Enemy of an enemy relation

Question

I'm going through the book "How to prove it" by Velleman and I have come across the following question: "Let E = {(p, q) ∈ P × P | the person p is an enemy of the person q}, and F = {(p, q) ∈ P × P | the person p is a friend of the person q}, where P is the set of all people. What does the saying "an enemy of one’s enemy is one’s friend" mean about the relations E and F?"

In attempting to solve this problem I let E be the set {(J, M), (B, M)}. Where J, M and B are people named John, Bill and Mark. I assume this would then mean that E^-1 ∘ E = F. But the answer key gives a different solution:

E ∘ E ⊆ F (answer key)

While I agree that a subset of F is correct (one can have friends that are not enemies of enemies). I find the choice of E ∘ E peculiar, after some further reasoning this only seems to be true if both the pairs (J, M) and (M, J) are included. But it's certainly possible to be one's enemy without that being the opposite way around unless I am misunderstanding something here?

Any input is appreciated.

Answer 1

I think this follows directly, using the definition on page 173 in the second edition of the book you refer to:
By that definition we have $$E\circ E:=\{(a,c)\in P\times P\mid\exists b\in P\colon (a,b)\in E\land(b,c)\in E\}$$ and this is exactly the relation "$a$ is an enemy of an enemy of $c$":

• Let $(a,c)\in E\circ E$, then $a$ is by definition an enemy of some $b\in P$ and $b$ is an enemy of $c$, so $a$ is an enemy of an enemy of $c$.
• Let $a$ be an enemy of an enemy of $c$. Call this enemy $b$. Then $(a,b)\in E$ and $(b,c)\in E$, hence $(a,c)\in E\circ E$.