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 = ( -g𝐺 )
subgsubcld.s ( 𝜑𝑆 ∈ ( SubGrp ‘ 𝐺 ) )
subgsubcld.x ( 𝜑𝑋𝑆 )
subgsubcld.y ( 𝜑𝑌𝑆 )
Assertion subgsubcld ( 𝜑 → ( 𝑋 𝑌 ) ∈ 𝑆 )

Proof

Step Hyp Ref Expression
1 subgsubcld.m = ( -g𝐺 )
2 subgsubcld.s ( 𝜑𝑆 ∈ ( SubGrp ‘ 𝐺 ) )
3 subgsubcld.x ( 𝜑𝑋𝑆 )
4 subgsubcld.y ( 𝜑𝑌𝑆 )
5 1 subgsubcl ( ( 𝑆 ∈ ( SubGrp ‘ 𝐺 ) ∧ 𝑋𝑆𝑌𝑆 ) → ( 𝑋 𝑌 ) ∈ 𝑆 )
6 2 3 4 5 syl3anc ( 𝜑 → ( 𝑋 𝑌 ) ∈ 𝑆 )