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 ) ≈ 2o ) )

Proof

Step Hyp Ref Expression
1 fveq2 ( 𝑔 = 𝐺 → ( NrmSGrp ‘ 𝑔 ) = ( NrmSGrp ‘ 𝐺 ) )
2 1 breq1d ( 𝑔 = 𝐺 → ( ( NrmSGrp ‘ 𝑔 ) ≈ 2o ↔ ( NrmSGrp ‘ 𝐺 ) ≈ 2o ) )
3 df-simpg SimpGrp = { 𝑔 ∈ Grp ∣ ( NrmSGrp ‘ 𝑔 ) ≈ 2o }
4 2 3 elrab2 ( 𝐺 ∈ SimpGrp ↔ ( 𝐺 ∈ Grp ∧ ( NrmSGrp ‘ 𝐺 ) ≈ 2o ) )