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 ∧ 𝐴𝑋 ) → ( 𝑁𝐴 ) ∈ 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 ∧ 𝐴𝑋 ) → ( 𝑁𝐴 ) = ( 𝑦𝑋 ( 𝑦 𝐺 𝐴 ) = ( GId ‘ 𝐺 ) ) )
5 1 3 grpoinveu ( ( 𝐺 ∈ GrpOp ∧ 𝐴𝑋 ) → ∃! 𝑦𝑋 ( 𝑦 𝐺 𝐴 ) = ( GId ‘ 𝐺 ) )
6 riotacl ( ∃! 𝑦𝑋 ( 𝑦 𝐺 𝐴 ) = ( GId ‘ 𝐺 ) → ( 𝑦𝑋 ( 𝑦 𝐺 𝐴 ) = ( GId ‘ 𝐺 ) ) ∈ 𝑋 )
7 5 6 syl ( ( 𝐺 ∈ GrpOp ∧ 𝐴𝑋 ) → ( 𝑦𝑋 ( 𝑦 𝐺 𝐴 ) = ( GId ‘ 𝐺 ) ) ∈ 𝑋 )
8 4 7 eqeltrd ( ( 𝐺 ∈ GrpOp ∧ 𝐴𝑋 ) → ( 𝑁𝐴 ) ∈ 𝑋 )