Metamath Proof Explorer


Theorem imasetpreimafvbijlemfo

Description: Lemma for imasetpreimafvbij : the mapping H is a function onto the range of function F . (Contributed by AV, 22-Mar-2024)

Ref Expression
Hypotheses fundcmpsurinj.p P=z|xAz=F-1Fx
fundcmpsurinj.h H=pPFp
Assertion imasetpreimafvbijlemfo FFnAAVH:PontoFA

Proof

Step Hyp Ref Expression
1 fundcmpsurinj.p P=z|xAz=F-1Fx
2 fundcmpsurinj.h H=pPFp
3 1 2 imasetpreimafvbijlemf FFnAH:PFA
4 3 adantr FFnAAVH:PFA
5 1 preimafvelsetpreimafv FFnAAVaAF-1FaP
6 5 3expa FFnAAVaAF-1FaP
7 imaeq2 p=F-1FaFp=FF-1Fa
8 7 unieqd p=F-1FaFp=FF-1Fa
9 8 eqeq2d p=F-1FaFa=FpFa=FF-1Fa
10 9 adantl FFnAAVaAp=F-1FaFa=FpFa=FF-1Fa
11 uniimaprimaeqfv FFnAaAFF-1Fa=Fa
12 11 adantlr FFnAAVaAFF-1Fa=Fa
13 12 eqcomd FFnAAVaAFa=FF-1Fa
14 6 10 13 rspcedvd FFnAAVaApPFa=Fp
15 eqeq1 y=Fay=FpFa=Fp
16 15 eqcoms Fa=yy=FpFa=Fp
17 16 rexbidv Fa=ypPy=FppPFa=Fp
18 14 17 syl5ibrcom FFnAAVaAFa=ypPy=Fp
19 18 rexlimdva FFnAAVaAFa=ypPy=Fp
20 8 eqcomd p=F-1FaFF-1Fa=Fp
21 13 20 sylan9eq FFnAAVaAp=F-1FaFa=Fp
22 21 ex FFnAAVaAp=F-1FaFa=Fp
23 22 reximdva FFnAAVaAp=F-1FaaAFa=Fp
24 1 elsetpreimafv pPxAp=F-1Fx
25 fveq2 a=xFa=Fx
26 25 sneqd a=xFa=Fx
27 26 imaeq2d a=xF-1Fa=F-1Fx
28 27 eqeq2d a=xp=F-1Fap=F-1Fx
29 28 cbvrexvw aAp=F-1FaxAp=F-1Fx
30 24 29 sylibr pPaAp=F-1Fa
31 23 30 impel FFnAAVpPaAFa=Fp
32 eqeq2 y=FpFa=yFa=Fp
33 32 rexbidv y=FpaAFa=yaAFa=Fp
34 31 33 syl5ibrcom FFnAAVpPy=FpaAFa=y
35 34 rexlimdva FFnAAVpPy=FpaAFa=y
36 19 35 impbid FFnAAVaAFa=ypPy=Fp
37 36 abbidv FFnAAVy|aAFa=y=y|pPy=Fp
38 fnfun FFnAFunF
39 fndm FFnAdomF=A
40 eqimss2 domF=AAdomF
41 39 40 syl FFnAAdomF
42 38 41 jca FFnAFunFAdomF
43 42 adantr FFnAAVFunFAdomF
44 dfimafn FunFAdomFFA=y|aAFa=y
45 43 44 syl FFnAAVFA=y|aAFa=y
46 2 rnmpt ranH=y|pPy=Fp
47 46 a1i FFnAAVranH=y|pPy=Fp
48 37 45 47 3eqtr4rd FFnAAVranH=FA
49 dffo2 H:PontoFAH:PFAranH=FA
50 4 48 49 sylanbrc FFnAAVH:PontoFA