Metamath Proof Explorer


Theorem gexid

Description: Any element to the power of the group exponent. (Contributed by Mario Carneiro, 24-Apr-2016)

Ref Expression
Hypotheses gexcl.1 X=BaseG
gexcl.2 E=gExG
gexid.3 ·˙=G
gexid.4 0˙=0G
Assertion gexid AXE·˙A=0˙

Proof

Step Hyp Ref Expression
1 gexcl.1 X=BaseG
2 gexcl.2 E=gExG
3 gexid.3 ·˙=G
4 gexid.4 0˙=0G
5 oveq1 E=0E·˙A=0·˙A
6 1 4 3 mulg0 AX0·˙A=0˙
7 5 6 sylan9eqr AXE=0E·˙A=0˙
8 7 adantrr AXE=0y|xXy·˙x=0˙=E·˙A=0˙
9 oveq1 y=Ey·˙x=E·˙x
10 9 eqeq1d y=Ey·˙x=0˙E·˙x=0˙
11 10 ralbidv y=ExXy·˙x=0˙xXE·˙x=0˙
12 11 elrab Ey|xXy·˙x=0˙ExXE·˙x=0˙
13 12 simprbi Ey|xXy·˙x=0˙xXE·˙x=0˙
14 oveq2 x=AE·˙x=E·˙A
15 14 eqeq1d x=AE·˙x=0˙E·˙A=0˙
16 15 rspcva AXxXE·˙x=0˙E·˙A=0˙
17 13 16 sylan2 AXEy|xXy·˙x=0˙E·˙A=0˙
18 elfvex ABaseGGV
19 18 1 eleq2s AXGV
20 eqid y|xXy·˙x=0˙=y|xXy·˙x=0˙
21 1 3 4 2 20 gexlem1 GVE=0y|xXy·˙x=0˙=Ey|xXy·˙x=0˙
22 19 21 syl AXE=0y|xXy·˙x=0˙=Ey|xXy·˙x=0˙
23 8 17 22 mpjaodan AXE·˙A=0˙