This is an inofficial mirror of http://metamath.tirix.org for personal testing of a visualizer extension only.

Metamath Proof Explorer

Theorem grpidinv

Description: A group has a left and right identity element, and every member has a left and right inverse. (Contributed by NM, 14-Oct-2006) (Revised by AV, 1-Sep-2021)

		Ref	Expression
	Hypotheses	grpidinv.b	⊢ 𝐵 = ( Base ‘ 𝐺 )
		grpidinv.p	⊢ + = ( +_g ‘ 𝐺 )
	Assertion	grpidinv	⊢ ( 𝐺 ∈ Grp → ∃ 𝑢 ∈ 𝐵 ∀ 𝑥 ∈ 𝐵 ( ( ( 𝑢 + 𝑥 ) = 𝑥 ∧ ( 𝑥 + 𝑢 ) = 𝑥 ) ∧ ∃ 𝑦 ∈ 𝐵 ( ( 𝑦 + 𝑥 ) = 𝑢 ∧ ( 𝑥 + 𝑦 ) = 𝑢 ) ) )

Proof

Step	Hyp	Ref	Expression
1		grpidinv.b	⊢ 𝐵 = ( Base ‘ 𝐺 )
2		grpidinv.p	⊢ + = ( +_g ‘ 𝐺 )
3		eqid	⊢ ( 0_g ‘ 𝐺 ) = ( 0_g ‘ 𝐺 )
4	1 3	grpidcl	⊢ ( 𝐺 ∈ Grp → ( 0_g ‘ 𝐺 ) ∈ 𝐵 )
5		oveq1	⊢ ( 𝑢 = ( 0_g ‘ 𝐺 ) → ( 𝑢 + 𝑥 ) = ( ( 0_g ‘ 𝐺 ) + 𝑥 ) )
6	5	eqeq1d	⊢ ( 𝑢 = ( 0_g ‘ 𝐺 ) → ( ( 𝑢 + 𝑥 ) = 𝑥 ↔ ( ( 0_g ‘ 𝐺 ) + 𝑥 ) = 𝑥 ) )
7		oveq2	⊢ ( 𝑢 = ( 0_g ‘ 𝐺 ) → ( 𝑥 + 𝑢 ) = ( 𝑥 + ( 0_g ‘ 𝐺 ) ) )
8	7	eqeq1d	⊢ ( 𝑢 = ( 0_g ‘ 𝐺 ) → ( ( 𝑥 + 𝑢 ) = 𝑥 ↔ ( 𝑥 + ( 0_g ‘ 𝐺 ) ) = 𝑥 ) )
9	6 8	anbi12d	⊢ ( 𝑢 = ( 0_g ‘ 𝐺 ) → ( ( ( 𝑢 + 𝑥 ) = 𝑥 ∧ ( 𝑥 + 𝑢 ) = 𝑥 ) ↔ ( ( ( 0_g ‘ 𝐺 ) + 𝑥 ) = 𝑥 ∧ ( 𝑥 + ( 0_g ‘ 𝐺 ) ) = 𝑥 ) ) )
10		eqeq2	⊢ ( 𝑢 = ( 0_g ‘ 𝐺 ) → ( ( 𝑦 + 𝑥 ) = 𝑢 ↔ ( 𝑦 + 𝑥 ) = ( 0_g ‘ 𝐺 ) ) )
11		eqeq2	⊢ ( 𝑢 = ( 0_g ‘ 𝐺 ) → ( ( 𝑥 + 𝑦 ) = 𝑢 ↔ ( 𝑥 + 𝑦 ) = ( 0_g ‘ 𝐺 ) ) )
12	10 11	anbi12d	⊢ ( 𝑢 = ( 0_g ‘ 𝐺 ) → ( ( ( 𝑦 + 𝑥 ) = 𝑢 ∧ ( 𝑥 + 𝑦 ) = 𝑢 ) ↔ ( ( 𝑦 + 𝑥 ) = ( 0_g ‘ 𝐺 ) ∧ ( 𝑥 + 𝑦 ) = ( 0_g ‘ 𝐺 ) ) ) )
