Step | Hyp | Ref
| Expression |
1 | | simpl1 1189 |
. . . . . . 7
⊢ (((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) ∧ (𝑦 ∈ 𝑇 ∧ 𝑧 ∈ 𝑈)) → 𝐺 ∈ Abel) |
2 | | simpl2 1190 |
. . . . . . . 8
⊢ (((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) ∧ (𝑦 ∈ 𝑇 ∧ 𝑧 ∈ 𝑈)) → 𝑇 ⊆ 𝐵) |
3 | | simprl 767 |
. . . . . . . 8
⊢ (((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) ∧ (𝑦 ∈ 𝑇 ∧ 𝑧 ∈ 𝑈)) → 𝑦 ∈ 𝑇) |
4 | 2, 3 | sseldd 3918 |
. . . . . . 7
⊢ (((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) ∧ (𝑦 ∈ 𝑇 ∧ 𝑧 ∈ 𝑈)) → 𝑦 ∈ 𝐵) |
5 | | simpl3 1191 |
. . . . . . . 8
⊢ (((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) ∧ (𝑦 ∈ 𝑇 ∧ 𝑧 ∈ 𝑈)) → 𝑈 ⊆ 𝐵) |
6 | | simprr 769 |
. . . . . . . 8
⊢ (((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) ∧ (𝑦 ∈ 𝑇 ∧ 𝑧 ∈ 𝑈)) → 𝑧 ∈ 𝑈) |
7 | 5, 6 | sseldd 3918 |
. . . . . . 7
⊢ (((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) ∧ (𝑦 ∈ 𝑇 ∧ 𝑧 ∈ 𝑈)) → 𝑧 ∈ 𝐵) |
8 | | lsmcomx.v |
. . . . . . . 8
⊢ 𝐵 = (Base‘𝐺) |
9 | | eqid 2738 |
. . . . . . . 8
⊢
(+g‘𝐺) = (+g‘𝐺) |
10 | 8, 9 | ablcom 19319 |
. . . . . . 7
⊢ ((𝐺 ∈ Abel ∧ 𝑦 ∈ 𝐵 ∧ 𝑧 ∈ 𝐵) → (𝑦(+g‘𝐺)𝑧) = (𝑧(+g‘𝐺)𝑦)) |
11 | 1, 4, 7, 10 | syl3anc 1369 |
. . . . . 6
⊢ (((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) ∧ (𝑦 ∈ 𝑇 ∧ 𝑧 ∈ 𝑈)) → (𝑦(+g‘𝐺)𝑧) = (𝑧(+g‘𝐺)𝑦)) |
12 | 11 | eqeq2d 2749 |
. . . . 5
⊢ (((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) ∧ (𝑦 ∈ 𝑇 ∧ 𝑧 ∈ 𝑈)) → (𝑥 = (𝑦(+g‘𝐺)𝑧) ↔ 𝑥 = (𝑧(+g‘𝐺)𝑦))) |
13 | 12 | 2rexbidva 3227 |
. . . 4
⊢ ((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) → (∃𝑦 ∈ 𝑇 ∃𝑧 ∈ 𝑈 𝑥 = (𝑦(+g‘𝐺)𝑧) ↔ ∃𝑦 ∈ 𝑇 ∃𝑧 ∈ 𝑈 𝑥 = (𝑧(+g‘𝐺)𝑦))) |
14 | | rexcom 3281 |
. . . 4
⊢
(∃𝑦 ∈
𝑇 ∃𝑧 ∈ 𝑈 𝑥 = (𝑧(+g‘𝐺)𝑦) ↔ ∃𝑧 ∈ 𝑈 ∃𝑦 ∈ 𝑇 𝑥 = (𝑧(+g‘𝐺)𝑦)) |
15 | 13, 14 | bitrdi 286 |
. . 3
⊢ ((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) → (∃𝑦 ∈ 𝑇 ∃𝑧 ∈ 𝑈 𝑥 = (𝑦(+g‘𝐺)𝑧) ↔ ∃𝑧 ∈ 𝑈 ∃𝑦 ∈ 𝑇 𝑥 = (𝑧(+g‘𝐺)𝑦))) |
16 | | lsmcomx.s |
. . . 4
⊢ ⊕ =
(LSSum‘𝐺) |
17 | 8, 9, 16 | lsmelvalx 19160 |
. . 3
⊢ ((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) → (𝑥 ∈ (𝑇 ⊕ 𝑈) ↔ ∃𝑦 ∈ 𝑇 ∃𝑧 ∈ 𝑈 𝑥 = (𝑦(+g‘𝐺)𝑧))) |
18 | 8, 9, 16 | lsmelvalx 19160 |
. . . 4
⊢ ((𝐺 ∈ Abel ∧ 𝑈 ⊆ 𝐵 ∧ 𝑇 ⊆ 𝐵) → (𝑥 ∈ (𝑈 ⊕ 𝑇) ↔ ∃𝑧 ∈ 𝑈 ∃𝑦 ∈ 𝑇 𝑥 = (𝑧(+g‘𝐺)𝑦))) |
19 | 18 | 3com23 1124 |
. . 3
⊢ ((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) → (𝑥 ∈ (𝑈 ⊕ 𝑇) ↔ ∃𝑧 ∈ 𝑈 ∃𝑦 ∈ 𝑇 𝑥 = (𝑧(+g‘𝐺)𝑦))) |
20 | 15, 17, 19 | 3bitr4d 310 |
. 2
⊢ ((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) → (𝑥 ∈ (𝑇 ⊕ 𝑈) ↔ 𝑥 ∈ (𝑈 ⊕ 𝑇))) |
21 | 20 | eqrdv 2736 |
1
⊢ ((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) → (𝑇 ⊕ 𝑈) = (𝑈 ⊕ 𝑇)) |