Metamath Proof Explorer


Theorem funss

Description: Subclass theorem for function predicate. (Contributed by NM, 16-Aug-1994) (Proof shortened by Mario Carneiro, 24-Jun-2014)

Ref Expression
Assertion funss A B Fun B Fun A

Proof

Step Hyp Ref Expression
1 relss A B Rel B Rel A
2 coss1 A B A A -1 B A -1
3 cnvss A B A -1 B -1
4 coss2 A -1 B -1 B A -1 B B -1
5 3 4 syl A B B A -1 B B -1
6 2 5 sstrd A B A A -1 B B -1
7 sstr2 A A -1 B B -1 B B -1 I A A -1 I
8 6 7 syl A B B B -1 I A A -1 I
9 1 8 anim12d A B Rel B B B -1 I Rel A A A -1 I
10 df-fun Fun B Rel B B B -1 I
11 df-fun Fun A Rel A A A -1 I
12 9 10 11 3imtr4g A B Fun B Fun A