Metamath Proof Explorer

Theorem grpinvid2

Description: The inverse of a group element expressed in terms of the identity element. (Contributed by NM, 24-Aug-2011)

Ref Expression
Hypotheses grpinv.b ⊢ 𝐵 = ( Base ‘ 𝐺 )
grpinv.p ⊢ + = ( +g ‘ 𝐺 )
grpinv.u ⊢ 0 = ( 0g ‘ 𝐺 )
grpinv.n ⊢ 𝑁 = ( invg ‘ 𝐺 )
Assertion grpinvid2 ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑁 ‘ 𝑋 ) = 𝑌 ↔ ( 𝑌 + 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 oveq1 ⊢ ( ( 𝑁 ‘ 𝑋 ) = 𝑌 → ( ( 𝑁 ‘ 𝑋 ) + 𝑋 ) = ( 𝑌 + 𝑋 ) )
6 5 adantl ⊢ ( ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ∧ ( 𝑁 ‘ 𝑋 ) = 𝑌 ) → ( ( 𝑁 ‘ 𝑋 ) + 𝑋 ) = ( 𝑌 + 𝑋 ) )
7 1 2 3 4 grplinv ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( ( 𝑁 ‘ 𝑋 ) + 𝑋 ) = 0 )
8 7 3adant3 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑁 ‘ 𝑋 ) + 𝑋 ) = 0 )
9 8 adantr ⊢ ( ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ∧ ( 𝑁 ‘ 𝑋 ) = 𝑌 ) → ( ( 𝑁 ‘ 𝑋 ) + 𝑋 ) = 0 )
10 6 9 eqtr3d ⊢ ( ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ∧ ( 𝑁 ‘ 𝑋 ) = 𝑌 ) → ( 𝑌 + 𝑋 ) = 0 )
11 1 4 grpinvcl ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( 𝑁 ‘ 𝑋 ) ∈ 𝐵 )
12 1 2 3 grplid ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑁 ‘ 𝑋 ) ∈ 𝐵 ) → ( 0 + ( 𝑁 ‘ 𝑋 ) ) = ( 𝑁 ‘ 𝑋 ) )
13 11 12 syldan ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( 0 + ( 𝑁 ‘ 𝑋 ) ) = ( 𝑁 ‘ 𝑋 ) )
14 13 3adant3 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 0 + ( 𝑁 ‘ 𝑋 ) ) = ( 𝑁 ‘ 𝑋 ) )
15 14 eqcomd ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑁 ‘ 𝑋 ) = ( 0 + ( 𝑁 ‘ 𝑋 ) ) )
16 15 adantr ⊢ ( ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ∧ ( 𝑌 + 𝑋 ) = 0 ) → ( 𝑁 ‘ 𝑋 ) = ( 0 + ( 𝑁 ‘ 𝑋 ) ) )
17 oveq1 ⊢ ( ( 𝑌 + 𝑋 ) = 0 → ( ( 𝑌 + 𝑋 ) + ( 𝑁 ‘ 𝑋 ) ) = ( 0 + ( 𝑁 ‘ 𝑋 ) ) )
18 17 adantl ⊢ ( ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ∧ ( 𝑌 + 𝑋 ) = 0 ) → ( ( 𝑌 + 𝑋 ) + ( 𝑁 ‘ 𝑋 ) ) = ( 0 + ( 𝑁 ‘ 𝑋 ) ) )
19 simprr ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → 𝑌 ∈ 𝐵 )
20 simprl ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → 𝑋 ∈ 𝐵 )
21 11 adantrr ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → ( 𝑁 ‘ 𝑋 ) ∈ 𝐵 )
22 19 20 21 3jca ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → ( 𝑌 ∈ 𝐵 ∧ 𝑋 ∈ 𝐵 ∧ ( 𝑁 ‘ 𝑋 ) ∈ 𝐵 ) )
23 1 2 grpass ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑌 ∈ 𝐵 ∧ 𝑋 ∈ 𝐵 ∧ ( 𝑁 ‘ 𝑋 ) ∈ 𝐵 ) ) → ( ( 𝑌 + 𝑋 ) + ( 𝑁 ‘ 𝑋 ) ) = ( 𝑌 + ( 𝑋 + ( 𝑁 ‘ 𝑋 ) ) ) )
24 22 23 syldan ⊢ ( ( 𝐺 ∈ Grp ∧ ( 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ) → ( ( 𝑌 + 𝑋 ) + ( 𝑁 ‘ 𝑋 ) ) = ( 𝑌 + ( 𝑋 + ( 𝑁 ‘ 𝑋 ) ) ) )
25 24 3impb ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑌 + 𝑋 ) + ( 𝑁 ‘ 𝑋 ) ) = ( 𝑌 + ( 𝑋 + ( 𝑁 ‘ 𝑋 ) ) ) )
26 1 2 3 4 grprinv ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( 𝑋 + ( 𝑁 ‘ 𝑋 ) ) = 0 )
27 26 oveq2d ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ) → ( 𝑌 + ( 𝑋 + ( 𝑁 ‘ 𝑋 ) ) ) = ( 𝑌 + 0 ) )
28 27 3adant3 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑌 + ( 𝑋 + ( 𝑁 ‘ 𝑋 ) ) ) = ( 𝑌 + 0 ) )
29 1 2 3 grprid ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵 ) → ( 𝑌 + 0 ) = 𝑌 )
30 29 3adant2 ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑌 + 0 ) = 𝑌 )
31 25 28 30 3eqtrd ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑌 + 𝑋 ) + ( 𝑁 ‘ 𝑋 ) ) = 𝑌 )
32 31 adantr ⊢ ( ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ∧ ( 𝑌 + 𝑋 ) = 0 ) → ( ( 𝑌 + 𝑋 ) + ( 𝑁 ‘ 𝑋 ) ) = 𝑌 )
33 16 18 32 3eqtr2d ⊢ ( ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) ∧ ( 𝑌 + 𝑋 ) = 0 ) → ( 𝑁 ‘ 𝑋 ) = 𝑌 )
34 10 33 impbida ⊢ ( ( 𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑁 ‘ 𝑋 ) = 𝑌 ↔ ( 𝑌 + 𝑋 ) = 0 ) )