Metamath Proof Explorer


Theorem eq2tri

Description: A compound transitive inference for class equality. (Contributed by NM, 22-Jan-2004)

Ref Expression
Hypotheses eq2tri.1
|- ( A = C -> D = F )
eq2tri.2
|- ( B = D -> C = G )
Assertion eq2tri
|- ( ( A = C /\ B = F ) <-> ( B = D /\ A = G ) )

Proof

Step Hyp Ref Expression
1 eq2tri.1
 |-  ( A = C -> D = F )
2 eq2tri.2
 |-  ( B = D -> C = G )
3 ancom
 |-  ( ( A = C /\ B = D ) <-> ( B = D /\ A = C ) )
4 1 eqeq2d
 |-  ( A = C -> ( B = D <-> B = F ) )
5 4 pm5.32i
 |-  ( ( A = C /\ B = D ) <-> ( A = C /\ B = F ) )
6 2 eqeq2d
 |-  ( B = D -> ( A = C <-> A = G ) )
7 6 pm5.32i
 |-  ( ( B = D /\ A = C ) <-> ( B = D /\ A = G ) )
8 3 5 7 3bitr3i
 |-  ( ( A = C /\ B = F ) <-> ( B = D /\ A = G ) )