Metamath Proof Explorer


Theorem grpoinvcl

Description: A group element's inverse is a group element. (Contributed by NM, 27-Oct-2006) (Revised by Mario Carneiro, 15-Dec-2013) (New usage is discouraged.)

Ref Expression
Hypotheses grpinvcl.1 ⊢ 𝑋 = ran 𝐺
grpinvcl.2 ⊢ 𝑁 = ( inv ‘ 𝐺 )
Assertion grpoinvcl ( ( 𝐺 ∈ GrpOp ∧ 𝐴 ∈ 𝑋 ) → ( 𝑁 ‘ 𝐴 ) ∈ 𝑋 )

Proof

Step Hyp Ref Expression
1 grpinvcl.1 ⊢ 𝑋 = ran 𝐺
2 grpinvcl.2 ⊢ 𝑁 = ( inv ‘ 𝐺 )
3 eqid ⊢ ( GId ‘ 𝐺 ) = ( GId ‘ 𝐺 )
4 1 3 2 grpoinvval ⊢ ( ( 𝐺 ∈ GrpOp ∧ 𝐴 ∈ 𝑋 ) → ( 𝑁 ‘ 𝐴 ) = ( ℩ 𝑦 ∈ 𝑋 ( 𝑦 𝐺 𝐴 ) = ( GId ‘ 𝐺 ) ) )
5 1 3 grpoinveu ⊢ ( ( 𝐺 ∈ GrpOp ∧ 𝐴 ∈ 𝑋 ) → ∃! 𝑦 ∈ 𝑋 ( 𝑦 𝐺 𝐴 ) = ( GId ‘ 𝐺 ) )
6 riotacl ⊢ ( ∃! 𝑦 ∈ 𝑋 ( 𝑦 𝐺 𝐴 ) = ( GId ‘ 𝐺 ) → ( ℩ 𝑦 ∈ 𝑋 ( 𝑦 𝐺 𝐴 ) = ( GId ‘ 𝐺 ) ) ∈ 𝑋 )
7 5 6 syl ⊢ ( ( 𝐺 ∈ GrpOp ∧ 𝐴 ∈ 𝑋 ) → ( ℩ 𝑦 ∈ 𝑋 ( 𝑦 𝐺 𝐴 ) = ( GId ‘ 𝐺 ) ) ∈ 𝑋 )
8 4 7 eqeltrd ⊢ ( ( 𝐺 ∈ GrpOp ∧ 𝐴 ∈ 𝑋 ) → ( 𝑁 ‘ 𝐴 ) ∈ 𝑋 )