Binary Logical Connectives with Inverse

Theorem
Let $\circ$ be a binary logical connective.

Then there exists another binary logical connective $*$ such that:
 * $\forall p, q \in \left\{{F, T}\right\}: \left({p \circ q}\right) * q \dashv \vdash p \dashv \vdash q * \left({p \circ q}\right)$

iff $\circ$ is either:


 * $(1): \quad$ the exclusive or operator

or:
 * $(2): \quad$ the biconditional operator.

That is, the only truth functions that have an inverse operation are the exclusive or and the biconditional.

Necessary Condition
Let $\circ$ be a binary logical connective such that there exists $*$ such that:
 * $\left({p \circ q}\right) * q \dashv \vdash p$

That is, by definition (and minor abuse of notation):
 * $\forall p, q \in \left\{{F, T}\right\}: \left({p \circ q}\right) * q = p$

For reference purposes, let us list from Binary Truth Functions the complete truth table containing all of the binary logical connectives:

$\begin{array}{|r|cccc|} \hline p                                                  & T & T & F & F \\ q                                                  & T & F & T & F \\ \hline f_T \left({p, q}\right)                            & T & T & T & T \\ p \lor q                                           & T & T & T & F \\ p \impliedby q                             & T & T & F & T \\ \operatorname{pr}_1 \left({p, q}\right)            & T & T & F & F \\ p \implies q                                       & T & F & T & T \\ \operatorname{pr}_2 \left({p, q}\right)            & T & F & T & F \\ p \iff q                                           & T & F & F & T \\ p \land q                                          & T & F & F & F \\ p \uparrow q                                       & F & T & T & T \\ \neg \left({p \iff q}\right)                       & F & T & T & F \\ \overline {\operatorname{pr}_2} \left({p, q}\right) & F & T & F & T \\ \neg \left({p \implies q}\right)                   & F & T & F & F \\ \overline {\operatorname{pr}_1} \left({p, q}\right) & F & F & T & T \\ \neg \left({p \impliedby q}\right)         & F & F & T & F \\ p \downarrow q                                     & F & F & F & T \\ f_F \left({p, q}\right)                            & F & F & F & F \\ \hline \end{array}$

Suppose that for some $q \in \left\{{F, T}\right\}$:
 * $\left({p \circ q}\right)_{p = F} = \left({p \circ q}\right)_{p = T}$

Then:
 * $\left({\left({p \circ q}\right) * q}\right)_{p = F} = \left({\left({p \circ q}\right) * q}\right)_{p = T}$

and so either:
 * $\left({\left({p \circ q}\right) * q}\right)_{p = F} \ne p$

or:
 * $\left({\left({p \circ q}\right) * q}\right)_{p = T} \ne p$

Thus for $\circ$ to have an inverse operation it is necessary for $F \circ q \ne T \circ q$.

This eliminates:

The remaining connectives which may have inverses are:

$\begin{array}{|r|cccc|} \hline p                                                  & T & T & F & F \\ q                                                  & T & F & T & F \\ \hline \operatorname{pr}_1 \left({p, q}\right)            & T & T & F & F \\ p \iff q                                           & T & F & F & T \\ \neg \left({p \iff q}\right)                       & F & T & T & F \\ \overline {\operatorname{pr}_1} \left({p, q}\right) & F & F & T & T \\ \hline \end{array}$

Suppose that for some $p \in \left\{{F, T}\right\}$:
 * $\left({p \circ q}\right)_{q = F} = \left({p \circ q}\right)_{q = T}$

Then:
 * $\left({q * \left({p \circ q}\right)}\right)_{q = F} = \left({q * \left({p \circ q}\right)}\right)_{q = T}$

and so either:
 * $\left({q * \left({p \circ q}\right)}\right)_{q = F} \ne p$

or:
 * $\left({q * \left({p \circ q}\right)}\right)_{q = T} \ne p$

This eliminates:

We are left with exclusive or and the biconditional.

The result follows from Exclusive Or is Self-Inverse and Biconditional is Self-Inverse.

Sufficient Condition
Let $\circ$ be the exclusive or operator.

Then by Exclusive Or is Self-Inverse it follows that:
 * $\left({p \circ q}\right) \circ q \dashv \vdash p$

Thus $*$ is the inverse operation of the exclusive or operation.

Similarly, let $\circ$ be the biconditional operator.

Then by Biconditional is Self-Inverse it follows that:
 * $\left({p \circ q}\right) \circ q \dashv \vdash p$