Talk:Binary Logical Connectives with Inverse

I don't understand the rationale for this part:


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

why is that the case? --GFauxPas (talk) 13:52, 8 April 2013 (UTC)


 * We have supposed that


 * $\left({p \circ q}\right)_{p = F} = \left({p \circ q}\right)_{p = T}$
 * which means $T \circ q = F \circ q$
 * This holds, for example, when $\circ$ is $\implies$, and $q = T$: whatever $p$ is, $p \implies T$ is true.


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


 * by rules of equations: as $\left({p \circ q}\right)$ is the same whatever $p$ is, then applying $* q$ to both sides results in the same both sides.
 * Same as if $x = y$, then $x * n = y * n$


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


 * In case $1$ we have:
 * $F = \left({\left({p \circ q}\right) * q}\right)_{p = T}$
 * which is not equal to $p$ when $p = T$, or:
 * $\left({\left({p \circ q}\right) * q}\right)_{p = F} = T$
 * which is not equal to $p$ when $p = F$.

--prime mover (talk) 14:15, 8 April 2013 (UTC)


 * Got it, thanks. It was the very last part I was having trouble with. --GFauxPas (talk) 15:41, 8 April 2013 (UTC)