Theorem grpinvex 18931

Description: Every member of a group has a left inverse. (Contributed by NM, 16-Aug-2011.) (Revised by Mario Carneiro, 6-Jan-2015.)

Hypotheses

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

Assertion

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

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

Allowed substitution hints:   + (𝑦)   0 (𝑦)

Proof of Theorem grpinvex

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		grpcl.b	. . . 4 ⊢ 𝐵 = (Base‘𝐺)
2		grpcl.p	. . . 4 ⊢ + = (+g‘𝐺)
3		grpinvex.p	. . . 4 ⊢ 0 = (0g‘𝐺)
4	1, 2, 3	isgrp 18927	. . 3 ⊢ (𝐺 ∈ Grp ↔ (𝐺 ∈ Mnd ∧ ∀𝑥 ∈ 𝐵 ∃𝑦 ∈ 𝐵 (𝑦 + 𝑥) = 0 ))
5	4	simprbi 496	. 2 ⊢ (𝐺 ∈ Grp → ∀𝑥 ∈ 𝐵 ∃𝑦 ∈ 𝐵 (𝑦 + 𝑥) = 0 )
6		oveq2 7418	. . . . 5 ⊢ (𝑥 = 𝑋 → (𝑦 + 𝑥) = (𝑦 + 𝑋))
7	6	eqeq1d 2738	. . . 4 ⊢ (𝑥 = 𝑋 → ((𝑦 + 𝑥) = 0 ↔ (𝑦 + 𝑋) = 0 ))
8	7	rexbidv 3165	. . 3 ⊢ (𝑥 = 𝑋 → (∃𝑦 ∈ 𝐵 (𝑦 + 𝑥) = 0 ↔ ∃𝑦 ∈ 𝐵 (𝑦 + 𝑋) = 0 ))
9	8	rspccva 3605	. 2 ⊢ ((∀𝑥 ∈ 𝐵 ∃𝑦 ∈ 𝐵 (𝑦 + 𝑥) = 0 ∧ 𝑋 ∈ 𝐵) → ∃𝑦 ∈ 𝐵 (𝑦 + 𝑋) = 0 )
10	5, 9	sylan 580	1 ⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) → ∃𝑦 ∈ 𝐵 (𝑦 + 𝑋) = 0 )

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1540   ∈ wcel 2109  ∀wral 3052  ∃wrex 3061  ‘cfv 6536  (class class class)co 7410  Basecbs 17233  +_gcplusg 17276  0_gc0g 17458  Mndcmnd 18717  Grpcgrp 18921

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2715  df-cleq 2728  df-clel 2810  df-ral 3053  df-rex 3062  df-rab 3421  df-v 3466  df-dif 3934  df-un 3936  df-ss 3948  df-nul 4314  df-if 4506  df-sn 4607  df-pr 4609  df-op 4613  df-uni 4889  df-br 5125  df-iota 6489  df-fv 6544  df-ov 7413  df-grp 18924

This theorem is referenced by:  dfgrp2  18950  grprcan  18961  grpinveu  18962  grprinv  18978