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Metamath Proof Explorer

Theorem grplinv

Description: The left inverse of a group element. (Contributed by NM, 24-Aug-2011) (Revised by Mario Carneiro, 6-Jan-2015)

Ref Expression
Hypotheses grpinv.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
grpinv.p ⊢ + = ( +g ‘ 𝐺 )
grpinv.u ⊢ 0 = ( 0g ‘ 𝐺 )
grpinv.n ⊢ 𝑁 = ( invg ‘ 𝐺 )
Assertion grplinv ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( ( 𝑁 ‘ 𝑋 ) + 𝑋 ) = 0 )

Proof

Step Hyp Ref Expression
1 grpinv.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
2 grpinv.p ⊢ + = ( +g ‘ 𝐺 )
3 grpinv.u ⊢ 0 = ( 0g ‘ 𝐺 )
4 grpinv.n ⊢ 𝑁 = ( invg ‘ 𝐺 )
5 1 2 3 4 grpinvval ⊢ ( 𝑋 ∈ 𝐵 → ( 𝑁 ‘ 𝑋 ) = ( ℩ 𝑦 ∈ 𝐵 ( 𝑦 + 𝑋 ) = 0 ) )
6 5 adantl ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( 𝑁 ‘ 𝑋 ) = ( ℩ 𝑦 ∈ 𝐵 ( 𝑦 + 𝑋 ) = 0 ) )
7 1 2 3 grpinveu ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ∃! 𝑦 ∈ 𝐵 ( 𝑦 + 𝑋 ) = 0 )
8 riotacl2 ⊢ ( ∃! 𝑦 ∈ 𝐵 ( 𝑦 + 𝑋 ) = 0 → ( ℩ 𝑦 ∈ 𝐵 ( 𝑦 + 𝑋 ) = 0 ) ∈ { 𝑦 ∈ 𝐵 ∣ ( 𝑦 + 𝑋 ) = 0 } )
9 7 8 syl ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( ℩ 𝑦 ∈ 𝐵 ( 𝑦 + 𝑋 ) = 0 ) ∈ { 𝑦 ∈ 𝐵 ∣ ( 𝑦 + 𝑋 ) = 0 } )
10 6 9 eqeltrd ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( 𝑁 ‘ 𝑋 ) ∈ { 𝑦 ∈ 𝐵 ∣ ( 𝑦 + 𝑋 ) = 0 } )
11 oveq1 ⊢ ( 𝑦 = ( 𝑁 ‘ 𝑋 ) → ( 𝑦 + 𝑋 ) = ( ( 𝑁 ‘ 𝑋 ) + 𝑋 ) )
12 11 eqeq1d ⊢ ( 𝑦 = ( 𝑁 ‘ 𝑋 ) → ( ( 𝑦 + 𝑋 ) = 0 ↔ ( ( 𝑁 ‘ 𝑋 ) + 𝑋 ) = 0 ) )
13 12 elrab ⊢ ( ( 𝑁 ‘ 𝑋 ) ∈ { 𝑦 ∈ 𝐵 ∣ ( 𝑦 + 𝑋 ) = 0 } ↔ ( ( 𝑁 ‘ 𝑋 ) ∈ 𝐵 ∧ ( ( 𝑁 ‘ 𝑋 ) + 𝑋 ) = 0 ) )
14 13 simprbi ⊢ ( ( 𝑁 ‘ 𝑋 ) ∈ { 𝑦 ∈ 𝐵 ∣ ( 𝑦 + 𝑋 ) = 0 } → ( ( 𝑁 ‘ 𝑋 ) + 𝑋 ) = 0 )
15 10 14 syl ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( ( 𝑁 ‘ 𝑋 ) + 𝑋 ) = 0 )