Metamath Proof Explorer

Theorem eqtrd

Description: An equality transitivity deduction. (Contributed by NM, 21-Jun-1993)

Ref Expression
Hypotheses eqtrd.1 
|- ( ph -> A = B )
eqtrd.2 
|- ( ph -> B = C )
Assertion eqtrd 
|- ( ph -> A = C )

Proof

Step Hyp Ref Expression
1 eqtrd.1 
 |-  ( ph -> A = B )
2 eqtrd.2 
 |-  ( ph -> B = C )
3 2 eqeq2d 
 |-  ( ph -> ( A = B <-> A = C ) )
4 1 3 mpbid 
 |-  ( ph -> A = C )