Metamath Proof Explorer


Theorem gexlem1

Description: The group element order is either zero or a nonzero multiplier that annihilates the element. (Contributed by Mario Carneiro, 23-Apr-2016) (Proof shortened by AV, 26-Sep-2020)

Ref Expression
Hypotheses gexval.1 X=BaseG
gexval.2 ·˙=G
gexval.3 0˙=0G
gexval.4 E=gExG
gexval.i I=y|xXy·˙x=0˙
Assertion gexlem1 GVE=0I=EI

Proof

Step Hyp Ref Expression
1 gexval.1 X=BaseG
2 gexval.2 ·˙=G
3 gexval.3 0˙=0G
4 gexval.4 E=gExG
5 gexval.i I=y|xXy·˙x=0˙
6 1 2 3 4 5 gexval GVE=ifI=0supI<
7 eqeq2 0=ifI=0supI<E=0E=ifI=0supI<
8 7 imbi1d 0=ifI=0supI<E=0E=0I=EIE=ifI=0supI<E=0I=EI
9 eqeq2 supI<=ifI=0supI<E=supI<E=ifI=0supI<
10 9 imbi1d supI<=ifI=0supI<E=supI<E=0I=EIE=ifI=0supI<E=0I=EI
11 orc E=0I=E=0I=EI
12 11 expcom I=E=0E=0I=EI
13 12 adantl GVI=E=0E=0I=EI
14 ssrab2 y|xXy·˙x=0˙
15 nnuz =1
16 15 eqcomi 1=
17 14 5 16 3sstr4i I1
18 neqne ¬I=I
19 18 adantl GV¬I=I
20 infssuzcl I1IsupI<I
21 17 19 20 sylancr GV¬I=supI<I
22 eleq1a supI<IE=supI<EI
23 21 22 syl GV¬I=E=supI<EI
24 olc EIE=0I=EI
25 23 24 syl6 GV¬I=E=supI<E=0I=EI
26 8 10 13 25 ifbothda GVE=ifI=0supI<E=0I=EI
27 6 26 mpd GVE=0I=EI