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 ABFunBFunA

Proof

Step Hyp Ref Expression
1 relss ABRelBRelA
2 coss1 ABAA-1BA-1
3 cnvss ABA-1B-1
4 coss2 A-1B-1BA-1BB-1
5 3 4 syl ABBA-1BB-1
6 2 5 sstrd ABAA-1BB-1
7 sstr2 AA-1BB-1BB-1IAA-1I
8 6 7 syl ABBB-1IAA-1I
9 1 8 anim12d ABRelBBB-1IRelAAA-1I
10 df-fun FunBRelBBB-1I
11 df-fun FunARelAAA-1I
12 9 10 11 3imtr4g ABFunBFunA