Proof of Theorem xpcan
| Step | Hyp | Ref
| Expression |
| 1 | | xp11 6169 |
. . 3
⊢ ((𝐶 ≠ ∅ ∧ 𝐴 ≠ ∅) → ((𝐶 × 𝐴) = (𝐶 × 𝐵) ↔ (𝐶 = 𝐶 ∧ 𝐴 = 𝐵))) |
| 2 | | eqid 2736 |
. . . 4
⊢ 𝐶 = 𝐶 |
| 3 | 2 | biantrur 530 |
. . 3
⊢ (𝐴 = 𝐵 ↔ (𝐶 = 𝐶 ∧ 𝐴 = 𝐵)) |
| 4 | 1, 3 | bitr4di 289 |
. 2
⊢ ((𝐶 ≠ ∅ ∧ 𝐴 ≠ ∅) → ((𝐶 × 𝐴) = (𝐶 × 𝐵) ↔ 𝐴 = 𝐵)) |
| 5 | | nne 2937 |
. . . 4
⊢ (¬
𝐴 ≠ ∅ ↔ 𝐴 = ∅) |
| 6 | | simpr 484 |
. . . . 5
⊢ ((𝐶 ≠ ∅ ∧ 𝐴 = ∅) → 𝐴 = ∅) |
| 7 | | xpeq2 5680 |
. . . . . . . . . 10
⊢ (𝐴 = ∅ → (𝐶 × 𝐴) = (𝐶 × ∅)) |
| 8 | | xp0 6152 |
. . . . . . . . . 10
⊢ (𝐶 × ∅) =
∅ |
| 9 | 7, 8 | eqtrdi 2787 |
. . . . . . . . 9
⊢ (𝐴 = ∅ → (𝐶 × 𝐴) = ∅) |
| 10 | 9 | eqeq1d 2738 |
. . . . . . . 8
⊢ (𝐴 = ∅ → ((𝐶 × 𝐴) = (𝐶 × 𝐵) ↔ ∅ = (𝐶 × 𝐵))) |
| 11 | | eqcom 2743 |
. . . . . . . 8
⊢ (∅
= (𝐶 × 𝐵) ↔ (𝐶 × 𝐵) = ∅) |
| 12 | 10, 11 | bitrdi 287 |
. . . . . . 7
⊢ (𝐴 = ∅ → ((𝐶 × 𝐴) = (𝐶 × 𝐵) ↔ (𝐶 × 𝐵) = ∅)) |
| 13 | 12 | adantl 481 |
. . . . . 6
⊢ ((𝐶 ≠ ∅ ∧ 𝐴 = ∅) → ((𝐶 × 𝐴) = (𝐶 × 𝐵) ↔ (𝐶 × 𝐵) = ∅)) |
| 14 | | df-ne 2934 |
. . . . . . . 8
⊢ (𝐶 ≠ ∅ ↔ ¬ 𝐶 = ∅) |
| 15 | | xpeq0 6154 |
. . . . . . . . 9
⊢ ((𝐶 × 𝐵) = ∅ ↔ (𝐶 = ∅ ∨ 𝐵 = ∅)) |
| 16 | | orel1 888 |
. . . . . . . . 9
⊢ (¬
𝐶 = ∅ → ((𝐶 = ∅ ∨ 𝐵 = ∅) → 𝐵 = ∅)) |
| 17 | 15, 16 | biimtrid 242 |
. . . . . . . 8
⊢ (¬
𝐶 = ∅ → ((𝐶 × 𝐵) = ∅ → 𝐵 = ∅)) |
| 18 | 14, 17 | sylbi 217 |
. . . . . . 7
⊢ (𝐶 ≠ ∅ → ((𝐶 × 𝐵) = ∅ → 𝐵 = ∅)) |
| 19 | 18 | adantr 480 |
. . . . . 6
⊢ ((𝐶 ≠ ∅ ∧ 𝐴 = ∅) → ((𝐶 × 𝐵) = ∅ → 𝐵 = ∅)) |
| 20 | 13, 19 | sylbid 240 |
. . . . 5
⊢ ((𝐶 ≠ ∅ ∧ 𝐴 = ∅) → ((𝐶 × 𝐴) = (𝐶 × 𝐵) → 𝐵 = ∅)) |
| 21 | | eqtr3 2758 |
. . . . 5
⊢ ((𝐴 = ∅ ∧ 𝐵 = ∅) → 𝐴 = 𝐵) |
| 22 | 6, 20, 21 | syl6an 684 |
. . . 4
⊢ ((𝐶 ≠ ∅ ∧ 𝐴 = ∅) → ((𝐶 × 𝐴) = (𝐶 × 𝐵) → 𝐴 = 𝐵)) |
| 23 | 5, 22 | sylan2b 594 |
. . 3
⊢ ((𝐶 ≠ ∅ ∧ ¬ 𝐴 ≠ ∅) → ((𝐶 × 𝐴) = (𝐶 × 𝐵) → 𝐴 = 𝐵)) |
| 24 | | xpeq2 5680 |
. . 3
⊢ (𝐴 = 𝐵 → (𝐶 × 𝐴) = (𝐶 × 𝐵)) |
| 25 | 23, 24 | impbid1 225 |
. 2
⊢ ((𝐶 ≠ ∅ ∧ ¬ 𝐴 ≠ ∅) → ((𝐶 × 𝐴) = (𝐶 × 𝐵) ↔ 𝐴 = 𝐵)) |
| 26 | 4, 25 | pm2.61dan 812 |
1
⊢ (𝐶 ≠ ∅ → ((𝐶 × 𝐴) = (𝐶 × 𝐵) ↔ 𝐴 = 𝐵)) |