Metamath Proof Explorer

Theorem funeq

Description: Equality theorem for function predicate. (Contributed by NM, 16-Aug-1994)

Ref Expression
Assertion funeq 
|- ( A = B -> ( Fun A <-> Fun B ) )

Proof

Step Hyp Ref Expression
1 eqimss2 
 |-  ( A = B -> B C_ A )
2 funss 
 |-  ( B C_ A -> ( Fun A -> Fun B ) )
3 1 2 syl 
 |-  ( A = B -> ( Fun A -> Fun B ) )
4 eqimss 
 |-  ( A = B -> A C_ B )
5 funss 
 |-  ( A C_ B -> ( Fun B -> Fun A ) )
6 4 5 syl 
 |-  ( A = B -> ( Fun B -> Fun A ) )
7 3 6 impbid 
 |-  ( A = B -> ( Fun A <-> Fun B ) )