| Step | Hyp | Ref | Expression | 
|---|
| 1 |  | simpl1 1192 | . . . . . . 7
⊢ (((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) ∧ (𝑦 ∈ 𝑇 ∧ 𝑧 ∈ 𝑈)) → 𝐺 ∈ Abel) | 
| 2 |  | simpl2 1193 | . . . . . . . 8
⊢ (((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) ∧ (𝑦 ∈ 𝑇 ∧ 𝑧 ∈ 𝑈)) → 𝑇 ⊆ 𝐵) | 
| 3 |  | simprl 771 | . . . . . . . 8
⊢ (((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) ∧ (𝑦 ∈ 𝑇 ∧ 𝑧 ∈ 𝑈)) → 𝑦 ∈ 𝑇) | 
| 4 | 2, 3 | sseldd 3984 | . . . . . . 7
⊢ (((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) ∧ (𝑦 ∈ 𝑇 ∧ 𝑧 ∈ 𝑈)) → 𝑦 ∈ 𝐵) | 
| 5 |  | simpl3 1194 | . . . . . . . 8
⊢ (((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) ∧ (𝑦 ∈ 𝑇 ∧ 𝑧 ∈ 𝑈)) → 𝑈 ⊆ 𝐵) | 
| 6 |  | simprr 773 | . . . . . . . 8
⊢ (((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) ∧ (𝑦 ∈ 𝑇 ∧ 𝑧 ∈ 𝑈)) → 𝑧 ∈ 𝑈) | 
| 7 | 5, 6 | sseldd 3984 | . . . . . . 7
⊢ (((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) ∧ (𝑦 ∈ 𝑇 ∧ 𝑧 ∈ 𝑈)) → 𝑧 ∈ 𝐵) | 
| 8 |  | lsmcomx.v | . . . . . . . 8
⊢ 𝐵 = (Base‘𝐺) | 
| 9 |  | eqid 2737 | . . . . . . . 8
⊢
(+g‘𝐺) = (+g‘𝐺) | 
| 10 | 8, 9 | ablcom 19817 | . . . . . . 7
⊢ ((𝐺 ∈ Abel ∧ 𝑦 ∈ 𝐵 ∧ 𝑧 ∈ 𝐵) → (𝑦(+g‘𝐺)𝑧) = (𝑧(+g‘𝐺)𝑦)) | 
| 11 | 1, 4, 7, 10 | syl3anc 1373 | . . . . . 6
⊢ (((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) ∧ (𝑦 ∈ 𝑇 ∧ 𝑧 ∈ 𝑈)) → (𝑦(+g‘𝐺)𝑧) = (𝑧(+g‘𝐺)𝑦)) | 
| 12 | 11 | eqeq2d 2748 | . . . . 5
⊢ (((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) ∧ (𝑦 ∈ 𝑇 ∧ 𝑧 ∈ 𝑈)) → (𝑥 = (𝑦(+g‘𝐺)𝑧) ↔ 𝑥 = (𝑧(+g‘𝐺)𝑦))) | 
| 13 | 12 | 2rexbidva 3220 | . . . 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 3290 | . . . 4
⊢
(∃𝑦 ∈
𝑇 ∃𝑧 ∈ 𝑈 𝑥 = (𝑧(+g‘𝐺)𝑦) ↔ ∃𝑧 ∈ 𝑈 ∃𝑦 ∈ 𝑇 𝑥 = (𝑧(+g‘𝐺)𝑦)) | 
| 15 | 13, 14 | bitrdi 287 | . . 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 19658 | . . 3
⊢ ((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) → (𝑥 ∈ (𝑇 ⊕ 𝑈) ↔ ∃𝑦 ∈ 𝑇 ∃𝑧 ∈ 𝑈 𝑥 = (𝑦(+g‘𝐺)𝑧))) | 
| 18 | 8, 9, 16 | lsmelvalx 19658 | . . . 4
⊢ ((𝐺 ∈ Abel ∧ 𝑈 ⊆ 𝐵 ∧ 𝑇 ⊆ 𝐵) → (𝑥 ∈ (𝑈 ⊕ 𝑇) ↔ ∃𝑧 ∈ 𝑈 ∃𝑦 ∈ 𝑇 𝑥 = (𝑧(+g‘𝐺)𝑦))) | 
| 19 | 18 | 3com23 1127 | . . 3
⊢ ((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) → (𝑥 ∈ (𝑈 ⊕ 𝑇) ↔ ∃𝑧 ∈ 𝑈 ∃𝑦 ∈ 𝑇 𝑥 = (𝑧(+g‘𝐺)𝑦))) | 
| 20 | 15, 17, 19 | 3bitr4d 311 | . 2
⊢ ((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) → (𝑥 ∈ (𝑇 ⊕ 𝑈) ↔ 𝑥 ∈ (𝑈 ⊕ 𝑇))) | 
| 21 | 20 | eqrdv 2735 | 1
⊢ ((𝐺 ∈ Abel ∧ 𝑇 ⊆ 𝐵 ∧ 𝑈 ⊆ 𝐵) → (𝑇 ⊕ 𝑈) = (𝑈 ⊕ 𝑇)) |