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 = Base G
gexval.2 · ˙ = G
gexval.3 0 ˙ = 0 G
gexval.4 E = gEx G
gexval.i I = y | x X y · ˙ x = 0 ˙
Assertion gexlem1 G V E = 0 I = E I

Proof

Step Hyp Ref Expression
1 gexval.1 X = Base G
2 gexval.2 · ˙ = G
3 gexval.3 0 ˙ = 0 G
4 gexval.4 E = gEx G
5 gexval.i I = y | x X y · ˙ x = 0 ˙
6 1 2 3 4 5 gexval G V E = if I = 0 sup I <
7 eqeq2 0 = if I = 0 sup I < E = 0 E = if I = 0 sup I <
8 7 imbi1d 0 = if I = 0 sup I < E = 0 E = 0 I = E I E = if I = 0 sup I < E = 0 I = E I
9 eqeq2 sup I < = if I = 0 sup I < E = sup I < E = if I = 0 sup I <
10 9 imbi1d sup I < = if I = 0 sup I < E = sup I < E = 0 I = E I E = if I = 0 sup I < E = 0 I = E I
11 orc E = 0 I = E = 0 I = E I
12 11 expcom I = E = 0 E = 0 I = E I
13 12 adantl G V I = E = 0 E = 0 I = E I
14 ssrab2 y | x X y · ˙ x = 0 ˙
15 nnuz = 1
16 15 eqcomi 1 =
17 14 5 16 3sstr4i I 1
18 neqne ¬ I = I
19 18 adantl G V ¬ I = I
20 infssuzcl I 1 I sup I < I
21 17 19 20 sylancr G V ¬ I = sup I < I
22 eleq1a sup I < I E = sup I < E I
23 21 22 syl G V ¬ I = E = sup I < E I
24 olc E I E = 0 I = E I
25 23 24 syl6 G V ¬ I = E = sup I < E = 0 I = E I
26 8 10 13 25 ifbothda G V E = if I = 0 sup I < E = 0 I = E I
27 6 26 mpd G V E = 0 I = E I