Metamath Proof Explorer


Theorem syl6req

Description: An equality transitivity deduction. (Contributed by NM, 29-Mar-1998)

Ref Expression
Hypotheses syl6req.1 φ A = B
syl6req.2 B = C
Assertion syl6req φ C = A

Proof

Step Hyp Ref Expression
1 syl6req.1 φ A = B
2 syl6req.2 B = C
3 1 2 syl6eq φ A = C
4 3 eqcomd φ C = A