Metamath Proof Explorer


Theorem sylan9eq

Description: An equality transitivity deduction. (Contributed by NM, 8-May-1994) (Proof shortened by Andrew Salmon, 25-May-2011)

Ref Expression
Hypotheses sylan9eq.1 φA=B
sylan9eq.2 ψB=C
Assertion sylan9eq φψA=C

Proof

Step Hyp Ref Expression
1 sylan9eq.1 φA=B
2 sylan9eq.2 ψB=C
3 eqtr A=BB=CA=C
4 1 2 3 syl2an φψA=C