Proof of Theorem xp11
| Step | Hyp | Ref | Expression | 
|---|
| 1 |  | xpnz 6178 | . . 3
⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅) | 
| 2 |  | anidm 564 | . . . . . 6
⊢ (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐴 × 𝐵) ≠ ∅) ↔ (𝐴 × 𝐵) ≠ ∅) | 
| 3 |  | neeq1 3002 | . . . . . . 7
⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ ↔ (𝐶 × 𝐷) ≠ ∅)) | 
| 4 | 3 | anbi2d 630 | . . . . . 6
⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐴 × 𝐵) ≠ ∅) ↔ ((𝐴 × 𝐵) ≠ ∅ ∧ (𝐶 × 𝐷) ≠ ∅))) | 
| 5 | 2, 4 | bitr3id 285 | . . . . 5
⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ ↔ ((𝐴 × 𝐵) ≠ ∅ ∧ (𝐶 × 𝐷) ≠ ∅))) | 
| 6 |  | eqimss 4041 | . . . . . . . 8
⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 × 𝐵) ⊆ (𝐶 × 𝐷)) | 
| 7 |  | ssxpb 6193 | . . . . . . . 8
⊢ ((𝐴 × 𝐵) ≠ ∅ → ((𝐴 × 𝐵) ⊆ (𝐶 × 𝐷) ↔ (𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷))) | 
| 8 | 6, 7 | syl5ibcom 245 | . . . . . . 7
⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ → (𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷))) | 
| 9 |  | eqimss2 4042 | . . . . . . . 8
⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐶 × 𝐷) ⊆ (𝐴 × 𝐵)) | 
| 10 |  | ssxpb 6193 | . . . . . . . 8
⊢ ((𝐶 × 𝐷) ≠ ∅ → ((𝐶 × 𝐷) ⊆ (𝐴 × 𝐵) ↔ (𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵))) | 
| 11 | 9, 10 | syl5ibcom 245 | . . . . . . 7
⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐶 × 𝐷) ≠ ∅ → (𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵))) | 
| 12 | 8, 11 | anim12d 609 | . . . . . 6
⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐶 × 𝐷) ≠ ∅) → ((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷) ∧ (𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵)))) | 
| 13 |  | an4 656 | . . . . . . 7
⊢ (((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷) ∧ (𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵)) ↔ ((𝐴 ⊆ 𝐶 ∧ 𝐶 ⊆ 𝐴) ∧ (𝐵 ⊆ 𝐷 ∧ 𝐷 ⊆ 𝐵))) | 
| 14 |  | eqss 3998 | . . . . . . . 8
⊢ (𝐴 = 𝐶 ↔ (𝐴 ⊆ 𝐶 ∧ 𝐶 ⊆ 𝐴)) | 
| 15 |  | eqss 3998 | . . . . . . . 8
⊢ (𝐵 = 𝐷 ↔ (𝐵 ⊆ 𝐷 ∧ 𝐷 ⊆ 𝐵)) | 
| 16 | 14, 15 | anbi12i 628 | . . . . . . 7
⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ↔ ((𝐴 ⊆ 𝐶 ∧ 𝐶 ⊆ 𝐴) ∧ (𝐵 ⊆ 𝐷 ∧ 𝐷 ⊆ 𝐵))) | 
| 17 | 13, 16 | bitr4i 278 | . . . . . 6
⊢ (((𝐴 ⊆ 𝐶 ∧ 𝐵 ⊆ 𝐷) ∧ (𝐶 ⊆ 𝐴 ∧ 𝐷 ⊆ 𝐵)) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)) | 
| 18 | 12, 17 | imbitrdi 251 | . . . . 5
⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (((𝐴 × 𝐵) ≠ ∅ ∧ (𝐶 × 𝐷) ≠ ∅) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))) | 
| 19 | 5, 18 | sylbid 240 | . . . 4
⊢ ((𝐴 × 𝐵) = (𝐶 × 𝐷) → ((𝐴 × 𝐵) ≠ ∅ → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))) | 
| 20 | 19 | com12 32 | . . 3
⊢ ((𝐴 × 𝐵) ≠ ∅ → ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))) | 
| 21 | 1, 20 | sylbi 217 | . 2
⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))) | 
| 22 |  | xpeq12 5709 | . 2
⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → (𝐴 × 𝐵) = (𝐶 × 𝐷)) | 
| 23 | 21, 22 | impbid1 225 | 1
⊢ ((𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅) → ((𝐴 × 𝐵) = (𝐶 × 𝐷) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))) |