Theorem grpinvex 19011

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 19007	. . 3 ⊢ (𝐺 ∈ Grp ↔ (𝐺 ∈ Mnd ∧ ∀𝑥 ∈ 𝐵 ∃𝑦 ∈ 𝐵 (𝑦 + 𝑥) = 0 ))
5	4	simprbi 502	. 2 ⊢ (𝐺 ∈ Grp → ∀𝑥 ∈ 𝐵 ∃𝑦 ∈ 𝐵 (𝑦 + 𝑥) = 0 )
6		oveq2 7420	. . . . 5 ⊢ (𝑥 = 𝑋 → (𝑦 + 𝑥) = (𝑦 + 𝑋))
7	6	eqeq1d 2765	. . . 4 ⊢ (𝑥 = 𝑋 → ((𝑦 + 𝑥) = 0 ↔ (𝑦 + 𝑋) = 0 ))
8	7	rexbidv 3189	. . 3 ⊢ (𝑥 = 𝑋 → (∃𝑦 ∈ 𝐵 (𝑦 + 𝑥) = 0 ↔ ∃𝑦 ∈ 𝐵 (𝑦 + 𝑋) = 0 ))
9	8	rspccva 3581	. 2 ⊢ ((∀𝑥 ∈ 𝐵 ∃𝑦 ∈ 𝐵 (𝑦 + 𝑥) = 0 ∧ 𝑋 ∈ 𝐵) → ∃𝑦 ∈ 𝐵 (𝑦 + 𝑋) = 0 )
10	5, 9	sylan 591	1 ⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) → ∃𝑦 ∈ 𝐵 (𝑦 + 𝑋) = 0 )

Syntax hints:   → wi 4   ∧ wa 400   = wceq 1570   ∈ wcel 2143  ∀wral 3079  ∃wrex 3089  ‘cfv 6538  (class class class)co 7412  Basecbs 17270  +_gcplusg 17311  0_gc0g 17493  Mndcmnd 18793  Grpcgrp 19001

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-ral 3080  df-rex 3090  df-rab 3417  df-v 3457  df-dif 3909  df-un 3911  df-ss 3923  df-nul 4288  df-if 4489  df-sn 4591  df-pr 4593  df-op 4597  df-uni 4874  df-br 5111  df-iota 6494  df-fv 6546  df-ov 7415  df-grp 19004

This theorem is referenced by:  dfgrp2  19030  grprcan  19041  grpinveu  19042  grprinv  19058