Metamath Proof Explorer

Theorem fneq2i

Description: Equality inference for function predicate with domain. (Contributed by NM, 4-Sep-2011)

Ref Expression
Hypothesis fneq2i.1 
|- A = B
Assertion fneq2i 
|- ( F Fn A <-> F Fn B )

Proof

Step Hyp Ref Expression
1 fneq2i.1 
 |-  A = B
2 fneq2 
 |-  ( A = B -> ( F Fn A <-> F Fn B ) )
3 1 2 ax-mp 
 |-  ( F Fn A <-> F Fn B )