13	12	rexbidv	⊢ ( 𝑢 = ( 0_g ‘ 𝐺 ) → ( ∃ 𝑦 ∈ 𝐵 ( ( 𝑦 + 𝑥 ) = 𝑢 ∧ ( 𝑥 + 𝑦 ) = 𝑢 ) ↔ ∃ 𝑦 ∈ 𝐵 ( ( 𝑦 + 𝑥 ) = ( 0_g ‘ 𝐺 ) ∧ ( 𝑥 + 𝑦 ) = ( 0_g ‘ 𝐺 ) ) ) )
14	9 13	anbi12d	⊢ ( 𝑢 = ( 0_g ‘ 𝐺 ) → ( ( ( ( 𝑢 + 𝑥 ) = 𝑥 ∧ ( 𝑥 + 𝑢 ) = 𝑥 ) ∧ ∃ 𝑦 ∈ 𝐵 ( ( 𝑦 + 𝑥 ) = 𝑢 ∧ ( 𝑥 + 𝑦 ) = 𝑢 ) ) ↔ ( ( ( ( 0_g ‘ 𝐺 ) + 𝑥 ) = 𝑥 ∧ ( 𝑥 + ( 0_g ‘ 𝐺 ) ) = 𝑥 ) ∧ ∃ 𝑦 ∈ 𝐵 ( ( 𝑦 + 𝑥 ) = ( 0_g ‘ 𝐺 ) ∧ ( 𝑥 + 𝑦 ) = ( 0_g ‘ 𝐺 ) ) ) ) )
15	14	ralbidv	⊢ ( 𝑢 = ( 0_g ‘ 𝐺 ) → ( ∀ 𝑥 ∈ 𝐵 ( ( ( 𝑢 + 𝑥 ) = 𝑥 ∧ ( 𝑥 + 𝑢 ) = 𝑥 ) ∧ ∃ 𝑦 ∈ 𝐵 ( ( 𝑦 + 𝑥 ) = 𝑢 ∧ ( 𝑥 + 𝑦 ) = 𝑢 ) ) ↔ ∀ 𝑥 ∈ 𝐵 ( ( ( ( 0_g ‘ 𝐺 ) + 𝑥 ) = 𝑥 ∧ ( 𝑥 + ( 0_g ‘ 𝐺 ) ) = 𝑥 ) ∧ ∃ 𝑦 ∈ 𝐵 ( ( 𝑦 + 𝑥 ) = ( 0_g ‘ 𝐺 ) ∧ ( 𝑥 + 𝑦 ) = ( 0_g ‘ 𝐺 ) ) ) ) )
16	15	adantl	⊢ ( ( 𝐺 ∈ Grp ∧ 𝑢 = ( 0_g ‘ 𝐺 ) ) → ( ∀ 𝑥 ∈ 𝐵 ( ( ( 𝑢 + 𝑥 ) = 𝑥 ∧ ( 𝑥 + 𝑢 ) = 𝑥 ) ∧ ∃ 𝑦 ∈ 𝐵 ( ( 𝑦 + 𝑥 ) = 𝑢 ∧ ( 𝑥 + 𝑦 ) = 𝑢 ) ) ↔ ∀ 𝑥 ∈ 𝐵 ( ( ( ( 0_g ‘ 𝐺 ) + 𝑥 ) = 𝑥 ∧ ( 𝑥 + ( 0_g ‘ 𝐺 ) ) = 𝑥 ) ∧ ∃ 𝑦 ∈ 𝐵 ( ( 𝑦 + 𝑥 ) = ( 0_g ‘ 𝐺 ) ∧ ( 𝑥 + 𝑦 ) = ( 0_g ‘ 𝐺 ) ) ) ) )
17	1 2 3	grpidinv2	⊢ ( ( 𝐺 ∈ Grp ∧ 𝑥 ∈ 𝐵 ) → ( ( ( ( 0_g ‘ 𝐺 ) + 𝑥 ) = 𝑥 ∧ ( 𝑥 + ( 0_g ‘ 𝐺 ) ) = 𝑥 ) ∧ ∃ 𝑦 ∈ 𝐵 ( ( 𝑦 + 𝑥 ) = ( 0_g ‘ 𝐺 ) ∧ ( 𝑥 + 𝑦 ) = ( 0_g ‘ 𝐺 ) ) ) )
18	17	ralrimiva	⊢ ( 𝐺 ∈ Grp → ∀ 𝑥 ∈ 𝐵 ( ( ( ( 0_g ‘ 𝐺 ) + 𝑥 ) = 𝑥 ∧ ( 𝑥 + ( 0_g ‘ 𝐺 ) ) = 𝑥 ) ∧ ∃ 𝑦 ∈ 𝐵 ( ( 𝑦 + 𝑥 ) = ( 0_g ‘ 𝐺 ) ∧ ( 𝑥 + 𝑦 ) = ( 0_g ‘ 𝐺 ) ) ) )
19	4 16 18	rspcedvd	⊢ ( 𝐺 ∈ Grp → ∃ 𝑢 ∈ 𝐵 ∀ 𝑥 ∈ 𝐵 ( ( ( 𝑢 + 𝑥 ) = 𝑥 ∧ ( 𝑥 + 𝑢 ) = 𝑥 ) ∧ ∃ 𝑦 ∈ 𝐵 ( ( 𝑦 + 𝑥 ) = 𝑢 ∧ ( 𝑥 + 𝑦 ) = 𝑢 ) ) )