Theorem grpidinv 19209

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.)

Hypotheses

Ref	Expression
grpidinv.b	⊢ 𝐵 = (Base‘𝐺)
grpidinv.p	⊢ + = (+g‘𝐺)

Assertion

Ref	Expression
grpidinv	⊢ (𝐺 ∈ Grp → ∃𝑢 ∈ 𝐵 ∀𝑥 ∈ 𝐵 (((𝑢 + 𝑥) = 𝑥 ∧ (𝑥 + 𝑢) = 𝑥) ∧ ∃𝑦 ∈ 𝐵 ((𝑦 + 𝑥) = 𝑢 ∧ (𝑥 + 𝑦) = 𝑢)))

Distinct variable groups:   𝑢,𝐺,𝑥,𝑦   𝑢,𝐵,𝑦   𝑢, + ,𝑦

Allowed substitution hints:   𝐵(𝑥)   + (𝑥)

Proof of Theorem grpidinv

Step	Hyp	Ref	Expression
1		grpidinv.b	. . 3 ⊢ 𝐵 = (Base‘𝐺)
2		eqid 2761	. . 3 ⊢ (0g‘𝐺) = (0g‘𝐺)
3	1, 2	grpidcl 19176	. 2 ⊢ (𝐺 ∈ Grp → (0g‘𝐺) ∈ 𝐵)
4		oveq1 7427	. . . . . . 7 ⊢ (𝑢 = (0g‘𝐺) → (𝑢 + 𝑥) = ((0g‘𝐺) + 𝑥))
5	4	eqeq1d 2763	. . . . . 6 ⊢ (𝑢 = (0g‘𝐺) → ((𝑢 + 𝑥) = 𝑥 ↔ ((0g‘𝐺) + 𝑥) = 𝑥))
6		oveq2 7428	. . . . . . 7 ⊢ (𝑢 = (0g‘𝐺) → (𝑥 + 𝑢) = (𝑥 + (0g‘𝐺)))
7	6	eqeq1d 2763	. . . . . 6 ⊢ (𝑢 = (0g‘𝐺) → ((𝑥 + 𝑢) = 𝑥 ↔ (𝑥 + (0g‘𝐺)) = 𝑥))
8	5, 7	anbi12d 644	. . . . 5 ⊢ (𝑢 = (0g‘𝐺) → (((𝑢 + 𝑥) = 𝑥 ∧ (𝑥 + 𝑢) = 𝑥) ↔ (((0g‘𝐺) + 𝑥) = 𝑥 ∧ (𝑥 + (0g‘𝐺)) = 𝑥)))
9		eqeq2 2773	. . . . . . 7 ⊢ (𝑢 = (0g‘𝐺) → ((𝑦 + 𝑥) = 𝑢 ↔ (𝑦 + 𝑥) = (0g‘𝐺)))
10		eqeq2 2773	. . . . . . 7 ⊢ (𝑢 = (0g‘𝐺) → ((𝑥 + 𝑦) = 𝑢 ↔ (𝑥 + 𝑦) = (0g‘𝐺)))
11	9, 10	anbi12d 644	. . . . . 6 ⊢ (𝑢 = (0g‘𝐺) → (((𝑦 + 𝑥) = 𝑢 ∧ (𝑥 + 𝑦) = 𝑢) ↔ ((𝑦 + 𝑥) = (0g‘𝐺) ∧ (𝑥 + 𝑦) = (0g‘𝐺))))
12	11	rexbidv 3187	. . . . 5 ⊢ (𝑢 = (0g‘𝐺) → (∃𝑦 ∈ 𝐵 ((𝑦 + 𝑥) = 𝑢 ∧ (𝑥 + 𝑦) = 𝑢) ↔ ∃𝑦 ∈ 𝐵 ((𝑦 + 𝑥) = (0g‘𝐺) ∧ (𝑥 + 𝑦) = (0g‘𝐺))))
13	8, 12	anbi12d 644	. . . 4 ⊢ (𝑢 = (0g‘𝐺) → ((((𝑢 + 𝑥) = 𝑥 ∧ (𝑥 + 𝑢) = 𝑥) ∧ ∃𝑦 ∈ 𝐵 ((𝑦 + 𝑥) = 𝑢 ∧ (𝑥 + 𝑦) = 𝑢)) ↔ ((((0g‘𝐺) + 𝑥) = 𝑥 ∧ (𝑥 + (0g‘𝐺)) = 𝑥) ∧ ∃𝑦 ∈ 𝐵 ((𝑦 + 𝑥) = (0g‘𝐺) ∧ (𝑥 + 𝑦) = (0g‘𝐺)))))
14	13	ralbidv 3186	. . 3 ⊢ (𝑢 = (0g‘𝐺) → (∀𝑥 ∈ 𝐵 (((𝑢 + 𝑥) = 𝑥 ∧ (𝑥 + 𝑢) = 𝑥) ∧ ∃𝑦 ∈ 𝐵 ((𝑦 + 𝑥) = 𝑢 ∧ (𝑥 + 𝑦) = 𝑢)) ↔ ∀𝑥 ∈ 𝐵 ((((0g‘𝐺) + 𝑥) = 𝑥 ∧ (𝑥 + (0g‘𝐺)) = 𝑥) ∧ ∃𝑦 ∈ 𝐵 ((𝑦 + 𝑥) = (0g‘𝐺) ∧ (𝑥 + 𝑦) = (0g‘𝐺)))))
