Proof of Theorem xppss12
| Step | Hyp | Ref
| Expression |
| 1 | | pssss 4049 |
. . 3
⊢ (𝐴 ⊊ 𝐵 → 𝐴 ⊆ 𝐵) |
| 2 | | pssss 4049 |
. . 3
⊢ (𝐶 ⊊ 𝐷 → 𝐶 ⊆ 𝐷) |
| 3 | | xpss12 5674 |
. . 3
⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐶 ⊆ 𝐷) → (𝐴 × 𝐶) ⊆ (𝐵 × 𝐷)) |
| 4 | 1, 2, 3 | syl2an 608 |
. 2
⊢ ((𝐴 ⊊ 𝐵 ∧ 𝐶 ⊊ 𝐷) → (𝐴 × 𝐶) ⊆ (𝐵 × 𝐷)) |
| 5 | | simpl 488 |
. . . . 5
⊢ ((𝐴 ⊊ 𝐵 ∧ 𝐶 ⊊ 𝐷) → 𝐴 ⊊ 𝐵) |
| 6 | | pssne 4050 |
. . . . . 6
⊢ (𝐴 ⊊ 𝐵 → 𝐴 ≠ 𝐵) |
| 7 | 6 | necomd 3012 |
. . . . 5
⊢ (𝐴 ⊊ 𝐵 → 𝐵 ≠ 𝐴) |
| 8 | | neneq 2963 |
. . . . . 6
⊢ (𝐵 ≠ 𝐴 → ¬ 𝐵 = 𝐴) |
| 9 | 8 | intnanrd 495 |
. . . . 5
⊢ (𝐵 ≠ 𝐴 → ¬ (𝐵 = 𝐴 ∧ 𝐷 = 𝐶)) |
| 10 | 5, 7, 9 | 3syl 19 |
. . . 4
⊢ ((𝐴 ⊊ 𝐵 ∧ 𝐶 ⊊ 𝐷) → ¬ (𝐵 = 𝐴 ∧ 𝐷 = 𝐶)) |
| 11 | | pssn0 43105 |
. . . . 5
⊢ (𝐴 ⊊ 𝐵 → 𝐵 ≠ ∅) |
| 12 | | pssn0 43105 |
. . . . 5
⊢ (𝐶 ⊊ 𝐷 → 𝐷 ≠ ∅) |
| 13 | | xp11 6172 |
. . . . 5
⊢ ((𝐵 ≠ ∅ ∧ 𝐷 ≠ ∅) → ((𝐵 × 𝐷) = (𝐴 × 𝐶) ↔ (𝐵 = 𝐴 ∧ 𝐷 = 𝐶))) |
| 14 | 11, 12, 13 | syl2an 608 |
. . . 4
⊢ ((𝐴 ⊊ 𝐵 ∧ 𝐶 ⊊ 𝐷) → ((𝐵 × 𝐷) = (𝐴 × 𝐶) ↔ (𝐵 = 𝐴 ∧ 𝐷 = 𝐶))) |
| 15 | 10, 14 | mtbird 328 |
. . 3
⊢ ((𝐴 ⊊ 𝐵 ∧ 𝐶 ⊊ 𝐷) → ¬ (𝐵 × 𝐷) = (𝐴 × 𝐶)) |
| 16 | | neqne 2965 |
. . . 4
⊢ (¬
(𝐵 × 𝐷) = (𝐴 × 𝐶) → (𝐵 × 𝐷) ≠ (𝐴 × 𝐶)) |
| 17 | 16 | necomd 3012 |
. . 3
⊢ (¬
(𝐵 × 𝐷) = (𝐴 × 𝐶) → (𝐴 × 𝐶) ≠ (𝐵 × 𝐷)) |
| 18 | 15, 17 | syl 18 |
. 2
⊢ ((𝐴 ⊊ 𝐵 ∧ 𝐶 ⊊ 𝐷) → (𝐴 × 𝐶) ≠ (𝐵 × 𝐷)) |
| 19 | | df-pss 3922 |
. 2
⊢ ((𝐴 × 𝐶) ⊊ (𝐵 × 𝐷) ↔ ((𝐴 × 𝐶) ⊆ (𝐵 × 𝐷) ∧ (𝐴 × 𝐶) ≠ (𝐵 × 𝐷))) |
| 20 | 4, 18, 19 | sylanbrc 595 |
1
⊢ ((𝐴 ⊊ 𝐵 ∧ 𝐶 ⊊ 𝐷) → (𝐴 × 𝐶) ⊊ (𝐵 × 𝐷)) |