# Prove that $(A,B)\sim(P,Q)$ and $(C,D)\sim (P,Q)\implies (A,B)\sim (C,D)$?

I have the following laws:

And I did the following:

• $(A,B)\sim(P,Q)\wedge (C,D)\sim (P,Q) \stackrel{?}{\implies} (A,B)\sim (C,D)$
• $(A,B)\sim(P,Q)\wedge \stackrel{symmetry}{(P,Q)\sim (C,D)}\stackrel{?}{\implies} (A,B)\sim (C,D)$
• $(A,B)\sim(P,Q)\wedge (P,Q)\sim (C,D) \implies (A,B)\sim (C,D)$

I guess this is it. Am I missing something? Also, is transitivity actually needed? It seems to be only a variant of symmetry, I guess that only symmetry is needed to show transitivity but I may be wrong.

• Seems correct. You do need transitivity for the last step. – Nescrio Apr 12 '15 at 7:40