Let $R$ and $S$ be relations on $A$.
Proof of Proposition 7.3.1.1.4.
Item 1: Interaction With Inverses
Clear.
Item 2: Interaction With Composition
See [MSE 2096272].[1]
Footnotes
[1] Intuition: Transitivity for $R$ and $S$ fails to imply that of $S\mathbin {\diamond }R$ because the composition operation for relations intertwines $R$ and $S$ in an incompatible way:
- If $a\sim _{S\mathbin {\diamond }R}c$ and $c\sim _{S\mathbin {\diamond }r}e$, then:
- There is some $b\in A$ such that:
- $a\sim _{R}b$;
- $b\sim _{S}c$;
- There is some $d\in A$ such that:
- $c\sim _{R}d$;
- $d\sim _{S}e$.
- There is some $b\in A$ such that:
