Talk:Composite of Symmetric Relations is not necessarily Symmetric

(18/09/2023)kmm_WeltRene: Sorry, I'm unsure where to send this, but this proof is wrong, and this is untrue in general. Could this be removed?


 * Can you provide a counterexample? Incidentally, this is the link which tells you (how) to sign your posts. --prime mover (talk) 11:29, 18 September 2023 (UTC)

oh, thanks. I found one on Math StackExchange. Alternatively, consider R = {(1,2), (2,1)} and S = {(2,3),(3,2)}. Both are symmetric, but their composition, {(1,3)}, is not, since (3,1) is not it. --Kmm WeltRene (talk) 13:02, 18 September 2023 (UTC)


 * hmm ... wonder where my logic breaks down ... I'll have to study this. --prime mover (talk) 15:32, 18 September 2023 (UTC)


 * I found the mistake. You start with:
 * $\exists y \in A: \tuple {x, y} \in \SS^{-1} \land \tuple {y, z} \in \RR^{-1}$
 * and conclude:
 * $\tuple {x, y} \in \SS^{-1} \circ \RR^{-1}$
 * but the correct conclusion is:
 * $\tuple {x, y} \in \RR^{-1} \circ \SS^{-1}$
 * from which the result does not follow. --CircuitCraft (talk) 17:52, 18 September 2023 (UTC)


 * Yeah I sort of got there. My mind's not as sharp as it used to be. Age. --prime mover (talk) 17:59, 18 September 2023 (UTC)