Metamath Proof Explorer


Theorem coeq2d

Description: Equality deduction for composition of two classes. (Contributed by NM, 16-Nov-2000)

Ref Expression
Hypothesis coeq1d.1
|- ( ph -> A = B )
Assertion coeq2d
|- ( ph -> ( C o. A ) = ( C o. B ) )

Proof

Step Hyp Ref Expression
1 coeq1d.1
 |-  ( ph -> A = B )
2 coeq2
 |-  ( A = B -> ( C o. A ) = ( C o. B ) )
3 1 2 syl
 |-  ( ph -> ( C o. A ) = ( C o. B ) )