Theorem grpidinv 13866

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 2238	. . 3 ⊢ (0g‘𝐺) = (0g‘𝐺)
3	1, 2	grpidcl 13836	. 2 ⊢ (𝐺 ∈ Grp → (0g‘𝐺) ∈ 𝐵)
4		oveq1 6092	. . . . . . 7 ⊢ (𝑢 = (0g‘𝐺) → (𝑢 + 𝑥) = ((0g‘𝐺) + 𝑥))
5	4	eqeq1d 2247	. . . . . 6 ⊢ (𝑢 = (0g‘𝐺) → ((𝑢 + 𝑥) = 𝑥 ↔ ((0g‘𝐺) + 𝑥) = 𝑥))
6		oveq2 6093	. . . . . . 7 ⊢ (𝑢 = (0g‘𝐺) → (𝑥 + 𝑢) = (𝑥 + (0g‘𝐺)))
7	6	eqeq1d 2247	. . . . . 6 ⊢ (𝑢 = (0g‘𝐺) → ((𝑥 + 𝑢) = 𝑥 ↔ (𝑥 + (0g‘𝐺)) = 𝑥))
8	5, 7	anbi12d 477	. . . . 5 ⊢ (𝑢 = (0g‘𝐺) → (((𝑢 + 𝑥) = 𝑥 ∧ (𝑥 + 𝑢) = 𝑥) ↔ (((0g‘𝐺) + 𝑥) = 𝑥 ∧ (𝑥 + (0g‘𝐺)) = 𝑥)))
9		eqeq2 2248	. . . . . . 7 ⊢ (𝑢 = (0g‘𝐺) → ((𝑦 + 𝑥) = 𝑢 ↔ (𝑦 + 𝑥) = (0g‘𝐺)))
10		eqeq2 2248	. . . . . . 7 ⊢ (𝑢 = (0g‘𝐺) → ((𝑥 + 𝑦) = 𝑢 ↔ (𝑥 + 𝑦) = (0g‘𝐺)))
11	9, 10	anbi12d 477	. . . . . 6 ⊢ (𝑢 = (0g‘𝐺) → (((𝑦 + 𝑥) = 𝑢 ∧ (𝑥 + 𝑦) = 𝑢) ↔ ((𝑦 + 𝑥) = (0g‘𝐺) ∧ (𝑥 + 𝑦) = (0g‘𝐺))))
12	11	rexbidv 2551	. . . . 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 477	. . . 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 2550	. . 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 277	. 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 13865	. . 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 2623	. 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 2935	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   ∧ wa 104   ↔ wb 105   = wceq 1402   ∈ wcel 2209  ∀wral 2528  ∃wrex 2529  ‘cfv 5377  (class class class)co 6085  Basecbs 13354  +_gcplusg 13433  0_gc0g 13612  Grpcgrp 13807

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-14 2212  ax-ext 2220  ax-coll 4246  ax-sep 4249  ax-pow 4311  ax-pr 4346  ax-un 4578  ax-cnex 8270  ax-resscn 8271  ax-1re 8273  ax-addrcl 8276

This proof depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  df-nf 1514  df-sb 1816  df-eu 2089  df-mo 2090  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ral 2533  df-rex 2534  df-reu 2535  df-rmo 2536  df-rab 2537  df-v 2823  df-sbc 3052  df-csb 3148  df-un 3224  df-in 3226  df-ss 3233  df-pw 3690  df-sn 3715  df-pr 3716  df-op 3718  df-uni 3936  df-int 3971  df-iun 4014  df-br 4131  df-opab 4193  df-mpt 4194  df-id 4438  df-xp 4780  df-rel 4781  df-cnv 4782  df-co 4783  df-dm 4784  df-rn 4785  df-res 4786  df-ima 4787  df-iota 5337  df-fun 5379  df-fn 5380  df-f 5381  df-f1 5382  df-fo 5383  df-f1o 5384  df-fv 5385  df-riota 6038  df-ov 6088  df-inn 9306  df-2 9364  df-ndx 13357  df-slot 13358  df-base 13360  df-plusg 13446  df-0g 13614  df-mgm 13678  df-sgrp 13719  df-mnd 13732  df-grp 13810  df-minusg 13811

This theorem is used by: (None)