I have the following laws:

enter image description here

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.

  • 1
    $\begingroup$ Seems correct. You do need transitivity for the last step. $\endgroup$ – Nescrio Apr 12 '15 at 7:40

\begin{align} (A,B) \sim (P,Q) \text{ and } (C,D) \sim (P,Q) & \overset{(b)}\implies (A,B) \sim (P,Q) \text{ and } (P,Q) \sim (C,D) \\ & \overset{(c)}\implies (A,B) \sim (C,D) \end{align} In general, transitivity can't be shown from symmetry, see here.

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