Binary Logical Connectives with Inverse

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Theorem

Let $\circ$ be a binary logical connective.

Then there exists another binary logical connective $*$ such that:

$\forall p, q \in \set {\F, \T}: \paren {p \circ q} * q \dashv \vdash p \dashv \vdash q * \paren {p \circ q}$

if and only if $\circ$ 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:

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

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:

\(\displaystyle \) \(\) \(\displaystyle \map {f_\T} {p, q}\) 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 $q = \T$
\(\displaystyle \) \(\) \(\displaystyle \map {\pr_2} {p, q}\) 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 $q = \F$
\(\displaystyle \) \(\) \(\displaystyle \map {\overline {\pr_2} } {p, q}\) as $p \circ q = \T$ for $q = \F$ and also $p \circ q = \F$ for $q = \T$
\(\displaystyle \) \(\) \(\displaystyle \map \neg {p \implies q}\) as $p \circ q = \F$ for $q = \T$
\(\displaystyle \) \(\) \(\displaystyle \map \neg {p \impliedby q}\) as $p \circ q = \F$ for $q = \F$
\(\displaystyle \) \(\) \(\displaystyle p \downarrow q\) as $p \circ q = \F$ for $q = \T$
\(\displaystyle \) \(\) \(\displaystyle \map {f_\F} {p, q}\) as $p \circ q = \T$ for all values of $p$ and $q$


The remaining connectives which may have inverses are:

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


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:

\(\displaystyle \) \(\) \(\displaystyle \map {\pr_1} {p, q}\) as $p \circ q = \T$ for $p = \T$ and also $p \circ q = \F$ for $p = \F$
\(\displaystyle \) \(\) \(\displaystyle \map {\overline {\pr_1} } {p, q}\) 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:

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

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:

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

$\blacksquare$