Proof of Theorem eqgabl
Step | Hyp | Ref
| Expression |
1 | | eqgabl.x |
. . 3
⊢ 𝑋 = (Base‘𝐺) |
2 | | eqid 2739 |
. . 3
⊢
(invg‘𝐺) = (invg‘𝐺) |
3 | | eqid 2739 |
. . 3
⊢
(+g‘𝐺) = (+g‘𝐺) |
4 | | eqgabl.r |
. . 3
⊢ ∼ =
(𝐺 ~QG
𝑆) |
5 | 1, 2, 3, 4 | eqgval 18786 |
. 2
⊢ ((𝐺 ∈ Abel ∧ 𝑆 ⊆ 𝑋) → (𝐴 ∼ 𝐵 ↔ (𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ (((invg‘𝐺)‘𝐴)(+g‘𝐺)𝐵) ∈ 𝑆))) |
6 | | simpll 763 |
. . . . . . 7
⊢ (((𝐺 ∈ Abel ∧ 𝑆 ⊆ 𝑋) ∧ (𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋)) → 𝐺 ∈ Abel) |
7 | | ablgrp 19372 |
. . . . . . . . 9
⊢ (𝐺 ∈ Abel → 𝐺 ∈ Grp) |
8 | 7 | ad2antrr 722 |
. . . . . . . 8
⊢ (((𝐺 ∈ Abel ∧ 𝑆 ⊆ 𝑋) ∧ (𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋)) → 𝐺 ∈ Grp) |
9 | | simprl 767 |
. . . . . . . 8
⊢ (((𝐺 ∈ Abel ∧ 𝑆 ⊆ 𝑋) ∧ (𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋)) → 𝐴 ∈ 𝑋) |
10 | 1, 2 | grpinvcl 18608 |
. . . . . . . 8
⊢ ((𝐺 ∈ Grp ∧ 𝐴 ∈ 𝑋) → ((invg‘𝐺)‘𝐴) ∈ 𝑋) |
11 | 8, 9, 10 | syl2anc 583 |
. . . . . . 7
⊢ (((𝐺 ∈ Abel ∧ 𝑆 ⊆ 𝑋) ∧ (𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋)) → ((invg‘𝐺)‘𝐴) ∈ 𝑋) |
12 | | simprr 769 |
. . . . . . 7
⊢ (((𝐺 ∈ Abel ∧ 𝑆 ⊆ 𝑋) ∧ (𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋)) → 𝐵 ∈ 𝑋) |
13 | 1, 3 | ablcom 19385 |
. . . . . . 7
⊢ ((𝐺 ∈ Abel ∧
((invg‘𝐺)‘𝐴) ∈ 𝑋 ∧ 𝐵 ∈ 𝑋) → (((invg‘𝐺)‘𝐴)(+g‘𝐺)𝐵) = (𝐵(+g‘𝐺)((invg‘𝐺)‘𝐴))) |
14 | 6, 11, 12, 13 | syl3anc 1369 |
. . . . . 6
⊢ (((𝐺 ∈ Abel ∧ 𝑆 ⊆ 𝑋) ∧ (𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋)) → (((invg‘𝐺)‘𝐴)(+g‘𝐺)𝐵) = (𝐵(+g‘𝐺)((invg‘𝐺)‘𝐴))) |
15 | | eqgabl.n |
. . . . . . . 8
⊢ − =
(-g‘𝐺) |
16 | 1, 3, 2, 15 | grpsubval 18606 |
. . . . . . 7
⊢ ((𝐵 ∈ 𝑋 ∧ 𝐴 ∈ 𝑋) → (𝐵 − 𝐴) = (𝐵(+g‘𝐺)((invg‘𝐺)‘𝐴))) |
17 | 12, 9, 16 | syl2anc 583 |
. . . . . 6
⊢ (((𝐺 ∈ Abel ∧ 𝑆 ⊆ 𝑋) ∧ (𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋)) → (𝐵 − 𝐴) = (𝐵(+g‘𝐺)((invg‘𝐺)‘𝐴))) |
18 | 14, 17 | eqtr4d 2782 |
. . . . 5
⊢ (((𝐺 ∈ Abel ∧ 𝑆 ⊆ 𝑋) ∧ (𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋)) → (((invg‘𝐺)‘𝐴)(+g‘𝐺)𝐵) = (𝐵 − 𝐴)) |
19 | 18 | eleq1d 2824 |
. . . 4
⊢ (((𝐺 ∈ Abel ∧ 𝑆 ⊆ 𝑋) ∧ (𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋)) → ((((invg‘𝐺)‘𝐴)(+g‘𝐺)𝐵) ∈ 𝑆 ↔ (𝐵 − 𝐴) ∈ 𝑆)) |
20 | 19 | pm5.32da 578 |
. . 3
⊢ ((𝐺 ∈ Abel ∧ 𝑆 ⊆ 𝑋) → (((𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋) ∧ (((invg‘𝐺)‘𝐴)(+g‘𝐺)𝐵) ∈ 𝑆) ↔ ((𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋) ∧ (𝐵 − 𝐴) ∈ 𝑆))) |
21 | | df-3an 1087 |
. . 3
⊢ ((𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ (((invg‘𝐺)‘𝐴)(+g‘𝐺)𝐵) ∈ 𝑆) ↔ ((𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋) ∧ (((invg‘𝐺)‘𝐴)(+g‘𝐺)𝐵) ∈ 𝑆)) |
22 | | df-3an 1087 |
. . 3
⊢ ((𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ (𝐵 − 𝐴) ∈ 𝑆) ↔ ((𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋) ∧ (𝐵 − 𝐴) ∈ 𝑆)) |
23 | 20, 21, 22 | 3bitr4g 313 |
. 2
⊢ ((𝐺 ∈ Abel ∧ 𝑆 ⊆ 𝑋) → ((𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ (((invg‘𝐺)‘𝐴)(+g‘𝐺)𝐵) ∈ 𝑆) ↔ (𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ (𝐵 − 𝐴) ∈ 𝑆))) |
24 | 5, 23 | bitrd 278 |
1
⊢ ((𝐺 ∈ Abel ∧ 𝑆 ⊆ 𝑋) → (𝐴 ∼ 𝐵 ↔ (𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ∧ (𝐵 − 𝐴) ∈ 𝑆))) |