Non-Equivalence

Theorem

 * $$\neg \left ({p \iff q}\right) \dashv \vdash \left({\neg p \and q}\right) \or \left({p \and \neg q}\right)$$
 * $$\neg \left ({p \iff q}\right) \dashv \vdash \neg \left({p \implies q}\right) \or \neg \left({q \implies p}\right)$$
 * $$\neg \left ({p \iff q}\right) \dashv \vdash \left({p \or q} \right) \and \neg \left({p \and q}\right)$$

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

Proof by Natural Deduction
By the tableau method:

The argument is reversible:

The above reasoning is completely reversible.


 * $$\neg \left ({p \iff q}\right) \dashv \vdash \left({p \or q} \right) \and \neg \left({p \and q}\right)$$:

First, get this simple result:

... and its converse:

Now the main part of the proof:

The above argument is reversible:

Proof by Truth Table
Let $$v: \left\{{p, q}\right\} \to \left\{{T, F}\right\}$$ be an interpretation for a logical formula $$\phi$$ of two variables $$p, q$$.

We see that $$v \left({\neg \left ({p \iff q}\right)}\right) = v \left({\left({\neg p \and q}\right) \or \left({p \and \neg q}\right)}\right)$$ for all interpretations $$v$$.

Hence the result by the definition of interderivable.

We see that $$v \left({\neg \left ({p \iff q}\right)}\right) = v \left({\neg \left({p \implies q}\right) \or \neg \left({q \implies p}\right)}\right)$$ for all interpretations $$v$$.

Hence the result by the definition of interderivable.

We see that $$v \left({\neg \left ({p \iff q}\right)}\right) = v \left({\left({p \or q} \right) \and \neg \left({p \and q}\right)}\right)$$ for all interpretations $$v$$.

Hence the result by the definition of interderivable.