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 φ A = B
sylan9eq.2 ψ B = C
Assertion sylan9eq φ ψ A = C

Proof

Step Hyp Ref Expression
1 sylan9eq.1 φ A = B
2 sylan9eq.2 ψ B = C
3 eqtr A = B B = C A = C
4 1 2 3 syl2an φ ψ A = C