Metamath Proof Explorer


Theorem subgsubcld

Description: A subgroup is closed under group subtraction. (Contributed by Thierry Arnoux, 6-Jul-2025)

Ref Expression
Hypotheses subgsubcld.m ⊢ - ˙ = - G
subgsubcld.s ⊢ φ → S ∈ SubGrp ⁡ G
subgsubcld.x ⊢ φ → X ∈ S
subgsubcld.y ⊢ φ → Y ∈ S
Assertion subgsubcld ⊢ φ → X - ˙ Y ∈ S

Proof

Step Hyp Ref Expression
1 subgsubcld.m ⊢ - ˙ = - G
2 subgsubcld.s ⊢ φ → S ∈ SubGrp ⁡ G
3 subgsubcld.x ⊢ φ → X ∈ S
4 subgsubcld.y ⊢ φ → Y ∈ S
5 1 subgsubcl ⊢ S ∈ SubGrp ⁡ G ∧ X ∈ S ∧ Y ∈ S → X - ˙ Y ∈ S
6 2 3 4 5 syl3anc ⊢ φ → X - ˙ Y ∈ S