Theorem grpinvex 18684

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 18680	. . 3 ⊢ (𝐺 ∈ Grp ↔ (𝐺 ∈ Mnd ∧ ∀𝑥 ∈ 𝐵 ∃𝑦 ∈ 𝐵 (𝑦 + 𝑥) = 0 ))
5	4	simprbi 497	. 2 ⊢ (𝐺 ∈ Grp → ∀𝑥 ∈ 𝐵 ∃𝑦 ∈ 𝐵 (𝑦 + 𝑥) = 0 )
6		oveq2 7346	. . . . 5 ⊢ (𝑥 = 𝑋 → (𝑦 + 𝑥) = (𝑦 + 𝑋))
7	6	eqeq1d 2738	. . . 4 ⊢ (𝑥 = 𝑋 → ((𝑦 + 𝑥) = 0 ↔ (𝑦 + 𝑋) = 0 ))
8	7	rexbidv 3171	. . 3 ⊢ (𝑥 = 𝑋 → (∃𝑦 ∈ 𝐵 (𝑦 + 𝑥) = 0 ↔ ∃𝑦 ∈ 𝐵 (𝑦 + 𝑋) = 0 ))
9	8	rspccva 3569	. 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 396   = wceq 1540   ∈ wcel 2105  ∀wral 3061  ∃wrex 3070  ‘cfv 6480  (class class class)co 7338  Basecbs 17010  +_gcplusg 17060  0_gc0g 17248  Mndcmnd 18483  Grpcgrp 18674

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-ext 2707

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1781  df-sb 2067  df-clab 2714  df-cleq 2728  df-clel 2814  df-ral 3062  df-rex 3071  df-rab 3404  df-v 3443  df-dif 3901  df-un 3903  df-in 3905  df-ss 3915  df-nul 4271  df-if 4475  df-sn 4575  df-pr 4577  df-op 4581  df-uni 4854  df-br 5094  df-iota 6432  df-fv 6488  df-ov 7341  df-grp 18677

This theorem is referenced by:  dfgrp2  18701  grprcan  18710  grpinveu  18711  grprinv  18726