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Theorem dmrnxp 48695
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 2935 . . . . . . 7 𝐴 ≠ ∅ ↔ 𝐴 = ∅)
42, 3sylib 218 . . . . . 6 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → 𝐴 = ∅)
54xpeq1d 5680 . . . . 5 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → (𝐴 × 𝐵) = (∅ × 𝐵))
6 0xp 5750 . . . . 5 (∅ × 𝐵) = ∅
75, 6eqtrdi 2785 . . . 4 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → (𝐴 × 𝐵) = ∅)
81, 7eqtrd 2769 . . 3 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → 𝑅 = ∅)
98dmeqd 5882 . . . . . 6 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → dom 𝑅 = dom ∅)
10 dm0 5897 . . . . . 6 dom ∅ = ∅
119, 10eqtrdi 2785 . . . . 5 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → dom 𝑅 = ∅)
128rneqd 5915 . . . . . 6 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → ran 𝑅 = ran ∅)
13 rn0 5902 . . . . . 6 ran ∅ = ∅
1412, 13eqtrdi 2785 . . . . 5 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → ran 𝑅 = ∅)
1511, 14xpeq12d 5682 . . . 4 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → (dom 𝑅 × ran 𝑅) = (∅ × ∅))
16 0xp 5750 . . . 4 (∅ × ∅) = ∅
1715, 16eqtrdi 2785 . . 3 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → (dom 𝑅 × ran 𝑅) = ∅)
188, 17eqtr4d 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 = ∅)
2220, 21sylib 218 . . . . . 6 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → 𝐵 = ∅)
2322xpeq2d 5681 . . . . 5 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → (𝐴 × 𝐵) = (𝐴 × ∅))
24 xp0 6144 . . . . 5 (𝐴 × ∅) = ∅
2523, 24eqtrdi 2785 . . . 4 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → (𝐴 × 𝐵) = ∅)
2619, 25eqtrd 2769 . . 3 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → 𝑅 = ∅)
2726dmeqd 5882 . . . . . 6 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → dom 𝑅 = dom ∅)
2827, 10eqtrdi 2785 . . . . 5 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → dom 𝑅 = ∅)
2926rneqd 5915 . . . . . 6 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → ran 𝑅 = ran ∅)
3029, 13eqtrdi 2785 . . . . 5 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → ran 𝑅 = ∅)
3128, 30xpeq12d 5682 . . . 4 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → (dom 𝑅 × ran 𝑅) = (∅ × ∅))
3231, 16eqtrdi 2785 . . 3 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → (dom 𝑅 × ran 𝑅) = ∅)
3326, 32eqtr4d 2772 . 2 ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → 𝑅 = (dom 𝑅 × ran 𝑅))
34 simpl 482 . . 3 ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → 𝑅 = (𝐴 × 𝐵))
3534dmeqd 5882 . . . . 5 ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → dom 𝑅 = dom (𝐴 × 𝐵))
36 dmxp 5905 . . . . . 6 (𝐵 ≠ ∅ → dom (𝐴 × 𝐵) = 𝐴)
3736ad2antll 729 . . . . 5 ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → dom (𝐴 × 𝐵) = 𝐴)
3835, 37eqtrd 2769 . . . 4 ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → dom 𝑅 = 𝐴)
3934rneqd 5915 . . . . 5 ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → ran 𝑅 = ran (𝐴 × 𝐵))
40 rnxp 6156 . . . . . 6 (𝐴 ≠ ∅ → ran (𝐴 × 𝐵) = 𝐵)
4140ad2antrl 728 . . . . 5 ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → ran (𝐴 × 𝐵) = 𝐵)
4239, 41eqtrd 2769 . . . 4 ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → ran 𝑅 = 𝐵)
4338, 42xpeq12d 5682 . . 3 ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → (dom 𝑅 × ran 𝑅) = (𝐴 × 𝐵))
4434, 43eqtr4d 2772 . 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 1539  wne 2931  c0 4306   × cxp 5649  dom cdm 5651  ran crn 5652
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-8 2109  ax-9 2117  ax-10 2140  ax-11 2156  ax-12 2176  ax-ext 2706  ax-sep 5263  ax-nul 5273  ax-pr 5399
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1542  df-fal 1552  df-ex 1779  df-nf 1783  df-sb 2064  df-clab 2713  df-cleq 2726  df-clel 2808  df-ne 2932  df-ral 3051  df-rex 3060  df-rab 3414  df-v 3459  df-dif 3927  df-un 3929  df-ss 3941  df-nul 4307  df-if 4499  df-sn 4600  df-pr 4602  df-op 4606  df-br 5117  df-opab 5179  df-xp 5657  df-rel 5658  df-cnv 5659  df-dm 5661  df-rn 5662
This theorem is referenced by: (None)
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