Metamath Proof Explorer


Theorem brgic

Description: The relation "is isomorphic to" for groups. (Contributed by Stefan O'Rear, 25-Jan-2015)

Ref Expression
Assertion brgic ( 𝑅 ≃𝑔 𝑆 ↔ ( 𝑅 GrpIso 𝑆 ) ≠ ∅ )

Proof

Step Hyp Ref Expression
1 df-gic ⊢ ≃𝑔 = ( ◡ GrpIso “ ( V ∖ 1o ) )
2 gimfn ⊢ GrpIso Fn ( Grp × Grp )
3 1 2 brwitnlem ⊢ ( 𝑅 ≃𝑔 𝑆 ↔ ( 𝑅 GrpIso 𝑆 ) ≠ ∅ )