I am having problem solving the right-to-left direction of the following iff proof:
Suppose A and B are two sets, show that for every relation R from A to B, $R \circ i_A = R$. (Note that $i_A = \{ (x,y)\in A \times A | x=y\} $)
A solution I found, which is the same as my left-to-right direction proof, is as follows:
Let x be an arbitrary element of A and z be an arbitrary element of B such that $ (x,z)\in R\circ i_A $.
So $ (x,z)\in R\circ i_A $ iff $\exists y\in A ((x,y)\in i_a \land (y,z)\in R)$
Since $ (x,y)\in i_a$ iff x=y
So, $ (x,x) \in i_A \land (x,z)\in R$ iff $(x,z)\in R$
Since (x, z) is arbitrary, so $R \circ i_A = R$
The problem I have with this, is that this seems only correct as the proof of left-to-right of $R \circ i_A = R$, not the other way.
If we do the right-to-left proof, we start off only with $(x,z)\in R$. But that seems to be it - we don't seem to be able to introduce $\exists y\in A ((x,y)\in i_a \land (y,z)\in R)$ because we are not told that (in this direction) (x, z) is a composition, let alone introducing the $i_A$.
Could anyone please help?
To show that $R\subseteq R\circ i_A$, let $\langle x,y\rangle\in R$. Then $x\in A$, so $\langle x,x\rangle\in i_A$. We now have $\langle x,\color{crimson}x\rangle\in i_A$ and $\langle\color{crimson}x,y\rangle\in R$, so by definition $\langle x,y\rangle\in R\circ i_A$, as desired.