Metamath Proof Explorer


Theorem nelsubginvcld

Description: The inverse of a non-subgroup-member is a non-subgroup-member. (Contributed by Steven Nguyen, 15-Apr-2023)

Ref Expression
Hypotheses nelsubginvcld.g ⊢ φ → G ∈ Grp
nelsubginvcld.s ⊢ φ → S ∈ SubGrp ⁡ G
nelsubginvcld.x ⊢ φ → X ∈ B ∖ S
nelsubginvcld.b ⊢ B = Base G
nelsubginvcld.p ⊢ N = inv g ⁡ G
Assertion nelsubginvcld ⊢ φ → N ⁡ X ∈ B ∖ S

Proof

Step Hyp Ref Expression
1 nelsubginvcld.g ⊢ φ → G ∈ Grp
2 nelsubginvcld.s ⊢ φ → S ∈ SubGrp ⁡ G
3 nelsubginvcld.x ⊢ φ → X ∈ B ∖ S
4 nelsubginvcld.b ⊢ B = Base G
5 nelsubginvcld.p ⊢ N = inv g ⁡ G
6 3 eldifad ⊢ φ → X ∈ B
7 4 5 grpinvcl ⊢ G ∈ Grp ∧ X ∈ B → N ⁡ X ∈ B
8 1 6 7 syl2anc ⊢ φ → N ⁡ X ∈ B
9 3 eldifbd ⊢ φ → ¬ X ∈ S
10 4 5 grpinvinv ⊢ G ∈ Grp ∧ X ∈ B → N ⁡ N ⁡ X = X
11 1 6 10 syl2anc ⊢ φ → N ⁡ N ⁡ X = X
12 11 adantr ⊢ φ ∧ N ⁡ X ∈ S → N ⁡ N ⁡ X = X
13 5 subginvcl ⊢ S ∈ SubGrp ⁡ G ∧ N ⁡ X ∈ S → N ⁡ N ⁡ X ∈ S
14 2 13 sylan ⊢ φ ∧ N ⁡ X ∈ S → N ⁡ N ⁡ X ∈ S
15 12 14 eqeltrrd ⊢ φ ∧ N ⁡ X ∈ S → X ∈ S
16 9 15 mtand ⊢ φ → ¬ N ⁡ X ∈ S
17 8 16 eldifd ⊢ φ → N ⁡ X ∈ B ∖ S