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Theorem grpinvex 18902

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 = (0_g‘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 = (0_g‘𝐺)
4	1, 2, 3	isgrp 18898	. . 3 ⊢ (𝐺 ∈ Grp ↔ (𝐺 ∈ Mnd ∧ ∀𝑥 ∈ 𝐵 ∃𝑦 ∈ 𝐵 (𝑦 + 𝑥) = 0 ))
5	4	simprbi 495	. 2 ⊢ (𝐺 ∈ Grp → ∀𝑥 ∈ 𝐵 ∃𝑦 ∈ 𝐵 (𝑦 + 𝑥) = 0 )
6		oveq2 7422	. . . . 5 ⊢ (𝑥 = 𝑋 → (𝑦 + 𝑥) = (𝑦 + 𝑋))
7	6	eqeq1d 2727	. . . 4 ⊢ (𝑥 = 𝑋 → ((𝑦 + 𝑥) = 0 ↔ (𝑦 + 𝑋) = 0 ))
8	7	rexbidv 3169	. . 3 ⊢ (𝑥 = 𝑋 → (∃𝑦 ∈ 𝐵 (𝑦 + 𝑥) = 0 ↔ ∃𝑦 ∈ 𝐵 (𝑦 + 𝑋) = 0 ))
9	8	rspccva 3600	. 2 ⊢ ((∀𝑥 ∈ 𝐵 ∃𝑦 ∈ 𝐵 (𝑦 + 𝑥) = 0 ∧ 𝑋 ∈ 𝐵) → ∃𝑦 ∈ 𝐵 (𝑦 + 𝑋) = 0 )
10	5, 9	sylan 578	1 ⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) → ∃𝑦 ∈ 𝐵 (𝑦 + 𝑋) = 0 )

Colors of variables: wff setvar class

Syntax hints: → wi 4 ∧ wa 394 = wceq 1533 ∈ wcel 2098 ∀wral 3051 ∃wrex 3060 ‘cfv 6541 (class class class)co 7414 Basecbs 17177 +_gcplusg 17230 0_gc0g 17418 Mndcmnd 18691 Grpcgrp 18892

This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1789 ax-4 1803 ax-5 1905 ax-6 1963 ax-7 2003 ax-8 2100 ax-9 2108 ax-ext 2696

This theorem depends on definitions: df-bi 206 df-an 395 df-or 846 df-3an 1086 df-tru 1536 df-fal 1546 df-ex 1774 df-sb 2060 df-clab 2703 df-cleq 2717 df-clel 2802 df-ral 3052 df-rex 3061 df-rab 3420 df-v 3465 df-dif 3942 df-un 3944 df-in 3946 df-ss 3956 df-nul 4317 df-if 4523 df-sn 4623 df-pr 4625 df-op 4629 df-uni 4902 df-br 5142 df-iota 6493 df-fv 6549 df-ov 7417 df-grp 18895

This theorem is referenced by: dfgrp2 18921 grprcan 18932 grpinveu 18933 grprinv 18949

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