Step | Hyp | Ref
| Expression |
1 | | eqid 2170 |
. . . 4
⊢ ran
(𝑥 ∈ 𝐵, 𝑦 ∈ 𝐷 ↦ (𝑥 × 𝑦)) = ran (𝑥 ∈ 𝐵, 𝑦 ∈ 𝐷 ↦ (𝑥 × 𝑦)) |
2 | 1 | txbasex 13051 |
. . 3
⊢ ((𝐵 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊) → ran (𝑥 ∈ 𝐵, 𝑦 ∈ 𝐷 ↦ (𝑥 × 𝑦)) ∈ V) |
3 | | resmpo 5951 |
. . . . . 6
⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐶 ⊆ 𝐷) → ((𝑥 ∈ 𝐵, 𝑦 ∈ 𝐷 ↦ (𝑥 × 𝑦)) ↾ (𝐴 × 𝐶)) = (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐶 ↦ (𝑥 × 𝑦))) |
4 | | resss 4915 |
. . . . . 6
⊢ ((𝑥 ∈ 𝐵, 𝑦 ∈ 𝐷 ↦ (𝑥 × 𝑦)) ↾ (𝐴 × 𝐶)) ⊆ (𝑥 ∈ 𝐵, 𝑦 ∈ 𝐷 ↦ (𝑥 × 𝑦)) |
5 | 3, 4 | eqsstrrdi 3200 |
. . . . 5
⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐶 ⊆ 𝐷) → (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐶 ↦ (𝑥 × 𝑦)) ⊆ (𝑥 ∈ 𝐵, 𝑦 ∈ 𝐷 ↦ (𝑥 × 𝑦))) |
6 | 5 | adantl 275 |
. . . 4
⊢ (((𝐵 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊) ∧ (𝐴 ⊆ 𝐵 ∧ 𝐶 ⊆ 𝐷)) → (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐶 ↦ (𝑥 × 𝑦)) ⊆ (𝑥 ∈ 𝐵, 𝑦 ∈ 𝐷 ↦ (𝑥 × 𝑦))) |
7 | | rnss 4841 |
. . . 4
⊢ ((𝑥 ∈ 𝐴, 𝑦 ∈ 𝐶 ↦ (𝑥 × 𝑦)) ⊆ (𝑥 ∈ 𝐵, 𝑦 ∈ 𝐷 ↦ (𝑥 × 𝑦)) → ran (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐶 ↦ (𝑥 × 𝑦)) ⊆ ran (𝑥 ∈ 𝐵, 𝑦 ∈ 𝐷 ↦ (𝑥 × 𝑦))) |
8 | 6, 7 | syl 14 |
. . 3
⊢ (((𝐵 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊) ∧ (𝐴 ⊆ 𝐵 ∧ 𝐶 ⊆ 𝐷)) → ran (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐶 ↦ (𝑥 × 𝑦)) ⊆ ran (𝑥 ∈ 𝐵, 𝑦 ∈ 𝐷 ↦ (𝑥 × 𝑦))) |
9 | | tgss 12857 |
. . 3
⊢ ((ran
(𝑥 ∈ 𝐵, 𝑦 ∈ 𝐷 ↦ (𝑥 × 𝑦)) ∈ V ∧ ran (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐶 ↦ (𝑥 × 𝑦)) ⊆ ran (𝑥 ∈ 𝐵, 𝑦 ∈ 𝐷 ↦ (𝑥 × 𝑦))) → (topGen‘ran (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐶 ↦ (𝑥 × 𝑦))) ⊆ (topGen‘ran (𝑥 ∈ 𝐵, 𝑦 ∈ 𝐷 ↦ (𝑥 × 𝑦)))) |
10 | 2, 8, 9 | syl2an2r 590 |
. 2
⊢ (((𝐵 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊) ∧ (𝐴 ⊆ 𝐵 ∧ 𝐶 ⊆ 𝐷)) → (topGen‘ran (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐶 ↦ (𝑥 × 𝑦))) ⊆ (topGen‘ran (𝑥 ∈ 𝐵, 𝑦 ∈ 𝐷 ↦ (𝑥 × 𝑦)))) |
11 | | ssexg 4128 |
. . . . 5
⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝑉) → 𝐴 ∈ V) |
12 | | ssexg 4128 |
. . . . 5
⊢ ((𝐶 ⊆ 𝐷 ∧ 𝐷 ∈ 𝑊) → 𝐶 ∈ V) |
13 | | eqid 2170 |
. . . . . 6
⊢ ran
(𝑥 ∈ 𝐴, 𝑦 ∈ 𝐶 ↦ (𝑥 × 𝑦)) = ran (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐶 ↦ (𝑥 × 𝑦)) |
14 | 13 | txval 13049 |
. . . . 5
⊢ ((𝐴 ∈ V ∧ 𝐶 ∈ V) → (𝐴 ×t 𝐶) = (topGen‘ran (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐶 ↦ (𝑥 × 𝑦)))) |
15 | 11, 12, 14 | syl2an 287 |
. . . 4
⊢ (((𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝑉) ∧ (𝐶 ⊆ 𝐷 ∧ 𝐷 ∈ 𝑊)) → (𝐴 ×t 𝐶) = (topGen‘ran (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐶 ↦ (𝑥 × 𝑦)))) |
16 | 15 | an4s 583 |
. . 3
⊢ (((𝐴 ⊆ 𝐵 ∧ 𝐶 ⊆ 𝐷) ∧ (𝐵 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊)) → (𝐴 ×t 𝐶) = (topGen‘ran (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐶 ↦ (𝑥 × 𝑦)))) |
17 | 16 | ancoms 266 |
. 2
⊢ (((𝐵 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊) ∧ (𝐴 ⊆ 𝐵 ∧ 𝐶 ⊆ 𝐷)) → (𝐴 ×t 𝐶) = (topGen‘ran (𝑥 ∈ 𝐴, 𝑦 ∈ 𝐶 ↦ (𝑥 × 𝑦)))) |
18 | 1 | txval 13049 |
. . 3
⊢ ((𝐵 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊) → (𝐵 ×t 𝐷) = (topGen‘ran (𝑥 ∈ 𝐵, 𝑦 ∈ 𝐷 ↦ (𝑥 × 𝑦)))) |
19 | 18 | adantr 274 |
. 2
⊢ (((𝐵 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊) ∧ (𝐴 ⊆ 𝐵 ∧ 𝐶 ⊆ 𝐷)) → (𝐵 ×t 𝐷) = (topGen‘ran (𝑥 ∈ 𝐵, 𝑦 ∈ 𝐷 ↦ (𝑥 × 𝑦)))) |
20 | 10, 17, 19 | 3sstr4d 3192 |
1
⊢ (((𝐵 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊) ∧ (𝐴 ⊆ 𝐵 ∧ 𝐶 ⊆ 𝐷)) → (𝐴 ×t 𝐶) ⊆ (𝐵 ×t 𝐷)) |