Metamath Proof Explorer


Theorem cocan2

Description: A surjection is right-cancelable. (Contributed by FL, 21-Nov-2011) (Proof shortened by Mario Carneiro, 21-Mar-2015)

Ref Expression
Assertion cocan2 F:AontoBHFnBKFnBHF=KFH=K

Proof

Step Hyp Ref Expression
1 fof F:AontoBF:AB
2 1 3ad2ant1 F:AontoBHFnBKFnBF:AB
3 fvco3 F:AByAHFy=HFy
4 2 3 sylan F:AontoBHFnBKFnByAHFy=HFy
5 fvco3 F:AByAKFy=KFy
6 2 5 sylan F:AontoBHFnBKFnByAKFy=KFy
7 4 6 eqeq12d F:AontoBHFnBKFnByAHFy=KFyHFy=KFy
8 7 ralbidva F:AontoBHFnBKFnByAHFy=KFyyAHFy=KFy
9 fveq2 Fy=xHFy=Hx
10 fveq2 Fy=xKFy=Kx
11 9 10 eqeq12d Fy=xHFy=KFyHx=Kx
12 11 cbvfo F:AontoByAHFy=KFyxBHx=Kx
13 12 3ad2ant1 F:AontoBHFnBKFnByAHFy=KFyxBHx=Kx
14 8 13 bitrd F:AontoBHFnBKFnByAHFy=KFyxBHx=Kx
15 simp2 F:AontoBHFnBKFnBHFnB
16 fnfco HFnBF:ABHFFnA
17 15 2 16 syl2anc F:AontoBHFnBKFnBHFFnA
18 simp3 F:AontoBHFnBKFnBKFnB
19 fnfco KFnBF:ABKFFnA
20 18 2 19 syl2anc F:AontoBHFnBKFnBKFFnA
21 eqfnfv HFFnAKFFnAHF=KFyAHFy=KFy
22 17 20 21 syl2anc F:AontoBHFnBKFnBHF=KFyAHFy=KFy
23 eqfnfv HFnBKFnBH=KxBHx=Kx
24 15 18 23 syl2anc F:AontoBHFnBKFnBH=KxBHx=Kx
25 14 22 24 3bitr4d F:AontoBHFnBKFnBHF=KFH=K