Non-Equivalence

Context
Natural deduction

Theorem

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


 * $$\lnot \left ({p \iff q}\right) \vdash \lnot \left({p \Longrightarrow q}\right) \lor \lnot \left({q \Longrightarrow p}\right)$$
 * $$\lnot \left({p \Longrightarrow q}\right) \lor \lnot \left({q \Longrightarrow p}\right) \vdash \lnot \left ({p \iff q}\right)$$


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

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

There is no standard symbol for this, but this one is common and intuitively obvious:

$$p \not \Leftrightarrow q$$ means "either $$p$$ is true or $$q$$ is true but not both."

The symbol $$ \not \Leftrightarrow$$ is called (from the Latin) "aut" (prounounced "out").

This usage of "or", that disallows the case where both disjuncts are true, is called "exclusive or".

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.

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

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

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: