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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 )