Metamath Proof Explorer

Theorem issimpg

Description: The predicate "is a simple group". (Contributed by Rohan Ridenour, 3-Aug-2023)

Ref Expression
Assertion issimpg ⊢ G ∈ SimpGrp ↔ G ∈ Grp ∧ NrmSGrp ⁡ G ≈ 2 𝑜

Proof

Step Hyp Ref Expression
1 fveq2 ⊢ g = G → NrmSGrp ⁡ g = NrmSGrp ⁡ G
2 1 breq1d ⊢ g = G → NrmSGrp ⁡ g ≈ 2 𝑜 ↔ NrmSGrp ⁡ G ≈ 2 𝑜
3 df-simpg ⊢ SimpGrp = g ∈ Grp | NrmSGrp ⁡ g ≈ 2 𝑜
4 2 3 elrab2 ⊢ G ∈ SimpGrp ↔ G ∈ Grp ∧ NrmSGrp ⁡ G ≈ 2 𝑜