Metamath Proof Explorer


Theorem feq2d

Description: Equality deduction for functions. (Contributed by Paul Chapman, 22-Jun-2011)

Ref Expression
Hypothesis feq2d.1
|- ( ph -> A = B )
Assertion feq2d
|- ( ph -> ( F : A --> C <-> F : B --> C ) )

Proof

Step Hyp Ref Expression
1 feq2d.1
 |-  ( ph -> A = B )
2 feq2
 |-  ( A = B -> ( F : A --> C <-> F : B --> C ) )
3 1 2 syl
 |-  ( ph -> ( F : A --> C <-> F : B --> C ) )