implies p}\right) \vdash \lnot \left ({p \iff q}\right)$$:

The above reasoning is completely reversible.

$$\lnot \left ({p \iff q}\right) \vdash \left({p \lor q} \right) \land \lnot \left({p \land q}\right)$$:

First, get this simple result:

$$p \land \lnot q \vdash \left({p \lor q}\right) \land \lnot q$$:

... and its converse:

$$\left({p \lor q}\right) \land \lnot q \vdash p \land \lnot q$$:

(The above may need tightening up. It's intuitively clear, but I believe it may not be logically watertight.)

Now the main part of the proof:

$$\left({p \lor q} \right) \land \lnot \left({p \land q}\right) \vdash \lnot \left ({p \iff q}\right)$$:

The above argument is basically reversible: