Theorem grpidinv 12759

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 2170	. . 3 ⊢ (0g‘𝐺) = (0g‘𝐺)
3	1, 2	grpidcl 12734	. 2 ⊢ (𝐺 ∈ Grp → (0g‘𝐺) ∈ 𝐵)
4		oveq1 5860	. . . . . . 7 ⊢ (𝑢 = (0g‘𝐺) → (𝑢 + 𝑥) = ((0g‘𝐺) + 𝑥))
5	4	eqeq1d 2179	. . . . . 6 ⊢ (𝑢 = (0g‘𝐺) → ((𝑢 + 𝑥) = 𝑥 ↔ ((0g‘𝐺) + 𝑥) = 𝑥))
6		oveq2 5861	. . . . . . 7 ⊢ (𝑢 = (0g‘𝐺) → (𝑥 + 𝑢) = (𝑥 + (0g‘𝐺)))
7	6	eqeq1d 2179	. . . . . 6 ⊢ (𝑢 = (0g‘𝐺) → ((𝑥 + 𝑢) = 𝑥 ↔ (𝑥 + (0g‘𝐺)) = 𝑥))
8	5, 7	anbi12d 470	. . . . 5 ⊢ (𝑢 = (0g‘𝐺) → (((𝑢 + 𝑥) = 𝑥 ∧ (𝑥 + 𝑢) = 𝑥) ↔ (((0g‘𝐺) + 𝑥) = 𝑥 ∧ (𝑥 + (0g‘𝐺)) = 𝑥)))
9		eqeq2 2180	. . . . . . 7 ⊢ (𝑢 = (0g‘𝐺) → ((𝑦 + 𝑥) = 𝑢 ↔ (𝑦 + 𝑥) = (0g‘𝐺)))
10		eqeq2 2180	. . . . . . 7 ⊢ (𝑢 = (0g‘𝐺) → ((𝑥 + 𝑦) = 𝑢 ↔ (𝑥 + 𝑦) = (0g‘𝐺)))
11	9, 10	anbi12d 470	. . . . . 6 ⊢ (𝑢 = (0g‘𝐺) → (((𝑦 + 𝑥) = 𝑢 ∧ (𝑥 + 𝑦) = 𝑢) ↔ ((𝑦 + 𝑥) = (0g‘𝐺) ∧ (𝑥 + 𝑦) = (0g‘𝐺))))
12	11	rexbidv 2471	. . . . 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 470	. . . 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 2470	. . 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 275	. 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 12758	. . 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 2543	. 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 2840	1 ⊢ (𝐺 ∈ Grp → ∃𝑢 ∈ 𝐵 ∀𝑥 ∈ 𝐵 (((𝑢 + 𝑥) = 𝑥 ∧ (𝑥 + 𝑢) = 𝑥) ∧ ∃𝑦 ∈ 𝐵 ((𝑦 + 𝑥) = 𝑢 ∧ (𝑥 + 𝑦) = 𝑢)))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104   = wceq 1348   ∈ wcel 2141  ∀wral 2448  ∃wrex 2449  ‘cfv 5198  (class class class)co 5853  Basecbs 12416  +_gcplusg 12480  0_gc0g 12596  Grpcgrp 12708

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-13 2143  ax-14 2144  ax-ext 2152  ax-coll 4104  ax-sep 4107  ax-pow 4160  ax-pr 4194  ax-un 4418  ax-cnex 7865  ax-resscn 7866  ax-1re 7868  ax-addrcl 7871

This theorem depends on definitions:  df-bi 116  df-3an 975  df-tru 1351  df-nf 1454  df-sb 1756  df-eu 2022  df-mo 2023  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-ral 2453  df-rex 2454  df-reu 2455  df-rmo 2456  df-rab 2457  df-v 2732  df-sbc 2956  df-csb 3050  df-un 3125  df-in 3127  df-ss 3134  df-pw 3568  df-sn 3589  df-pr 3590  df-op 3592  df-uni 3797  df-int 3832  df-iun 3875  df-br 3990  df-opab 4051  df-mpt 4052  df-id 4278  df-xp 4617  df-rel 4618  df-cnv 4619  df-co 4620  df-dm 4621  df-rn 4622  df-res 4623  df-ima 4624  df-iota 5160  df-fun 5200  df-fn 5201  df-f 5202  df-f1 5203  df-fo 5204  df-f1o 5205  df-fv 5206  df-riota 5809  df-ov 5856  df-inn 8879  df-2 8937  df-ndx 12419  df-slot 12420  df-base 12422  df-plusg 12493  df-0g 12598  df-mgm 12610  df-sgrp 12643  df-mnd 12653  df-grp 12711  df-minusg 12712

This theorem is referenced by: (None)