Proofing an uncertain theorem to verify it's truth?

[pandaBee]

Homework Statement

Suppose R is a relation from A to B and S and T are relations from
B to C.
Must it be true that (S \ T ) ◦ R ⊆ (S ◦ R) \ (T ◦ R)?

The Attempt at a Solution

I assume that (S \ T ) ◦ R ⊆ (S ◦ R) \ (T ◦ R) is true and attempt to prove it (if I run into a contradiction I know that it is false)

Part 1
let (a,c) \in (S\T ) ◦ R then, we can choose a b \in B such that,

\exists b \in B\ ((a,b) \in R\land(b,c) \in (S\T))

Therefore,
(a,b) \in R
(b,c) \in S
(b,c) \notin T

But this means that (a,c) \in (S\circ R)

Part 2
To prove that (a,c) \notin (T\circ R) we employ proof by contradiction:

Suppose (a,c) \in (T\circ R) Then we instantiate a b \in B, say, b' such that:
(a,b') \in R
(b',c) \in T
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I've hit a dead-end here and I'm not sure how to progress any further. As an exercise I wanted to refrain from plugging-and-chugging random scenarios into the theorem to prove it's veracity and decided instead to reach a contradiction within a contradiction to prove that the theorem is in fact, false.

However, I haven't (or at least I don't believe) that I have reached a contradiction in the proof, I'm just at a dead end (or at least that's how it seems).

Any insight would be much appreciated.

 

[HallsofIvy]

What does "R\S" mean? "Relations" are sets of pairs so I would assume that "R\S" is the set operation "all members of R that are not in S" but that can't be true here because NO member of R is in S.