Metamath Proof Explorer


Theorem grpinvex

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

Ref Expression
Hypotheses grpcl.b 𝐵 = ( Base ‘ 𝐺 )
grpcl.p + = ( +g𝐺 )
grpinvex.p 0 = ( 0g𝐺 )
Assertion grpinvex ( ( 𝐺 ∈ Grp ∧ 𝑋𝐵 ) → ∃ 𝑦𝐵 ( 𝑦 + 𝑋 ) = 0 )

Proof

Step Hyp Ref Expression
1 grpcl.b 𝐵 = ( Base ‘ 𝐺 )
2 grpcl.p + = ( +g𝐺 )
3 grpinvex.p 0 = ( 0g𝐺 )
4 1 2 3 isgrp ( 𝐺 ∈ Grp ↔ ( 𝐺 ∈ Mnd ∧ ∀ 𝑥𝐵𝑦𝐵 ( 𝑦 + 𝑥 ) = 0 ) )
5 4 simprbi ( 𝐺 ∈ Grp → ∀ 𝑥𝐵𝑦𝐵 ( 𝑦 + 𝑥 ) = 0 )
6 oveq2 ( 𝑥 = 𝑋 → ( 𝑦 + 𝑥 ) = ( 𝑦 + 𝑋 ) )
7 6 eqeq1d ( 𝑥 = 𝑋 → ( ( 𝑦 + 𝑥 ) = 0 ↔ ( 𝑦 + 𝑋 ) = 0 ) )
8 7 rexbidv ( 𝑥 = 𝑋 → ( ∃ 𝑦𝐵 ( 𝑦 + 𝑥 ) = 0 ↔ ∃ 𝑦𝐵 ( 𝑦 + 𝑋 ) = 0 ) )
9 8 rspccva ( ( ∀ 𝑥𝐵𝑦𝐵 ( 𝑦 + 𝑥 ) = 0𝑋𝐵 ) → ∃ 𝑦𝐵 ( 𝑦 + 𝑋 ) = 0 )
10 5 9 sylan ( ( 𝐺 ∈ Grp ∧ 𝑋𝐵 ) → ∃ 𝑦𝐵 ( 𝑦 + 𝑋 ) = 0 )