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Metamath Proof Explorer

Theorem subgrcl

Description: Reverse closure for the subgroup predicate. (Contributed by Mario Carneiro, 2-Dec-2014)

Ref Expression
Assertion subgrcl ⊢ S ∈ SubGrp ⁡ G → G ∈ Grp

Proof

Step Hyp Ref Expression
1 eqid ⊢ Base G = Base G
2 1 issubg ⊢ S ∈ SubGrp ⁡ G ↔ G ∈ Grp ∧ S ⊆ Base G ∧ G ↾ 𝑠 S ∈ Grp
3 2 simp1bi ⊢ S ∈ SubGrp ⁡ G → G ∈ Grp