Binary Logical Connectives with Inverse

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

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

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:
 * $\paren {p \circ q} * q \dashv \vdash p$

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

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

$\begin{array}{|r|cccc|} \hline p                               & \T & \T & \F & \F \\ q                               & \T & \F & \T & \F \\ \hline \map {f_\T} {p, q}              & \T & \T & \T & \T \\ p \lor q                        & \T & \T & \T & \F \\ p \impliedby q                  & \T & \T & \F & \T \\ \map {\pr_1} {p, q}             & \T & \T & \F & \F \\ p \implies q                    & \T & \F & \T & \T \\ \map {\pr_2} {p, q}             & \T & \F & \T & \F \\ p \iff q                        & \T & \F & \F & \T \\ p \land q                       & \T & \F & \F & \F \\ p \uparrow q                    & \F & \T & \T & \T \\ \map \neg {p \iff q}            & \F & \T & \T & \F \\ \map {\overline {\pr_2} } {p, q} & \F & \T & \F & \T \\ \map \neg {p \implies q}        & \F & \T & \F & \F \\ \map {\overline {\pr_1} } {p, q} & \F & \F & \T & \T \\ \map \neg {p \impliedby q}      & \F & \F & \T & \F \\ p \downarrow q                  & \F & \F & \F & \T \\ \map {f_\F} {p, q}              & \F & \F & \F & \F \\ \hline \end{array}$

Suppose that for some $q \in \set {\F, \T}$:
 * $\paren {p \circ q}_{p = \F} = \paren {p \circ q}_{p = \T}$

Then:
 * $\paren {\paren {p \circ q} * q}_{p = \F} = \paren {\paren {p \circ q} * q}_{p = \T}$

and so either:
 * $\paren {\paren {p \circ q} * q}_{p = \F} \ne p$

or:
 * $\paren {\paren {p \circ q} * q}_{p = \T} \ne p$

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:

$\begin{array}{|r|cccc|} \hline p                               & \T & \T & \F & \F \\ q                               & \T & \F & \T & \F \\ \hline \map {\pr_1} {p, q}             & \T & \T & \F & \F \\ p \iff q                        & \T & \F & \F & \T \\ \map \neg {p \iff q}            & \F & \T & \T & \F \\ \map {\overline {\pr_1} } {p, q} & \F & \F & \T & \T \\ \hline \end{array}$

Suppose that for some $p \in \set {\F, \T}$:
 * $\paren {p \circ q}_{q = \F} = \paren {p \circ q}_{q = \T}$

Then:
 * $\paren {q * \paren {p \circ q} }_{q = \F} = \paren {q * \paren {p \circ q} }_{q = \T}$

and so either:
 * $\paren {q * \paren {p \circ q} }_{q = \F} \ne p$

or:
 * $\paren {q * \paren {p \circ q} }_{q = \T} \ne p$

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:
 * $\paren {p \circ q} \circ q \dashv \vdash p$

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:
 * $\paren {p \circ q} \circ q \dashv \vdash p$