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Theorem dmrnxp 48818
Description: A Cartesian product is the Cartesian product of its domain and range. (Contributed by Zhi Wang, 30-Oct-2025.)
Assertion
Ref Expression
dmrnxp (𝑅 = (𝐴 × 𝐵) → 𝑅 = (dom 𝑅 × ran 𝑅))

Proof of Theorem dmrnxp
StepHypRef Expression
1 simpl 482 . . . 4 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → 𝑅 = (𝐴 × 𝐵))
2 simpr 484 . . . . . . 7 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → ¬ 𝐴 ≠ ∅)
3 nne 2929 . . . . . . 7 𝐴 ≠ ∅ ↔ 𝐴 = ∅)
42, 3sylib 218 . . . . . 6 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → 𝐴 = ∅)
54xpeq1d 5660 . . . . 5 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → (𝐴 × 𝐵) = (∅ × 𝐵))
6 0xp 5729 . . . . 5 (∅ × 𝐵) = ∅
75, 6eqtrdi 2780 . . . 4 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → (𝐴 × 𝐵) = ∅)
81, 7eqtrd 2764 . . 3 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → 𝑅 = ∅)
98dmeqd 5859 . . . . . 6 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → dom 𝑅 = dom ∅)
10 dm0 5874 . . . . . 6 dom ∅ = ∅
119, 10eqtrdi 2780 . . . . 5 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → dom 𝑅 = ∅)
128rneqd 5891 . . . . . 6 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → ran 𝑅 = ran ∅)
13 rn0 5879 . . . . . 6 ran ∅ = ∅
1412, 13eqtrdi 2780 . . . . 5 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → ran 𝑅 = ∅)
1511, 14xpeq12d 5662 . . . 4 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → (dom 𝑅 × ran 𝑅) = (∅ × ∅))
16 0xp 5729 . . . 4 (∅ × ∅) = ∅
1715, 16eqtrdi 2780 . . 3 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → (dom 𝑅 × ran 𝑅) = ∅)
188, 17eqtr4d 2767 . 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 2929 . . . . . . 7 𝐵 ≠ ∅ ↔ 𝐵 = ∅)
2220, 21sylib 218 . . . . . 6 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → 𝐵 = ∅)
2322xpeq2d 5661 . . . . 5 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → (𝐴 × 𝐵) = (𝐴 × ∅))
24 xp0 6119 . . . . 5 (𝐴 × ∅) = ∅
2523, 24eqtrdi 2780 . . . 4 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → (𝐴 × 𝐵) = ∅)
2619, 25eqtrd 2764 . . 3 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → 𝑅 = ∅)
2726dmeqd 5859 . . . . . 6 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → dom 𝑅 = dom ∅)
2827, 10eqtrdi 2780 . . . . 5 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → dom 𝑅 = ∅)
2926rneqd 5891 . . . . . 6 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → ran 𝑅 = ran ∅)
3029, 13eqtrdi 2780 . . . . 5 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → ran 𝑅 = ∅)
3128, 30xpeq12d 5662 . . . 4 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → (dom 𝑅 × ran 𝑅) = (∅ × ∅))
3231, 16eqtrdi 2780 . . 3 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → (dom 𝑅 × ran 𝑅) = ∅)
3326, 32eqtr4d 2767 . 2 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → 𝑅 = (dom 𝑅 × ran 𝑅))
34 simpl 482 . . 3 ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → 𝑅 = (𝐴 × 𝐵))
3534dmeqd 5859 . . . . 5 ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → dom 𝑅 = dom (𝐴 × 𝐵))
36 dmxp 5882 . . . . . 6 (𝐵 ≠ ∅ → dom (𝐴 × 𝐵) = 𝐴)
3736ad2antll 729 . . . . 5 ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → dom (𝐴 × 𝐵) = 𝐴)
3835, 37eqtrd 2764 . . . 4 ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → dom 𝑅 = 𝐴)
3934rneqd 5891 . . . . 5 ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → ran 𝑅 = ran (𝐴 × 𝐵))
40 rnxp 6131 . . . . . 6 (𝐴 ≠ ∅ → ran (𝐴 × 𝐵) = 𝐵)
4140ad2antrl 728 . . . . 5 ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → ran (𝐴 × 𝐵) = 𝐵)
4239, 41eqtrd 2764 . . . 4 ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → ran 𝑅 = 𝐵)
4338, 42xpeq12d 5662 . . 3 ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → (dom 𝑅 × ran 𝑅) = (𝐴 × 𝐵))
4434, 43eqtr4d 2767 . 2 ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → 𝑅 = (dom 𝑅 × ran 𝑅))
4518, 33, 44pm2.61dda 814 1 (𝑅 = (𝐴 × 𝐵) → 𝑅 = (dom 𝑅 × ran 𝑅))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 395   = wceq 1540  wne 2925  c0 4292   × cxp 5629  dom cdm 5631  ran crn 5632
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-10 2142  ax-11 2158  ax-12 2178  ax-ext 2701  ax-sep 5246  ax-nul 5256  ax-pr 5382
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2066  df-clab 2708  df-cleq 2721  df-clel 2803  df-ne 2926  df-ral 3045  df-rex 3054  df-rab 3403  df-v 3446  df-dif 3914  df-un 3916  df-ss 3928  df-nul 4293  df-if 4485  df-sn 4586  df-pr 4588  df-op 4592  df-br 5103  df-opab 5165  df-xp 5637  df-rel 5638  df-cnv 5639  df-dm 5641  df-rn 5642
This theorem is referenced by: (None)
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