Proof of Theorem grpsubadd
Step | Hyp | Ref
| Expression |
1 | | grpsubadd.b |
. . . . . . 7
⊢ 𝐵 = (Base‘𝐺) |
2 | | grpsubadd.p |
. . . . . . 7
⊢ + =
(+g‘𝐺) |
3 | | eqid 2738 |
. . . . . . 7
⊢
(invg‘𝐺) = (invg‘𝐺) |
4 | | grpsubadd.m |
. . . . . . 7
⊢ − =
(-g‘𝐺) |
5 | 1, 2, 3, 4 | grpsubval 18540 |
. . . . . 6
⊢ ((𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) → (𝑋 − 𝑌) = (𝑋 +
((invg‘𝐺)‘𝑌))) |
6 | 5 | 3adant3 1130 |
. . . . 5
⊢ ((𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵) → (𝑋 − 𝑌) = (𝑋 +
((invg‘𝐺)‘𝑌))) |
7 | 6 | adantl 481 |
. . . 4
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵)) → (𝑋 − 𝑌) = (𝑋 +
((invg‘𝐺)‘𝑌))) |
8 | 7 | eqeq1d 2740 |
. . 3
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵)) → ((𝑋 − 𝑌) = 𝑍 ↔ (𝑋 +
((invg‘𝐺)‘𝑌)) = 𝑍)) |
9 | | simpl 482 |
. . . 4
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵)) → 𝐺 ∈ Grp) |
10 | | simpr1 1192 |
. . . . 5
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵)) → 𝑋 ∈ 𝐵) |
11 | 1, 3 | grpinvcl 18542 |
. . . . . 6
⊢ ((𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵) → ((invg‘𝐺)‘𝑌) ∈ 𝐵) |
12 | 11 | 3ad2antr2 1187 |
. . . . 5
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵)) → ((invg‘𝐺)‘𝑌) ∈ 𝐵) |
13 | 1, 2 | grpcl 18500 |
. . . . 5
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵 ∧ ((invg‘𝐺)‘𝑌) ∈ 𝐵) → (𝑋 +
((invg‘𝐺)‘𝑌)) ∈ 𝐵) |
14 | 9, 10, 12, 13 | syl3anc 1369 |
. . . 4
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵)) → (𝑋 +
((invg‘𝐺)‘𝑌)) ∈ 𝐵) |
15 | | simpr3 1194 |
. . . 4
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵)) → 𝑍 ∈ 𝐵) |
16 | | simpr2 1193 |
. . . 4
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵)) → 𝑌 ∈ 𝐵) |
17 | 1, 2 | grprcan 18528 |
. . . 4
⊢ ((𝐺 ∈ Grp ∧ ((𝑋 +
((invg‘𝐺)‘𝑌)) ∈ 𝐵 ∧ 𝑍 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵)) → (((𝑋 +
((invg‘𝐺)‘𝑌)) + 𝑌) = (𝑍 + 𝑌) ↔ (𝑋 +
((invg‘𝐺)‘𝑌)) = 𝑍)) |
18 | 9, 14, 15, 16, 17 | syl13anc 1370 |
. . 3
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵)) → (((𝑋 +
((invg‘𝐺)‘𝑌)) + 𝑌) = (𝑍 + 𝑌) ↔ (𝑋 +
((invg‘𝐺)‘𝑌)) = 𝑍)) |
19 | 1, 2 | grpass 18501 |
. . . . . 6
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ ((invg‘𝐺)‘𝑌) ∈ 𝐵 ∧ 𝑌 ∈ 𝐵)) → ((𝑋 +
((invg‘𝐺)‘𝑌)) + 𝑌) = (𝑋 +
(((invg‘𝐺)‘𝑌) + 𝑌))) |
20 | 9, 10, 12, 16, 19 | syl13anc 1370 |
. . . . 5
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵)) → ((𝑋 +
((invg‘𝐺)‘𝑌)) + 𝑌) = (𝑋 +
(((invg‘𝐺)‘𝑌) + 𝑌))) |
21 | | eqid 2738 |
. . . . . . . 8
⊢
(0g‘𝐺) = (0g‘𝐺) |
22 | 1, 2, 21, 3 | grplinv 18543 |
. . . . . . 7
⊢ ((𝐺 ∈ Grp ∧ 𝑌 ∈ 𝐵) → (((invg‘𝐺)‘𝑌) + 𝑌) = (0g‘𝐺)) |
23 | 22 | 3ad2antr2 1187 |
. . . . . 6
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵)) → (((invg‘𝐺)‘𝑌) + 𝑌) = (0g‘𝐺)) |
24 | 23 | oveq2d 7271 |
. . . . 5
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵)) → (𝑋 +
(((invg‘𝐺)‘𝑌) + 𝑌)) = (𝑋 + (0g‘𝐺))) |
25 | 1, 2, 21 | grprid 18525 |
. . . . . 6
⊢ ((𝐺 ∈ Grp ∧ 𝑋 ∈ 𝐵) → (𝑋 + (0g‘𝐺)) = 𝑋) |
26 | 25 | 3ad2antr1 1186 |
. . . . 5
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵)) → (𝑋 + (0g‘𝐺)) = 𝑋) |
27 | 20, 24, 26 | 3eqtrd 2782 |
. . . 4
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵)) → ((𝑋 +
((invg‘𝐺)‘𝑌)) + 𝑌) = 𝑋) |
28 | 27 | eqeq1d 2740 |
. . 3
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵)) → (((𝑋 +
((invg‘𝐺)‘𝑌)) + 𝑌) = (𝑍 + 𝑌) ↔ 𝑋 = (𝑍 + 𝑌))) |
29 | 8, 18, 28 | 3bitr2d 306 |
. 2
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵)) → ((𝑋 − 𝑌) = 𝑍 ↔ 𝑋 = (𝑍 + 𝑌))) |
30 | | eqcom 2745 |
. 2
⊢ (𝑋 = (𝑍 + 𝑌) ↔ (𝑍 + 𝑌) = 𝑋) |
31 | 29, 30 | bitrdi 286 |
1
⊢ ((𝐺 ∈ Grp ∧ (𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵 ∧ 𝑍 ∈ 𝐵)) → ((𝑋 − 𝑌) = 𝑍 ↔ (𝑍 + 𝑌) = 𝑋)) |