Talk:Binary Logical Connective is Self-Inverse iff Exclusive Or

The term is involution, but it does not exactly apply. An observation is that $\oplus$ induces a boolean group with identity $\bot$. However, other connectives ($\implies$ not even being associative) behave very differently. I don't know how to name this result appropriately. --Lord_Farin (talk) 10:10, 21 February 2013 (UTC)


 * I suspect that it may be necessary to define an "inverse" $*$ of a general operation $\circ$ on a general alg struct as: $\forall p, q \in S: (p \circ q) * q = p$, and also to note that (as per Ben-Ari) the only boolean operator having an inverse, let alone having a self-inverse, is $\oplus$. This entry is part of a digression at the end of section 2.1 where he gives a general overview of the boolean operations from the point of view of their truth-tables. I vaguely remember seeing something along those lines in Seth Warner ages ago but can't find it now. It's a way of extending the definition of an inverse to a structure which is non-associative and does not necessarily have an identity. --prime mover (talk) 14:10, 21 February 2013 (UTC)


 * This digression has been omitted in the third edition. Definition sounds good. Though I imagine the applications will be marginal it could possibly be extended to operators of higher arity but that's probably beyond everybody's interests and uses. --Lord_Farin (talk) 14:19, 21 February 2013 (UTC)


 * It'd be cool if we could name this "inverse". It occurs to me that it is precisely the mapping occurring in the One-Step Subgroup Test, which is nice. Who knows what generalizations bring? --Lord_Farin (talk) 14:33, 21 February 2013 (UTC)


 * I appear to have got my idea from Wikipedia after all: Inverse Semigroup. And it wasn't what I remember it as being.


 * I think it's still worth defining an "inverse operation" (with left and right versions) as one which exhibits the behaviours above, noting that in a group the "inverse operation" of $\circ$ is $\ast$ defined as $a \ast b := a \circ b^{-1}$. But I'm not sure enough of my ground to be able to proceed without making up a whole load of stuff. That may or may not be appropriate. --prime mover (talk) 20:35, 21 February 2013 (UTC)

The result's wrong anyway:

$\begin{array}{|ccccc||c|} \hline (p & \iff & q) & \iff & q & p \\ \hline F & T & F & F & F & F \\ F & F & T & F & T & F \\ T & F & F & T & F & T \\ T & T & T & T & T & T \\ \hline \end{array}$

... as can be seen, $\iff$ has the same property here as $\oplus$.

--prime mover (talk) 22:52, 21 February 2013 (UTC)