Metamath Proof Explorer


Theorem sylan9eqr

Description: An equality transitivity deduction. (Contributed by NM, 8-May-1994)

Ref Expression
Hypotheses sylan9eqr.1
|- ( ph -> A = B )
sylan9eqr.2
|- ( ps -> B = C )
Assertion sylan9eqr
|- ( ( ps /\ ph ) -> A = C )

Proof

Step Hyp Ref Expression
1 sylan9eqr.1
 |-  ( ph -> A = B )
2 sylan9eqr.2
 |-  ( ps -> B = C )
3 1 2 sylan9eq
 |-  ( ( ph /\ ps ) -> A = C )
4 3 ancoms
 |-  ( ( ps /\ ph ) -> A = C )