Metamath Proof Explorer

Theorem feq2i

Description: Equality inference for functions. (Contributed by NM, 5-Sep-2011)

Ref Expression
Hypothesis feq2i.1 
|- A = B
Assertion feq2i 
|- ( F : A --> C <-> F : B --> C )

Proof

Step Hyp Ref Expression
1 feq2i.1 
 |-  A = B
2 feq2 
 |-  ( A = B -> ( F : A --> C <-> F : B --> C ) )
3 1 2 ax-mp 
 |-  ( F : A --> C <-> F : B --> C )