Metamath Proof Explorer


Theorem 2nsgsimpgd

Description: If any normal subgroup of a nontrivial group is either the trivial subgroup or the whole group, the group is simple. (Contributed by Rohan Ridenour, 3-Aug-2023)

Ref Expression
Hypotheses 2nsgsimpgd.1 ⊢ B = Base G
2nsgsimpgd.2 ⊢ 0 ˙ = 0 G
2nsgsimpgd.3 ⊢ φ → G ∈ Grp
2nsgsimpgd.4 ⊢ φ → ¬ 0 ˙ = B
2nsgsimpgd.5 ⊢ φ ∧ x ∈ NrmSGrp ⁡ G → x = 0 ˙ ∨ x = B
Assertion 2nsgsimpgd ⊢ φ → G ∈ SimpGrp

Proof

Step Hyp Ref Expression
1 2nsgsimpgd.1 ⊢ B = Base G
2 2nsgsimpgd.2 ⊢ 0 ˙ = 0 G
3 2nsgsimpgd.3 ⊢ φ → G ∈ Grp
4 2nsgsimpgd.4 ⊢ φ → ¬ 0 ˙ = B
5 2nsgsimpgd.5 ⊢ φ ∧ x ∈ NrmSGrp ⁡ G → x = 0 ˙ ∨ x = B
6 elprg ⊢ x ∈ NrmSGrp ⁡ G → x ∈ 0 ˙ B ↔ x = 0 ˙ ∨ x = B
7 6 adantl ⊢ φ ∧ x ∈ NrmSGrp ⁡ G → x ∈ 0 ˙ B ↔ x = 0 ˙ ∨ x = B
8 5 7 mpbird ⊢ φ ∧ x ∈ NrmSGrp ⁡ G → x ∈ 0 ˙ B
9 simpr ⊢ φ ∧ x = 0 ˙ → x = 0 ˙
10 2 0nsg ⊢ G ∈ Grp → 0 ˙ ∈ NrmSGrp ⁡ G
11 3 10 syl ⊢ φ → 0 ˙ ∈ NrmSGrp ⁡ G
12 11 adantr ⊢ φ ∧ x = 0 ˙ → 0 ˙ ∈ NrmSGrp ⁡ G
13 9 12 eqeltrd ⊢ φ ∧ x = 0 ˙ → x ∈ NrmSGrp ⁡ G
14 13 adantlr ⊢ φ ∧ x ∈ 0 ˙ B ∧ x = 0 ˙ → x ∈ NrmSGrp ⁡ G
15 simpr ⊢ φ ∧ x = B → x = B
16 1 nsgid ⊢ G ∈ Grp → B ∈ NrmSGrp ⁡ G
17 3 16 syl ⊢ φ → B ∈ NrmSGrp ⁡ G
18 17 adantr ⊢ φ ∧ x = B → B ∈ NrmSGrp ⁡ G
19 15 18 eqeltrd ⊢ φ ∧ x = B → x ∈ NrmSGrp ⁡ G
20 19 adantlr ⊢ φ ∧ x ∈ 0 ˙ B ∧ x = B → x ∈ NrmSGrp ⁡ G
21 elpri ⊢ x ∈ 0 ˙ B → x = 0 ˙ ∨ x = B
22 21 adantl ⊢ φ ∧ x ∈ 0 ˙ B → x = 0 ˙ ∨ x = B
23 14 20 22 mpjaodan ⊢ φ ∧ x ∈ 0 ˙ B → x ∈ NrmSGrp ⁡ G
24 8 23 impbida ⊢ φ → x ∈ NrmSGrp ⁡ G ↔ x ∈ 0 ˙ B
25 24 eqrdv ⊢ φ → NrmSGrp ⁡ G = 0 ˙ B
26 snex ⊢ 0 ˙ ∈ V
27 26 a1i ⊢ φ → 0 ˙ ∈ V
28 1 fvexi ⊢ B ∈ V
29 28 a1i ⊢ φ → B ∈ V
30 27 29 4 enpr2d ⊢ φ → 0 ˙ B ≈ 2 𝑜
31 25 30 eqbrtrd ⊢ φ → NrmSGrp ⁡ G ≈ 2 𝑜
32 3 31 issimpgd ⊢ φ → G ∈ SimpGrp