=== 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: {{begin-eqn}} {{eqn | o =       | r = \map {f_\T} {p, q}       | c = as $p \circ q = \T$ for all values of $p$ and $q$ }} {{eqn | o =       | r = p \lor q       | c = as $p \circ q = \T$ for $q = \T$ }} {{eqn | o =       | r = p \impliedby q       | c = as $p \circ q = \T$ for $q = \F$ }} {{eqn | o =       | r = p \implies q       | c = as $p \circ q = \T$ for $q = \T$ }} {{eqn | o =       | r = \map {\pr_2} {p, q}       | c = as $p \circ q = \T$ for $q = \T$ and also $p \circ q = \F$ for $q = \F$ }} {{eqn | o =       | r = p \land q       | c = as $p \circ q = \F$ for $q = \F$ }} {{eqn | o =       | r = p \uparrow q       | c = as $p \circ q = \T$ for $q = \F$ }} {{eqn | o =       | r = \map {\overline {\pr_2} } {p, q}       | c = as $p \circ q = \T$ for $q = \F$ and also $p \circ q = \F$ for $q = \T$ }} {{eqn | o =       | r = \map \neg {p \implies q}       | c = as $p \circ q = \F$ for $q = \T$ }} {{eqn | o =       | r = \map \neg {p \impliedby q}       | c = as $p \circ q = \F$ for $q = \F$ }} {{eqn | o =       | r = p \downarrow q       | c = as $p \circ q = \F$ for $q = \T$ }} {{eqn | o =       | r = \map {f_\F} {p, q}       | c = as $p \circ q = \T$ for all values of $p$ and $q$ }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | o =       | r = \map {\pr_1} {p, q}       | c = as $p \circ q = \T$ for $p = \T$ and also $p \circ q = \F$ for $p = \F$ }} {{eqn | o =       | r = \map {\overline {\pr_1} } {p, q}       | c = as $p \circ q = \T$ for $p = \F$ and also $p \circ q = \F$ for $p = \T$ }} {{end-eqn}} We are left with exclusive or and the biconditional. The result follows from Exclusive Or is Self-Inverse and Biconditional is Self-Inverse. {{qed}} === 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: :$\paren {p \circ q} \circ q \dashv \vdash p$ {{qed}} Category:Exclusive Or Category:Biconditional Category:Propositional Logic kbykk6g9uanajr5qrrir9gyc2juer2k