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.

Proof

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:

 $\displaystyle$  $\displaystyle f_T \left({p, q}\right)$ as $p \circ q = T$ for all values of $p$ and $q$ $\displaystyle$  $\displaystyle p \lor q$ as $p \circ q = T$ for $q = T$ $\displaystyle$  $\displaystyle p \impliedby q$ as $p \circ q = T$ for $q = F$ $\displaystyle$  $\displaystyle p \implies q$ as $p \circ q = T$ for $q = T$ $\displaystyle$  $\displaystyle \operatorname{pr}_2 \left({p, q}\right)$ as $p \circ q = T$ for $q = T$ and also $p \circ q = F$ for $q = F$ $\displaystyle$  $\displaystyle p \land q$ as $p \circ q = F$ for $q = F$ $\displaystyle$  $\displaystyle p \uparrow q$ as $p \circ q = T$ for $q = F$ $\displaystyle$  $\displaystyle \overline {\operatorname{pr}_2} \left({p, q}\right)$ as $p \circ q = T$ for $q = F$ and also $p \circ q = F$ for $q = T$ $\displaystyle$  $\displaystyle \neg \left({p \implies q}\right)$ as $p \circ q = F$ for $q = T$ $\displaystyle$  $\displaystyle \neg \left({p \impliedby q}\right)$ as $p \circ q = F$ for $q = F$ $\displaystyle$  $\displaystyle p \downarrow q$ as $p \circ q = F$ for $q = T$ $\displaystyle$  $\displaystyle f_F \left({p, q}\right)$ as $p \circ q = T$ for all values of $p$ and $q$

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:

 $\displaystyle$  $\displaystyle \operatorname{pr}_1 \left({p, q}\right)$ as $p \circ q = T$ for $p = T$ and also $p \circ q = F$ for $p = F$ $\displaystyle$  $\displaystyle \overline {\operatorname{pr}_1} \left({p, q}\right)$ as $p \circ q = T$ for $p = F$ and also $p \circ q = F$ for $p = T$

We are left with exclusive or and the biconditional.

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

$\blacksquare$

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:

$\left({p \circ q}\right) \circ q \dashv \vdash p$

$\blacksquare$