Talk:Binary Logical Connectives with Inverse
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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:
- $\left({\left({p \circ q}\right) * q}\right)_{p = F} = \left({\left({p \circ q}\right) * q}\right)_{p = T}$
- Thus:
- 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:
- $\left({\left({p \circ q}\right) * q}\right)_{p = F} = T = \left({\left({p \circ q}\right) * q}\right)_{p = F}$
- So either:
- 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 $p = F$.
- In case $1$ we have:
--prime mover (talk) 14:15, 8 April 2013 (UTC)