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 ∈ ℕ | ∀ 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 ∈ ℕ | ∀ x ∈ X y · ˙ x = 0 ˙
6 1 2 3 4 5 gexval ⊢ G ∈ V → E = if I = ∅ 0 inf I ℝ <
7 eqeq2 ⊢ 0 = if I = ∅ 0 inf I ℝ < → E = 0 ↔ E = if I = ∅ 0 inf I ℝ <
8 7 imbi1d ⊢ 0 = if I = ∅ 0 inf I ℝ < → E = 0 → E = 0 ∧ I = ∅ ∨ E ∈ I ↔ E = if I = ∅ 0 inf I ℝ < → E = 0 ∧ I = ∅ ∨ E ∈ I
9 eqeq2 ⊢ inf I ℝ < = if I = ∅ 0 inf I ℝ < → E = inf I ℝ < ↔ E = if I = ∅ 0 inf I ℝ <
10 9 imbi1d ⊢ inf I ℝ < = if I = ∅ 0 inf I ℝ < → E = inf I ℝ < → E = 0 ∧ I = ∅ ∨ E ∈ I ↔ E = if I = ∅ 0 inf I ℝ < → 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 ∈ ℕ | ∀ x ∈ X y · ˙ x = 0 ˙ ⊆ ℕ
15 nnuz ⊢ ℕ = ℤ ≥ 1
16 15 eqcomi ⊢ ℤ ≥ 1 = ℕ
17 14 5 16 3sstr4i ⊢ I ⊆ ℤ ≥ 1
18 neqne ⊢ ¬ I = ∅ → I ≠ ∅
19 18 adantl ⊢ G ∈ V ∧ ¬ I = ∅ → I ≠ ∅
20 infssuzcl ⊢ I ⊆ ℤ ≥ 1 ∧ I ≠ ∅ → inf I ℝ < ∈ I
21 17 19 20 sylancr ⊢ G ∈ V ∧ ¬ I = ∅ → inf I ℝ < ∈ I
22 eleq1a ⊢ inf I ℝ < ∈ I → E = inf I ℝ < → E ∈ I
23 21 22 syl ⊢ G ∈ V ∧ ¬ I = ∅ → E = inf I ℝ < → E ∈ I
24 olc ⊢ E ∈ I → E = 0 ∧ I = ∅ ∨ E ∈ I
25 23 24 syl6 ⊢ G ∈ V ∧ ¬ I = ∅ → E = inf I ℝ < → E = 0 ∧ I = ∅ ∨ E ∈ I
26 8 10 13 25 ifbothda ⊢ G ∈ V → E = if I = ∅ 0 inf I ℝ < → E = 0 ∧ I = ∅ ∨ E ∈ I
27 6 26 mpd ⊢ G ∈ V → E = 0 ∧ I = ∅ ∨ E ∈ I