Proof of Theorem grpinvadd
Step | Hyp | Ref
| Expression |
1 | | simp1 1134 |
. . . 4
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → 𝐺 ∈ Grp) |
2 | | simp2 1135 |
. . . 4
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → 𝑋 ∈ 𝐵) |
3 | | simp3 1136 |
. . . 4
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → 𝑌 ∈ 𝐵) |
4 | | grpinvadd.b |
. . . . . . 7
⊢ 𝐵 = (Base‘𝐺) |
5 | | grpinvadd.n |
. . . . . . 7
⊢ 𝑁 = (invg‘𝐺) |
6 | 4, 5 | grpinvcl 18608 |
. . . . . 6
⊢ ((𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵) → (𝑁‘𝑌) ∈ 𝐵) |
7 | 6 | 3adant2 1129 |
. . . . 5
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑁‘𝑌) ∈ 𝐵) |
8 | 4, 5 | grpinvcl 18608 |
. . . . . 6
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) → (𝑁‘𝑋) ∈ 𝐵) |
9 | 8 | 3adant3 1130 |
. . . . 5
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑁‘𝑋) ∈ 𝐵) |
10 | | grpinvadd.p |
. . . . . 6
⊢ + =
(+g‘𝐺) |
11 | 4, 10 | grpcl 18566 |
. . . . 5
⊢ ((𝐺 ∈ Grp ∧ (𝑁‘𝑌) ∈ 𝐵 ∧ (𝑁‘𝑋) ∈ 𝐵) → ((𝑁‘𝑌) + (𝑁‘𝑋)) ∈ 𝐵) |
12 | 1, 7, 9, 11 | syl3anc 1369 |
. . . 4
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → ((𝑁‘𝑌) + (𝑁‘𝑋)) ∈ 𝐵) |
13 | 4, 10 | grpass 18567 |
. . . 4
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ ((𝑁‘𝑌) + (𝑁‘𝑋)) ∈ 𝐵)) → ((𝑋 + 𝑌) + ((𝑁‘𝑌) + (𝑁‘𝑋))) = (𝑋 + (𝑌 + ((𝑁‘𝑌) + (𝑁‘𝑋))))) |
14 | 1, 2, 3, 12, 13 | syl13anc 1370 |
. . 3
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → ((𝑋 + 𝑌) + ((𝑁‘𝑌) + (𝑁‘𝑋))) = (𝑋 + (𝑌 + ((𝑁‘𝑌) + (𝑁‘𝑋))))) |
15 | | eqid 2739 |
. . . . . . . 8
⊢
(0g‘𝐺) = (0g‘𝐺) |
16 | 4, 10, 15, 5 | grprinv 18610 |
. . . . . . 7
⊢ ((𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵) → (𝑌 + (𝑁‘𝑌)) = (0g‘𝐺)) |
17 | 16 | 3adant2 1129 |
. . . . . 6
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑌 + (𝑁‘𝑌)) = (0g‘𝐺)) |
18 | 17 | oveq1d 7283 |
. . . . 5
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → ((𝑌 + (𝑁‘𝑌)) + (𝑁‘𝑋)) = ((0g‘𝐺) + (𝑁‘𝑋))) |
19 | 4, 10 | grpass 18567 |
. . . . . 6
⊢ ((𝐺 ∈ Grp ∧ (𝑌 ∈ 𝐵 ∧ (𝑁‘𝑌) ∈ 𝐵 ∧ (𝑁‘𝑋) ∈ 𝐵)) → ((𝑌 + (𝑁‘𝑌)) + (𝑁‘𝑋)) = (𝑌 + ((𝑁‘𝑌) + (𝑁‘𝑋)))) |
20 | 1, 3, 7, 9, 19 | syl13anc 1370 |
. . . . 5
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → ((𝑌 + (𝑁‘𝑌)) + (𝑁‘𝑋)) = (𝑌 + ((𝑁‘𝑌) + (𝑁‘𝑋)))) |
21 | 4, 10, 15 | grplid 18590 |
. . . . . 6
⊢ ((𝐺 ∈ Grp ∧ (𝑁‘𝑋) ∈ 𝐵) → ((0g‘𝐺) + (𝑁‘𝑋)) = (𝑁‘𝑋)) |
22 | 1, 9, 21 | syl2anc 583 |
. . . . 5
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → ((0g‘𝐺) + (𝑁‘𝑋)) = (𝑁‘𝑋)) |
23 | 18, 20, 22 | 3eqtr3d 2787 |
. . . 4
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑌 + ((𝑁‘𝑌) + (𝑁‘𝑋))) = (𝑁‘𝑋)) |
24 | 23 | oveq2d 7284 |
. . 3
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑋 + (𝑌 + ((𝑁‘𝑌) + (𝑁‘𝑋)))) = (𝑋 + (𝑁‘𝑋))) |
25 | 4, 10, 15, 5 | grprinv 18610 |
. . . 4
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) → (𝑋 + (𝑁‘𝑋)) = (0g‘𝐺)) |
26 | 25 | 3adant3 1130 |
. . 3
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑋 + (𝑁‘𝑋)) = (0g‘𝐺)) |
27 | 14, 24, 26 | 3eqtrd 2783 |
. 2
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → ((𝑋 + 𝑌) + ((𝑁‘𝑌) + (𝑁‘𝑋))) = (0g‘𝐺)) |
28 | 4, 10 | grpcl 18566 |
. . 3
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑋 + 𝑌) ∈ 𝐵) |
29 | 4, 10, 15, 5 | grpinvid1 18611 |
. . 3
⊢ ((𝐺 ∈ Grp ∧ (𝑋 + 𝑌) ∈ 𝐵 ∧ ((𝑁‘𝑌) + (𝑁‘𝑋)) ∈ 𝐵) → ((𝑁‘(𝑋 + 𝑌)) = ((𝑁‘𝑌) + (𝑁‘𝑋)) ↔ ((𝑋 + 𝑌) + ((𝑁‘𝑌) + (𝑁‘𝑋))) = (0g‘𝐺))) |
30 | 1, 28, 12, 29 | syl3anc 1369 |
. 2
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → ((𝑁‘(𝑋 + 𝑌)) = ((𝑁‘𝑌) + (𝑁‘𝑋)) ↔ ((𝑋 + 𝑌) + ((𝑁‘𝑌) + (𝑁‘𝑋))) = (0g‘𝐺))) |
31 | 27, 30 | mpbird 256 |
1
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑁‘(𝑋 + 𝑌)) = ((𝑁‘𝑌) + (𝑁‘𝑋))) |