Proof of Theorem grpinvid2
| Step | Hyp | Ref
 | Expression | 
| 1 |   | oveq1 5929 | 
. . . 4
⊢ ((𝑁‘𝑋) = 𝑌 → ((𝑁‘𝑋) + 𝑋) = (𝑌 + 𝑋)) | 
| 2 | 1 | adantl 277 | 
. . 3
⊢ (((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) ∧ (𝑁‘𝑋) = 𝑌) → ((𝑁‘𝑋) + 𝑋) = (𝑌 + 𝑋)) | 
| 3 |   | grpinv.b | 
. . . . . 6
⊢ 𝐵 = (Base‘𝐺) | 
| 4 |   | grpinv.p | 
. . . . . 6
⊢  + =
(+g‘𝐺) | 
| 5 |   | grpinv.u | 
. . . . . 6
⊢  0 =
(0g‘𝐺) | 
| 6 |   | grpinv.n | 
. . . . . 6
⊢ 𝑁 = (invg‘𝐺) | 
| 7 | 3, 4, 5, 6 | grplinv 13182 | 
. . . . 5
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) → ((𝑁‘𝑋) + 𝑋) = 0 ) | 
| 8 | 7 | 3adant3 1019 | 
. . . 4
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → ((𝑁‘𝑋) + 𝑋) = 0 ) | 
| 9 | 8 | adantr 276 | 
. . 3
⊢ (((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) ∧ (𝑁‘𝑋) = 𝑌) → ((𝑁‘𝑋) + 𝑋) = 0 ) | 
| 10 | 2, 9 | eqtr3d 2231 | 
. 2
⊢ (((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) ∧ (𝑁‘𝑋) = 𝑌) → (𝑌 + 𝑋) = 0 ) | 
| 11 | 3, 6 | grpinvcl 13180 | 
. . . . . . 7
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) → (𝑁‘𝑋) ∈ 𝐵) | 
| 12 | 3, 4, 5 | grplid 13163 | 
. . . . . . 7
⊢ ((𝐺 ∈ Grp ∧ (𝑁‘𝑋) ∈ 𝐵) → ( 0 + (𝑁‘𝑋)) = (𝑁‘𝑋)) | 
| 13 | 11, 12 | syldan 282 | 
. . . . . 6
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) → ( 0 + (𝑁‘𝑋)) = (𝑁‘𝑋)) | 
| 14 | 13 | 3adant3 1019 | 
. . . . 5
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → ( 0 + (𝑁‘𝑋)) = (𝑁‘𝑋)) | 
| 15 | 14 | eqcomd 2202 | 
. . . 4
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑁‘𝑋) = ( 0 + (𝑁‘𝑋))) | 
| 16 | 15 | adantr 276 | 
. . 3
⊢ (((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) ∧ (𝑌 + 𝑋) = 0 ) → (𝑁‘𝑋) = ( 0 + (𝑁‘𝑋))) | 
| 17 |   | oveq1 5929 | 
. . . 4
⊢ ((𝑌 + 𝑋) = 0 → ((𝑌 + 𝑋) + (𝑁‘𝑋)) = ( 0 + (𝑁‘𝑋))) | 
| 18 | 17 | adantl 277 | 
. . 3
⊢ (((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) ∧ (𝑌 + 𝑋) = 0 ) → ((𝑌 + 𝑋) + (𝑁‘𝑋)) = ( 0 + (𝑁‘𝑋))) | 
| 19 |   | simprr 531 | 
. . . . . . . 8
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵)) → 𝑌 ∈ 𝐵) | 
| 20 |   | simprl 529 | 
. . . . . . . 8
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵)) → 𝑋 ∈ 𝐵) | 
| 21 | 11 | adantrr 479 | 
. . . . . . . 8
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵)) → (𝑁‘𝑋) ∈ 𝐵) | 
| 22 | 19, 20, 21 | 3jca 1179 | 
. . . . . . 7
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵)) → (𝑌 ∈ 𝐵 ∧ 𝑋 ∈ 𝐵 ∧ (𝑁‘𝑋) ∈ 𝐵)) | 
| 23 | 3, 4 | grpass 13141 | 
. . . . . . 7
⊢ ((𝐺 ∈ Grp ∧ (𝑌 ∈ 𝐵 ∧ 𝑋 ∈ 𝐵 ∧ (𝑁‘𝑋) ∈ 𝐵)) → ((𝑌 + 𝑋) + (𝑁‘𝑋)) = (𝑌 + (𝑋 + (𝑁‘𝑋)))) | 
| 24 | 22, 23 | syldan 282 | 
. . . . . 6
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵)) → ((𝑌 + 𝑋) + (𝑁‘𝑋)) = (𝑌 + (𝑋 + (𝑁‘𝑋)))) | 
| 25 | 24 | 3impb 1201 | 
. . . . 5
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → ((𝑌 + 𝑋) + (𝑁‘𝑋)) = (𝑌 + (𝑋 + (𝑁‘𝑋)))) | 
| 26 | 3, 4, 5, 6 | grprinv 13183 | 
. . . . . . 7
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) → (𝑋 + (𝑁‘𝑋)) = 0 ) | 
| 27 | 26 | oveq2d 5938 | 
. . . . . 6
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) → (𝑌 + (𝑋 + (𝑁‘𝑋))) = (𝑌 + 0 )) | 
| 28 | 27 | 3adant3 1019 | 
. . . . 5
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑌 + (𝑋 + (𝑁‘𝑋))) = (𝑌 + 0 )) | 
| 29 | 3, 4, 5 | grprid 13164 | 
. . . . . 6
⊢ ((𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵) → (𝑌 + 0 ) = 𝑌) | 
| 30 | 29 | 3adant2 1018 | 
. . . . 5
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑌 + 0 ) = 𝑌) | 
| 31 | 25, 28, 30 | 3eqtrd 2233 | 
. . . 4
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → ((𝑌 + 𝑋) + (𝑁‘𝑋)) = 𝑌) | 
| 32 | 31 | adantr 276 | 
. . 3
⊢ (((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) ∧ (𝑌 + 𝑋) = 0 ) → ((𝑌 + 𝑋) + (𝑁‘𝑋)) = 𝑌) | 
| 33 | 16, 18, 32 | 3eqtr2d 2235 | 
. 2
⊢ (((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) ∧ (𝑌 + 𝑋) = 0 ) → (𝑁‘𝑋) = 𝑌) | 
| 34 | 10, 33 | impbida 596 | 
1
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → ((𝑁‘𝑋) = 𝑌 ↔ (𝑌 + 𝑋) = 0 )) |