Metamath Proof Explorer


Theorem dprdf11

Description: Two group sums over a direct product that give the same value are equal as functions. (Contributed by Mario Carneiro, 25-Apr-2016) (Revised by AV, 14-Jul-2019)

Ref Expression
Hypotheses eldprdi.0 0˙=0G
eldprdi.w W=hiISi|finSupp0˙h
eldprdi.1 φGdomDProdS
eldprdi.2 φdomS=I
eldprdi.3 φFW
dprdf11.4 φHW
Assertion dprdf11 φGF=GHF=H

Proof

Step Hyp Ref Expression
1 eldprdi.0 0˙=0G
2 eldprdi.w W=hiISi|finSupp0˙h
3 eldprdi.1 φGdomDProdS
4 eldprdi.2 φdomS=I
5 eldprdi.3 φFW
6 dprdf11.4 φHW
7 eqid BaseG=BaseG
8 2 3 4 5 7 dprdff φF:IBaseG
9 8 ffnd φFFnI
10 2 3 4 6 7 dprdff φH:IBaseG
11 10 ffnd φHFnI
12 eqfnfv FFnIHFnIF=HxIFx=Hx
13 9 11 12 syl2anc φF=HxIFx=Hx
14 eqid -G=-G
15 1 2 3 4 5 6 14 dprdfsub φF-GfHWGF-GfH=GF-GGH
16 15 simpld φF-GfHW
17 1 2 3 4 16 dprdfeq0 φGF-GfH=0˙F-GfH=xI0˙
18 15 simprd φGF-GfH=GF-GGH
19 18 eqeq1d φGF-GfH=0˙GF-GGH=0˙
20 3 4 dprddomcld φIV
21 fvexd φxIFxV
22 fvexd φxIHxV
23 8 feqmptd φF=xIFx
24 10 feqmptd φH=xIHx
25 20 21 22 23 24 offval2 φF-GfH=xIFx-GHx
26 25 eqeq1d φF-GfH=xI0˙xIFx-GHx=xI0˙
27 ovex Fx-GHxV
28 27 rgenw xIFx-GHxV
29 mpteqb xIFx-GHxVxIFx-GHx=xI0˙xIFx-GHx=0˙
30 28 29 ax-mp xIFx-GHx=xI0˙xIFx-GHx=0˙
31 dprdgrp GdomDProdSGGrp
32 3 31 syl φGGrp
33 32 adantr φxIGGrp
34 8 ffvelcdmda φxIFxBaseG
35 10 ffvelcdmda φxIHxBaseG
36 7 1 14 grpsubeq0 GGrpFxBaseGHxBaseGFx-GHx=0˙Fx=Hx
37 33 34 35 36 syl3anc φxIFx-GHx=0˙Fx=Hx
38 37 ralbidva φxIFx-GHx=0˙xIFx=Hx
39 30 38 bitrid φxIFx-GHx=xI0˙xIFx=Hx
40 26 39 bitrd φF-GfH=xI0˙xIFx=Hx
41 17 19 40 3bitr3d φGF-GGH=0˙xIFx=Hx
42 7 dprdssv GDProdSBaseG
43 1 2 3 4 5 eldprdi φGFGDProdS
44 42 43 sselid φGFBaseG
45 1 2 3 4 6 eldprdi φGHGDProdS
46 42 45 sselid φGHBaseG
47 7 1 14 grpsubeq0 GGrpGFBaseGGHBaseGGF-GGH=0˙GF=GH
48 32 44 46 47 syl3anc φGF-GGH=0˙GF=GH
49 13 41 48 3bitr2rd φGF=GHF=H