Metamath Proof Explorer


Theorem gsummulglem

Description: Lemma for gsummulg and gsummulgz . (Contributed by Mario Carneiro, 7-Jan-2015) (Revised by AV, 6-Jun-2019)

Ref Expression
Hypotheses gsummulg.b B=BaseG
gsummulg.z 0˙=0G
gsummulg.t ·˙=G
gsummulg.a φAV
gsummulg.f φkAXB
gsummulg.w φfinSupp0˙kAX
gsummulglem.g φGCMnd
gsummulglem.n φN
gsummulglem.o φGAbelN0
Assertion gsummulglem φGkAN·˙X=N·˙GkAX

Proof

Step Hyp Ref Expression
1 gsummulg.b B=BaseG
2 gsummulg.z 0˙=0G
3 gsummulg.t ·˙=G
4 gsummulg.a φAV
5 gsummulg.f φkAXB
6 gsummulg.w φfinSupp0˙kAX
7 gsummulglem.g φGCMnd
8 gsummulglem.n φN
9 gsummulglem.o φGAbelN0
10 cmnmnd GCMndGMnd
11 7 10 syl φGMnd
12 1 3 mulgghm GAbelNxBN·˙xGGrpHomG
13 ghmmhm xBN·˙xGGrpHomGxBN·˙xGMndHomG
14 12 13 syl GAbelNxBN·˙xGMndHomG
15 14 expcom NGAbelxBN·˙xGMndHomG
16 8 15 syl φGAbelxBN·˙xGMndHomG
17 1 3 mulgmhm GCMndN0xBN·˙xGMndHomG
18 17 ex GCMndN0xBN·˙xGMndHomG
19 7 18 syl φN0xBN·˙xGMndHomG
20 16 19 9 mpjaod φxBN·˙xGMndHomG
21 oveq2 x=XN·˙x=N·˙X
22 oveq2 x=GkAXN·˙x=N·˙GkAX
23 1 2 7 11 4 20 5 6 21 22 gsummhm2 φGkAN·˙X=N·˙GkAX