| Step | Hyp | Ref
 | Expression | 
| 1 |   | xpm 5091 | 
. . . . . . . 8
⊢
((∃𝑎 𝑎 ∈ 𝐴 ∧ ∃𝑏 𝑏 ∈ 𝐵) ↔ ∃𝑥 𝑥 ∈ (𝐴 × 𝐵)) | 
| 2 |   | dmxpm 4886 | 
. . . . . . . . 9
⊢
(∃𝑏 𝑏 ∈ 𝐵 → dom (𝐴 × 𝐵) = 𝐴) | 
| 3 | 2 | adantl 277 | 
. . . . . . . 8
⊢
((∃𝑎 𝑎 ∈ 𝐴 ∧ ∃𝑏 𝑏 ∈ 𝐵) → dom (𝐴 × 𝐵) = 𝐴) | 
| 4 | 1, 3 | sylbir 135 | 
. . . . . . 7
⊢
(∃𝑥 𝑥 ∈ (𝐴 × 𝐵) → dom (𝐴 × 𝐵) = 𝐴) | 
| 5 | 4 | adantr 276 | 
. . . . . 6
⊢
((∃𝑥 𝑥 ∈ (𝐴 × 𝐵) ∧ (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷)) → dom (𝐴 × 𝐵) = 𝐴) | 
| 6 |   | dmss 4865 | 
. . . . . . 7
⊢ ((𝐴 × 𝐵) ⊆ (𝐶 × 𝐷) → dom (𝐴 × 𝐵) ⊆ dom (𝐶 × 𝐷)) | 
| 7 | 6 | adantl 277 | 
. . . . . 6
⊢
((∃𝑥 𝑥 ∈ (𝐴 × 𝐵) ∧ (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷)) → dom (𝐴 × 𝐵) ⊆ dom (𝐶 × 𝐷)) | 
| 8 | 5, 7 | eqsstrrd 3220 | 
. . . . 5
⊢
((∃𝑥 𝑥 ∈ (𝐴 × 𝐵) ∧ (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷)) → 𝐴 ⊆ dom (𝐶 × 𝐷)) | 
| 9 |   | dmxpss 5100 | 
. . . . 5
⊢ dom
(𝐶 × 𝐷) ⊆ 𝐶 | 
| 10 | 8, 9 | sstrdi 3195 | 
. . . 4
⊢
((∃𝑥 𝑥 ∈ (𝐴 × 𝐵) ∧ (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷)) → 𝐴 ⊆ 𝐶) | 
| 11 |   | rnxpm 5099 | 
. . . . . . . . 9
⊢
(∃𝑎 𝑎 ∈ 𝐴 → ran (𝐴 × 𝐵) = 𝐵) | 
| 12 | 11 | adantr 276 | 
. . . . . . . 8
⊢
((∃𝑎 𝑎 ∈ 𝐴 ∧ ∃𝑏 𝑏 ∈ 𝐵) → ran (𝐴 × 𝐵) = 𝐵) | 
| 13 | 1, 12 | sylbir 135 | 
. . . . . . 7
⊢
(∃𝑥 𝑥 ∈ (𝐴 × 𝐵) → ran (𝐴 × 𝐵) = 𝐵) | 
| 14 | 13 | adantr 276 | 
. . . . . 6
⊢
((∃𝑥 𝑥 ∈ (𝐴 × 𝐵) ∧ (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷)) → ran (𝐴 × 𝐵) = 𝐵) | 
| 15 |   | rnss 4896 | 
. . . . . . 7
⊢ ((𝐴 × 𝐵) ⊆ (𝐶 × 𝐷) → ran (𝐴 × 𝐵) ⊆ ran (𝐶 × 𝐷)) | 
| 16 | 15 | adantl 277 | 
. . . . . 6
⊢
((∃𝑥 𝑥 ∈ (𝐴 × 𝐵) ∧ (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷)) → ran (𝐴 × 𝐵) ⊆ ran (𝐶 × 𝐷)) | 
| 17 | 14, 16 | eqsstrrd 3220 | 
. . . . 5
⊢
((∃𝑥 𝑥 ∈ (𝐴 × 𝐵) ∧ (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷)) → 𝐵 ⊆ ran (𝐶 × 𝐷)) | 
| 18 |   | rnxpss 5101 | 
. . . . 5
⊢ ran
(𝐶 × 𝐷) ⊆ 𝐷 | 
| 19 | 17, 18 | sstrdi 3195 | 
. . . 4
⊢
((∃𝑥 𝑥 ∈ (𝐴 × 𝐵) ∧ (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷)) → 𝐵 ⊆ 𝐷) | 
| 20 | 10, 19 | jca 306 | 
. . 3
⊢
((∃𝑥 𝑥 ∈ (𝐴 × 𝐵) ∧ (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷)) → (𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷)) | 
| 21 | 20 | ex 115 | 
. 2
⊢
(∃𝑥 𝑥 ∈ (𝐴 × 𝐵) → ((𝐴 × 𝐵) ⊆ (𝐶 × 𝐷) → (𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷))) | 
| 22 |   | xpss12 4770 | 
. 2
⊢ ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷) → (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷)) | 
| 23 | 21, 22 | impbid1 142 | 
1
⊢
(∃𝑥 𝑥 ∈ (𝐴 × 𝐵) → ((𝐴 × 𝐵) ⊆ (𝐶 × 𝐷) ↔ (𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷))) |