Theorem grpidinv 13817

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 2234	. . 3 ⊢ (0g‘𝐺) = (0g‘𝐺)
3	1, 2	grpidcl 13787	. 2 ⊢ (𝐺 ∈ Grp → (0g‘𝐺) ∈ 𝐵)
4		oveq1 6065	. . . . . . 7 ⊢ (𝑢 = (0g‘𝐺) → (𝑢 + 𝑥) = ((0g‘𝐺) + 𝑥))
5	4	eqeq1d 2243	. . . . . 6 ⊢ (𝑢 = (0g‘𝐺) → ((𝑢 + 𝑥) = 𝑥 ↔ ((0g‘𝐺) + 𝑥) = 𝑥))
6		oveq2 6066	. . . . . . 7 ⊢ (𝑢 = (0g‘𝐺) → (𝑥 + 𝑢) = (𝑥 + (0g‘𝐺)))
7	6	eqeq1d 2243	. . . . . 6 ⊢ (𝑢 = (0g‘𝐺) → ((𝑥 + 𝑢) = 𝑥 ↔ (𝑥 + (0g‘𝐺)) = 𝑥))
8	5, 7	anbi12d 473	. . . . 5 ⊢ (𝑢 = (0g‘𝐺) → (((𝑢 + 𝑥) = 𝑥 ∧ (𝑥 + 𝑢) = 𝑥) ↔ (((0g‘𝐺) + 𝑥) = 𝑥 ∧ (𝑥 + (0g‘𝐺)) = 𝑥)))
9		eqeq2 2244	. . . . . . 7 ⊢ (𝑢 = (0g‘𝐺) → ((𝑦 + 𝑥) = 𝑢 ↔ (𝑦 + 𝑥) = (0g‘𝐺)))
10		eqeq2 2244	. . . . . . 7 ⊢ (𝑢 = (0g‘𝐺) → ((𝑥 + 𝑦) = 𝑢 ↔ (𝑥 + 𝑦) = (0g‘𝐺)))
11	9, 10	anbi12d 473	. . . . . 6 ⊢ (𝑢 = (0g‘𝐺) → (((𝑦 + 𝑥) = 𝑢 ∧ (𝑥 + 𝑦) = 𝑢) ↔ ((𝑦 + 𝑥) = (0g‘𝐺) ∧ (𝑥 + 𝑦) = (0g‘𝐺))))
12	11	rexbidv 2545	. . . . 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 473	. . . 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 2544	. . 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 13816	. . 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 2617	. 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 2929	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 104   ↔ wb 105   = wceq 1398   ∈ wcel 2205  ∀wral 2522  ∃wrex 2523  ‘cfv 5357  (class class class)co 6058  Basecbs 13299  +_gcplusg 13377  0_gc0g 13556  Grpcgrp 13758

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-13 2207  ax-14 2208  ax-ext 2216  ax-coll 4230  ax-sep 4233  ax-pow 4292  ax-pr 4327  ax-un 4559  ax-cnex 8234  ax-resscn 8235  ax-1re 8237  ax-addrcl 8240

This theorem depends on definitions:  df-bi 117  df-3an 1007  df-tru 1401  df-nf 1510  df-sb 1812  df-eu 2085  df-mo 2086  df-clab 2221  df-cleq 2227  df-clel 2230  df-nfc 2375  df-ral 2527  df-rex 2528  df-reu 2529  df-rmo 2530  df-rab 2531  df-v 2817  df-sbc 3046  df-csb 3142  df-un 3218  df-in 3220  df-ss 3227  df-pw 3676  df-sn 3700  df-pr 3701  df-op 3703  df-uni 3920  df-int 3955  df-iun 3998  df-br 4115  df-opab 4177  df-mpt 4178  df-id 4419  df-xp 4760  df-rel 4761  df-cnv 4762  df-co 4763  df-dm 4764  df-rn 4765  df-res 4766  df-ima 4767  df-iota 5317  df-fun 5359  df-fn 5360  df-f 5361  df-f1 5362  df-fo 5363  df-f1o 5364  df-fv 5365  df-riota 6011  df-ov 6061  df-inn 9258  df-2 9316  df-ndx 13302  df-slot 13303  df-base 13305  df-plusg 13390  df-0g 13558  df-mgm 13622  df-sgrp 13668  df-mnd 13681  df-grp 13761  df-minusg 13762

This theorem is referenced by: (None)