Proof of Theorem dmrnxp
| Step | Hyp | Ref
| Expression |
| 1 | | simpl 482 |
. . . 4
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → 𝑅 = (𝐴 × 𝐵)) |
| 2 | | simpr 484 |
. . . . . . 7
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → ¬ 𝐴 ≠ ∅) |
| 3 | | nne 2935 |
. . . . . . 7
⊢ (¬
𝐴 ≠ ∅ ↔ 𝐴 = ∅) |
| 4 | 2, 3 | sylib 218 |
. . . . . 6
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → 𝐴 = ∅) |
| 5 | 4 | xpeq1d 5680 |
. . . . 5
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → (𝐴 × 𝐵) = (∅ × 𝐵)) |
| 6 | | 0xp 5750 |
. . . . 5
⊢ (∅
× 𝐵) =
∅ |
| 7 | 5, 6 | eqtrdi 2785 |
. . . 4
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → (𝐴 × 𝐵) = ∅) |
| 8 | 1, 7 | eqtrd 2769 |
. . 3
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → 𝑅 = ∅) |
| 9 | 8 | dmeqd 5882 |
. . . . . 6
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → dom 𝑅 = dom ∅) |
| 10 | | dm0 5897 |
. . . . . 6
⊢ dom
∅ = ∅ |
| 11 | 9, 10 | eqtrdi 2785 |
. . . . 5
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → dom 𝑅 = ∅) |
| 12 | 8 | rneqd 5915 |
. . . . . 6
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → ran 𝑅 = ran ∅) |
| 13 | | rn0 5902 |
. . . . . 6
⊢ ran
∅ = ∅ |
| 14 | 12, 13 | eqtrdi 2785 |
. . . . 5
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → ran 𝑅 = ∅) |
| 15 | 11, 14 | xpeq12d 5682 |
. . . 4
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → (dom 𝑅 × ran 𝑅) = (∅ ×
∅)) |
| 16 | | 0xp 5750 |
. . . 4
⊢ (∅
× ∅) = ∅ |
| 17 | 15, 16 | eqtrdi 2785 |
. . 3
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → (dom 𝑅 × ran 𝑅) = ∅) |
| 18 | 8, 17 | eqtr4d 2772 |
. 2
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → 𝑅 = (dom 𝑅 × ran 𝑅)) |
| 19 | | simpl 482 |
. . . 4
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → 𝑅 = (𝐴 × 𝐵)) |
| 20 | | simpr 484 |
. . . . . . 7
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → ¬ 𝐵 ≠ ∅) |
| 21 | | nne 2935 |
. . . . . . 7
⊢ (¬
𝐵 ≠ ∅ ↔ 𝐵 = ∅) |
| 22 | 20, 21 | sylib 218 |
. . . . . 6
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → 𝐵 = ∅) |
| 23 | 22 | xpeq2d 5681 |
. . . . 5
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → (𝐴 × 𝐵) = (𝐴 × ∅)) |
| 24 | | xp0 6144 |
. . . . 5
⊢ (𝐴 × ∅) =
∅ |
| 25 | 23, 24 | eqtrdi 2785 |
. . . 4
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → (𝐴 × 𝐵) = ∅) |
| 26 | 19, 25 | eqtrd 2769 |
. . 3
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → 𝑅 = ∅) |
| 27 | 26 | dmeqd 5882 |
. . . . . 6
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → dom 𝑅 = dom ∅) |
| 28 | 27, 10 | eqtrdi 2785 |
. . . . 5
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → dom 𝑅 = ∅) |
| 29 | 26 | rneqd 5915 |
. . . . . 6
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → ran 𝑅 = ran ∅) |
| 30 | 29, 13 | eqtrdi 2785 |
. . . . 5
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → ran 𝑅 = ∅) |
| 31 | 28, 30 | xpeq12d 5682 |
. . . 4
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → (dom 𝑅 × ran 𝑅) = (∅ ×
∅)) |
| 32 | 31, 16 | eqtrdi 2785 |
. . 3
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → (dom 𝑅 × ran 𝑅) = ∅) |
| 33 | 26, 32 | eqtr4d 2772 |
. 2
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → 𝑅 = (dom 𝑅 × ran 𝑅)) |
| 34 | | simpl 482 |
. . 3
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → 𝑅 = (𝐴 × 𝐵)) |
| 35 | 34 | dmeqd 5882 |
. . . . 5
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → dom 𝑅 = dom (𝐴 × 𝐵)) |
| 36 | | dmxp 5905 |
. . . . . 6
⊢ (𝐵 ≠ ∅ → dom (𝐴 × 𝐵) = 𝐴) |
| 37 | 36 | ad2antll 729 |
. . . . 5
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → dom (𝐴 × 𝐵) = 𝐴) |
| 38 | 35, 37 | eqtrd 2769 |
. . . 4
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → dom 𝑅 = 𝐴) |
| 39 | 34 | rneqd 5915 |
. . . . 5
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → ran 𝑅 = ran (𝐴 × 𝐵)) |
| 40 | | rnxp 6156 |
. . . . . 6
⊢ (𝐴 ≠ ∅ → ran (𝐴 × 𝐵) = 𝐵) |
| 41 | 40 | ad2antrl 728 |
. . . . 5
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → ran (𝐴 × 𝐵) = 𝐵) |
| 42 | 39, 41 | eqtrd 2769 |
. . . 4
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → ran 𝑅 = 𝐵) |
| 43 | 38, 42 | xpeq12d 5682 |
. . 3
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → (dom 𝑅 × ran 𝑅) = (𝐴 × 𝐵)) |
| 44 | 34, 43 | eqtr4d 2772 |
. 2
⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → 𝑅 = (dom 𝑅 × ran 𝑅)) |
| 45 | 18, 33, 44 | pm2.61dda 814 |
1
⊢ (𝑅 = (𝐴 × 𝐵) → 𝑅 = (dom 𝑅 × ran 𝑅)) |