Theorem grpinveu 19235

Description: The left inverse element of a group is unique. Lemma 2.2.1(b) of [Herstein] p. 55. (Contributed by NM, 24-Aug-2011.)

Hypotheses

Ref	Expression
grpinveu.b	⊢ 𝐵 = (Base‘𝐺)
grpinveu.p	⊢ + = (+g‘𝐺)
grpinveu.o	⊢ 0 = (0g‘𝐺)

Assertion

Ref	Expression
grpinveu	⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) → ∃!𝑦 ∈ 𝐵 (𝑦 + 𝑋) = 0 )

Distinct variable groups:   𝑦,𝐵   𝑦,𝐺   𝑦, +   𝑦, 0   𝑦,𝑋

Proof of Theorem grpinveu

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		grpinveu.b	. . . 4 ⊢ 𝐵 = (Base‘𝐺)
2		grpinveu.p	. . . 4 ⊢ + = (+g‘𝐺)
3		grpinveu.o	. . . 4 ⊢ 0 = (0g‘𝐺)
4	1, 2, 3	grpinvex 19204	. . 3 ⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) → ∃𝑦 ∈ 𝐵 (𝑦 + 𝑋) = 0 )
5		eqtr3 2785	. . . . . . . . . . . 12 ⊢ (((𝑦 + 𝑋) = 0 ∧ (𝑧 + 𝑋) = 0 ) → (𝑦 + 𝑋) = (𝑧 + 𝑋))
6	1, 2	grprcan 19234	. . . . . . . . . . . 12 ⊢ ((𝐺 ∈ Grp ∧ (𝑦 ∈ 𝐵 ∧ 𝑧 ∈ 𝐵 ∧ 𝑋 ∈ 𝐵)) → ((𝑦 + 𝑋) = (𝑧 + 𝑋) ↔ 𝑦 = 𝑧))
7	5, 6	imbitrid 247	. . . . . . . . . . 11 ⊢ ((𝐺 ∈ Grp ∧ (𝑦 ∈ 𝐵 ∧ 𝑧 ∈ 𝐵 ∧ 𝑋 ∈ 𝐵)) → (((𝑦 + 𝑋) = 0 ∧ (𝑧 + 𝑋) = 0 ) → 𝑦 = 𝑧))
8	7	3exp2 1373	. . . . . . . . . 10 ⊢ (𝐺 ∈ Grp → (𝑦 ∈ 𝐵 → (𝑧 ∈ 𝐵 → (𝑋 ∈ 𝐵 → (((𝑦 + 𝑋) = 0 ∧ (𝑧 + 𝑋) = 0 ) → 𝑦 = 𝑧)))))
9	8	com24 96	. . . . . . . . 9 ⊢ (𝐺 ∈ Grp → (𝑋 ∈ 𝐵 → (𝑧 ∈ 𝐵 → (𝑦 ∈ 𝐵 → (((𝑦 + 𝑋) = 0 ∧ (𝑧 + 𝑋) = 0 ) → 𝑦 = 𝑧)))))
10	9	imp41 431	. . . . . . . 8 ⊢ ((((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) ∧ 𝑧 ∈ 𝐵) ∧ 𝑦 ∈ 𝐵) → (((𝑦 + 𝑋) = 0 ∧ (𝑧 + 𝑋) = 0 ) → 𝑦 = 𝑧))
11	10	an32s 665	. . . . . . 7 ⊢ ((((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) ∧ 𝑦 ∈ 𝐵) ∧ 𝑧 ∈ 𝐵) → (((𝑦 + 𝑋) = 0 ∧ (𝑧 + 𝑋) = 0 ) → 𝑦 = 𝑧))
12	11	expd 421	. . . . . 6 ⊢ ((((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) ∧ 𝑦 ∈ 𝐵) ∧ 𝑧 ∈ 𝐵) → ((𝑦 + 𝑋) = 0 → ((𝑧 + 𝑋) = 0 → 𝑦 = 𝑧)))
13	12	ralrimdva 3166	. . . . 5 ⊢ (((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) ∧ 𝑦 ∈ 𝐵) → ((𝑦 + 𝑋) = 0 → ∀𝑧 ∈ 𝐵 ((𝑧 + 𝑋) = 0 → 𝑦 = 𝑧)))
14	13	ancld 560	. . . 4 ⊢ (((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) ∧ 𝑦 ∈ 𝐵) → ((𝑦 + 𝑋) = 0 → ((𝑦 + 𝑋) = 0 ∧ ∀𝑧 ∈ 𝐵 ((𝑧 + 𝑋) = 0 → 𝑦 = 𝑧))))
15	14	reximdva 3179	. . 3 ⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) → (∃𝑦 ∈ 𝐵 (𝑦 + 𝑋) = 0 → ∃𝑦 ∈ 𝐵 ((𝑦 + 𝑋) = 0 ∧ ∀𝑧 ∈ 𝐵 ((𝑧 + 𝑋) = 0 → 𝑦 = 𝑧))))
16	4, 15	mpd 16	. 2 ⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) → ∃𝑦 ∈ 𝐵 ((𝑦 + 𝑋) = 0 ∧ ∀𝑧 ∈ 𝐵 ((𝑧 + 𝑋) = 0 → 𝑦 = 𝑧)))
17		oveq1 7458	. . . 4 ⊢ (𝑦 = 𝑧 → (𝑦 + 𝑋) = (𝑧 + 𝑋))
18	17	eqeq1d 2765	. . 3 ⊢ (𝑦 = 𝑧 → ((𝑦 + 𝑋) = 0 ↔ (𝑧 + 𝑋) = 0 ))
19	18	reu8 3694	. 2 ⊢ (∃!𝑦 ∈ 𝐵 (𝑦 + 𝑋) = 0 ↔ ∃𝑦 ∈ 𝐵 ((𝑦 + 𝑋) = 0 ∧ ∀𝑧 ∈ 𝐵 ((𝑧 + 𝑋) = 0 → 𝑦 = 𝑧)))
20	16, 19	sylibr 237	1 ⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) → ∃!𝑦 ∈ 𝐵 (𝑦 + 𝑋) = 0 )

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   ∧ w3a 1103   = wceq 1570   ∈ wcel 2147  ∀wral 3080  ∃wrex 3090  ∃!wreu 3367  ‘cfv 6566  (class class class)co 7451  Basecbs 17431  +_gcplusg 17472  0_gc0g 17660  Grpcgrp 19194

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735  ax-sep 5255  ax-nul 5266  ax-pr 5398

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 2102  df-mo 2567  df-eu 2597  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-ne 2960  df-ral 3081  df-rex 3091  df-rmo 3369  df-reu 3370  df-rab 3417  df-v 3456  df-sbc 3743  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4841  df-uni 4873  df-br 5109  df-opab 5174  df-mpt 5193  df-id 5553  df-xp 5664  df-rel 5665  df-cnv 5666  df-co 5667  df-dm 5668  df-iota 6520  df-fun 6568  df-fv 6574  df-riota 7408  df-ov 7454  df-0g 17662  df-mgm 18866  df-sgrp 18958  df-mnd 18974  df-grp 19197

This theorem is used by:  grpinvf  19247  grplinv  19250  isgrpinv  19254