Metamath Proof Explorer


Theorem fprb

Description: A condition for functionhood over a pair. (Contributed by Scott Fenton, 16-Sep-2013)

Ref Expression
Hypotheses fprb.1 AV
fprb.2 BV
Assertion fprb ABF:ABRxRyRF=AxBy

Proof

Step Hyp Ref Expression
1 fprb.1 AV
2 fprb.2 BV
3 1 prid1 AAB
4 ffvelcdm F:ABRAABFAR
5 3 4 mpan2 F:ABRFAR
6 5 adantr F:ABRABFAR
7 2 prid2 BAB
8 ffvelcdm F:ABRBABFBR
9 7 8 mpan2 F:ABRFBR
10 9 adantr F:ABRABFBR
11 fvex FAV
12 1 11 fvpr1 ABAFABFBA=FA
13 fvex FBV
14 2 13 fvpr2 ABAFABFBB=FB
15 fveq2 x=AFx=FA
16 fveq2 x=AAFABFBx=AFABFBA
17 15 16 eqeq12d x=AFx=AFABFBxFA=AFABFBA
18 eqcom FA=AFABFBAAFABFBA=FA
19 17 18 bitrdi x=AFx=AFABFBxAFABFBA=FA
20 fveq2 x=BFx=FB
21 fveq2 x=BAFABFBx=AFABFBB
22 20 21 eqeq12d x=BFx=AFABFBxFB=AFABFBB
23 eqcom FB=AFABFBBAFABFBB=FB
24 22 23 bitrdi x=BFx=AFABFBxAFABFBB=FB
25 1 2 19 24 ralpr xABFx=AFABFBxAFABFBA=FAAFABFBB=FB
26 12 14 25 sylanbrc ABxABFx=AFABFBx
27 26 adantl F:ABRABxABFx=AFABFBx
28 ffn F:ABRFFnAB
29 1 2 11 13 fpr ABAFABFB:ABFAFB
30 29 ffnd ABAFABFBFnAB
31 eqfnfv FFnABAFABFBFnABF=AFABFBxABFx=AFABFBx
32 28 30 31 syl2an F:ABRABF=AFABFBxABFx=AFABFBx
33 27 32 mpbird F:ABRABF=AFABFB
34 opeq2 x=FAAx=AFA
35 34 preq1d x=FAAxBy=AFABy
36 35 eqeq2d x=FAF=AxByF=AFABy
37 opeq2 y=FBBy=BFB
38 37 preq2d y=FBAFABy=AFABFB
39 38 eqeq2d y=FBF=AFAByF=AFABFB
40 36 39 rspc2ev FARFBRF=AFABFBxRyRF=AxBy
41 6 10 33 40 syl3anc F:ABRABxRyRF=AxBy
42 41 expcom ABF:ABRxRyRF=AxBy
43 vex xV
44 vex yV
45 1 2 43 44 fpr ABAxBy:ABxy
46 prssi xRyRxyR
47 fss AxBy:ABxyxyRAxBy:ABR
48 45 46 47 syl2an ABxRyRAxBy:ABR
49 48 ex ABxRyRAxBy:ABR
50 feq1 F=AxByF:ABRAxBy:ABR
51 50 biimprcd AxBy:ABRF=AxByF:ABR
52 49 51 syl6 ABxRyRF=AxByF:ABR
53 52 rexlimdvv ABxRyRF=AxByF:ABR
54 42 53 impbid ABF:ABRxRyRF=AxBy