Metamath Proof Explorer


Theorem eqtr2OLD

Description: Obsolete version of eqtr2 as of 24-Oct-2024. (Contributed by NM, 20-May-2005) (Proof shortened by Andrew Salmon, 25-May-2011) (New usage is discouraged.) (Proof modification is discouraged.)

Ref Expression
Assertion eqtr2OLD
|- ( ( A = B /\ A = C ) -> B = C )

Proof

Step Hyp Ref Expression
1 eqcom
 |-  ( A = B <-> B = A )
2 eqtr
 |-  ( ( B = A /\ A = C ) -> B = C )
3 1 2 sylanb
 |-  ( ( A = B /\ A = C ) -> B = C )