15	14	adantl 487	. 2 ⊢ ((𝐺 ∈ Grp ∧ 𝑢 = (0g‘𝐺)) → (∀𝑥 ∈ 𝐵 (((𝑢 + 𝑥) = 𝑥 ∧ (𝑥 + 𝑢) = 𝑥) ∧ ∃𝑦 ∈ 𝐵 ((𝑦 + 𝑥) = 𝑢 ∧ (𝑥 + 𝑦) = 𝑢)) ↔ ∀𝑥 ∈ 𝐵 ((((0g‘𝐺) + 𝑥) = 𝑥 ∧ (𝑥 + (0g‘𝐺)) = 𝑥) ∧ ∃𝑦 ∈ 𝐵 ((𝑦 + 𝑥) = (0g‘𝐺) ∧ (𝑥 + 𝑦) = (0g‘𝐺)))))
16		grpidinv.p	. . . 4 ⊢ + = (+g‘𝐺)
17	1, 16, 2	grpidinv2 19208	. . 3 ⊢ ((𝐺 ∈ Grp ∧ 𝑥 ∈ 𝐵) → ((((0g‘𝐺) + 𝑥) = 𝑥 ∧ (𝑥 + (0g‘𝐺)) = 𝑥) ∧ ∃𝑦 ∈ 𝐵 ((𝑦 + 𝑥) = (0g‘𝐺) ∧ (𝑥 + 𝑦) = (0g‘𝐺))))
18	17	ralrimiva 3155	. 2 ⊢ (𝐺 ∈ Grp → ∀𝑥 ∈ 𝐵 ((((0g‘𝐺) + 𝑥) = 𝑥 ∧ (𝑥 + (0g‘𝐺)) = 𝑥) ∧ ∃𝑦 ∈ 𝐵 ((𝑦 + 𝑥) = (0g‘𝐺) ∧ (𝑥 + 𝑦) = (0g‘𝐺))))
19	3, 15, 18	rspcedvd 3579	1 ⊢ (𝐺 ∈ Grp → ∃𝑢 ∈ 𝐵 ∀𝑥 ∈ 𝐵 (((𝑢 + 𝑥) = 𝑥 ∧ (𝑥 + 𝑢) = 𝑥) ∧ ∃𝑦 ∈ 𝐵 ((𝑦 + 𝑥) = 𝑢 ∧ (𝑥 + 𝑦) = 𝑢)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2145  ∀wral 3077  ∃wrex 3087  ‘cfv 6538  (class class class)co 7420  Basecbs 17387  +_gcplusg 17428  0_gc0g 17610  Grpcgrp 19144

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2733  ax-sep 5249  ax-nul 5260  ax-pow 5327  ax-pr 5391  ax-un 7751

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-mo 2565  df-eu 2595  df-clab 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-ne 2957  df-ral 3078  df-rex 3088  df-rmo 3366  df-reu 3367  df-rab 3414  df-v 3453  df-sbc 3740  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-pw 4559  df-sn 4585  df-pr 4587  df-op 4591  df-uni 4868  df-br 5104  df-opab 5168  df-mpt 5187  df-id 5546  df-xp 5657  df-rel 5658  df-cnv 5659  df-co 5660  df-dm 5661  df-rn 5662  df-res 5663  df-ima 5664  df-iota 6494  df-fun 6540  df-fn 6541  df-f 6542  df-fv 6546  df-riota 7377  df-ov 7423  df-0g 17612  df-mgm 18816  df-sgrp 18908  df-mnd 18924  df-grp 19147  df-minusg 19148

This theorem is used by: (None)