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 = Base G
gexcl.2 ⊢ E = gEx ⁡ G
gexid.3 ⊢ · ˙ = ⋅ G
gexid.4 ⊢ 0 ˙ = 0 G
Assertion gexid ⊢ A ∈ X → E · ˙ A = 0 ˙

Proof

Step Hyp Ref Expression
1 gexcl.1 ⊢ X = Base G
2 gexcl.2 ⊢ E = gEx ⁡ G
3 gexid.3 ⊢ · ˙ = ⋅ G
4 gexid.4 ⊢ 0 ˙ = 0 G
5 oveq1 ⊢ E = 0 → E · ˙ A = 0 · ˙ A
6 1 4 3 mulg0 ⊢ A ∈ X → 0 · ˙ A = 0 ˙
7 5 6 sylan9eqr ⊢ A ∈ X ∧ E = 0 → E · ˙ A = 0 ˙
8 7 adantrr ⊢ A ∈ X ∧ E = 0 ∧ y ∈ ℕ | ∀ x ∈ X y · ˙ x = 0 ˙ = ∅ → E · ˙ A = 0 ˙
9 oveq1 ⊢ y = E → y · ˙ x = E · ˙ x
10 9 eqeq1d ⊢ y = E → y · ˙ x = 0 ˙ ↔ E · ˙ x = 0 ˙
11 10 ralbidv ⊢ y = E → ∀ x ∈ X y · ˙ x = 0 ˙ ↔ ∀ x ∈ X E · ˙ x = 0 ˙
12 11 elrab ⊢ E ∈ y ∈ ℕ | ∀ x ∈ X y · ˙ x = 0 ˙ ↔ E ∈ ℕ ∧ ∀ x ∈ X E · ˙ x = 0 ˙
13 12 simprbi ⊢ E ∈ y ∈ ℕ | ∀ x ∈ X y · ˙ x = 0 ˙ → ∀ x ∈ X E · ˙ x = 0 ˙
14 oveq2 ⊢ x = A → E · ˙ x = E · ˙ A
15 14 eqeq1d ⊢ x = A → E · ˙ x = 0 ˙ ↔ E · ˙ A = 0 ˙
16 15 rspcva ⊢ A ∈ X ∧ ∀ x ∈ X E · ˙ x = 0 ˙ → E · ˙ A = 0 ˙
17 13 16 sylan2 ⊢ A ∈ X ∧ E ∈ y ∈ ℕ | ∀ x ∈ X y · ˙ x = 0 ˙ → E · ˙ A = 0 ˙
18 elfvex ⊢ A ∈ Base G → G ∈ V
19 18 1 eleq2s ⊢ A ∈ X → G ∈ V
20 eqid ⊢ y ∈ ℕ | ∀ x ∈ X y · ˙ x = 0 ˙ = y ∈ ℕ | ∀ x ∈ X y · ˙ x = 0 ˙
21 1 3 4 2 20 gexlem1 ⊢ G ∈ V → E = 0 ∧ y ∈ ℕ | ∀ x ∈ X y · ˙ x = 0 ˙ = ∅ ∨ E ∈ y ∈ ℕ | ∀ x ∈ X y · ˙ x = 0 ˙
22 19 21 syl ⊢ A ∈ X → E = 0 ∧ y ∈ ℕ | ∀ x ∈ X y · ˙ x = 0 ˙ = ∅ ∨ E ∈ y ∈ ℕ | ∀ x ∈ X y · ˙ x = 0 ˙
23 8 17 22 mpjaodan ⊢ A ∈ X → E · ˙ A = 0 ˙