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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 ‘ 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 ‘ 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 )