Theorem grpidinv 13761

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 2232	. . 3 ⊢ (0g‘𝐺) = (0g‘𝐺)
3	1, 2	grpidcl 13731	. 2 ⊢ (𝐺 ∈ Grp → (0g‘𝐺) ∈ 𝐵)
4		oveq1 6056	. . . . . . 7 ⊢ (𝑢 = (0g‘𝐺) → (𝑢 + 𝑥) = ((0g‘𝐺) + 𝑥))
5	4	eqeq1d 2241	. . . . . 6 ⊢ (𝑢 = (0g‘𝐺) → ((𝑢 + 𝑥) = 𝑥 ↔ ((0g‘𝐺) + 𝑥) = 𝑥))
6		oveq2 6057	. . . . . . 7 ⊢ (𝑢 = (0g‘𝐺) → (𝑥 + 𝑢) = (𝑥 + (0g‘𝐺)))
7	6	eqeq1d 2241	. . . . . 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 2242	. . . . . . 7 ⊢ (𝑢 = (0g‘𝐺) → ((𝑦 + 𝑥) = 𝑢 ↔ (𝑦 + 𝑥) = (0g‘𝐺)))
10		eqeq2 2242	. . . . . . 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 2543	. . . . 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 2542	. . 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 13760	. . 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 2615	. 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 2926	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 2203  ∀wral 2520  ∃wrex 2521  ‘cfv 5351  (class class class)co 6049  Basecbs 13201  +_gcplusg 13279  0_gc0g 13458  Grpcgrp 13702

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 2205  ax-14 2206  ax-ext 2214  ax-coll 4224  ax-sep 4227  ax-pow 4286  ax-pr 4321  ax-un 4553  ax-cnex 8214  ax-resscn 8215  ax-1re 8217  ax-addrcl 8220

This theorem depends on definitions:  df-bi 117  df-3an 1007  df-tru 1401  df-nf 1510  df-sb 1812  df-eu 2083  df-mo 2084  df-clab 2219  df-cleq 2225  df-clel 2228  df-nfc 2373  df-ral 2525  df-rex 2526  df-reu 2527  df-rmo 2528  df-rab 2529  df-v 2814  df-sbc 3042  df-csb 3138  df-un 3214  df-in 3216  df-ss 3223  df-pw 3670  df-sn 3694  df-pr 3695  df-op 3697  df-uni 3914  df-int 3949  df-iun 3992  df-br 4109  df-opab 4171  df-mpt 4172  df-id 4413  df-xp 4754  df-rel 4755  df-cnv 4756  df-co 4757  df-dm 4758  df-rn 4759  df-res 4760  df-ima 4761  df-iota 5311  df-fun 5353  df-fn 5354  df-f 5355  df-f1 5356  df-fo 5357  df-f1o 5358  df-fv 5359  df-riota 6002  df-ov 6052  df-inn 9234  df-2 9292  df-ndx 13204  df-slot 13205  df-base 13207  df-plusg 13292  df-0g 13460  df-mgm 13558  df-sgrp 13604  df-mnd 13619  df-grp 13705  df-minusg 13706

This theorem is referenced by: (None)