Theorem dmrnxp 49500

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

Step	Hyp	Ref	Expression
1		simpl 487	. . . 4 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → 𝑅 = (𝐴 × 𝐵))
2		nne 2968	. . . . . . 7 ⊢ (¬ 𝐴 ≠ ∅ ↔ 𝐴 = ∅)
3	2	bilani 509	. . . . . 6 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → 𝐴 = ∅)
4	3	xpeq1d 5691	. . . . 5 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → (𝐴 × 𝐵) = (∅ × 𝐵))
5		0xp 5761	. . . . 5 ⊢ (∅ × 𝐵) = ∅
6	4, 5	eqtrdi 2820	. . . 4 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → (𝐴 × 𝐵) = ∅)
7	1, 6	eqtrd 2804	. . 3 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → 𝑅 = ∅)
8	7	dmeqd 5896	. . . . . 6 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → dom 𝑅 = dom ∅)
9		dm0 5911	. . . . . 6 ⊢ dom ∅ = ∅
10	8, 9	eqtrdi 2820	. . . . 5 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → dom 𝑅 = ∅)
11	7	rneqd 5929	. . . . . 6 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → ran 𝑅 = ran ∅)
12		rn0 5917	. . . . . 6 ⊢ ran ∅ = ∅
13	11, 12	eqtrdi 2820	. . . . 5 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → ran 𝑅 = ∅)
14	10, 13	xpeq12d 5693	. . . 4 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → (dom 𝑅 × ran 𝑅) = (∅ × ∅))
15		0xp 5761	. . . 4 ⊢ (∅ × ∅) = ∅
16	14, 15	eqtrdi 2820	. . 3 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → (dom 𝑅 × ran 𝑅) = ∅)
17	7, 16	eqtr4d 2807	. 2 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → 𝑅 = (dom 𝑅 × ran 𝑅))
18		simpl 487	. . . 4 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → 𝑅 = (𝐴 × 𝐵))
19		nne 2968	. . . . . . 7 ⊢ (¬ 𝐵 ≠ ∅ ↔ 𝐵 = ∅)
20	19	bilani 509	. . . . . 6 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → 𝐵 = ∅)
21	20	xpeq2d 5692	. . . . 5 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → (𝐴 × 𝐵) = (𝐴 × ∅))
22		xp0 5762	. . . . 5 ⊢ (𝐴 × ∅) = ∅
23	21, 22	eqtrdi 2820	. . . 4 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → (𝐴 × 𝐵) = ∅)
24	18, 23	eqtrd 2804	. . 3 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → 𝑅 = ∅)
25	24	dmeqd 5896	. . . . . 6 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → dom 𝑅 = dom ∅)
26	25, 9	eqtrdi 2820	. . . . 5 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → dom 𝑅 = ∅)
27	24	rneqd 5929	. . . . . 6 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → ran 𝑅 = ran ∅)
28	27, 12	eqtrdi 2820	. . . . 5 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → ran 𝑅 = ∅)
29	26, 28	xpeq12d 5693	. . . 4 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → (dom 𝑅 × ran 𝑅) = (∅ × ∅))
30	29, 15	eqtrdi 2820	. . 3 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → (dom 𝑅 × ran 𝑅) = ∅)
31	24, 30	eqtr4d 2807	. 2 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → 𝑅 = (dom 𝑅 × ran 𝑅))
32		simpl 487	. . 3 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → 𝑅 = (𝐴 × 𝐵))
33	32	dmeqd 5896	. . . . 5 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → dom 𝑅 = dom (𝐴 × 𝐵))
34		dmxp 5920	. . . . . 6 ⊢ (𝐵 ≠ ∅ → dom (𝐴 × 𝐵) = 𝐴)
35	34	ad2antll 741	. . . . 5 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → dom (𝐴 × 𝐵) = 𝐴)
36	33, 35	eqtrd 2804	. . . 4 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → dom 𝑅 = 𝐴)
37	32	rneqd 5929	. . . . 5 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → ran 𝑅 = ran (𝐴 × 𝐵))
38		rnxp 6169	. . . . . 6 ⊢ (𝐴 ≠ ∅ → ran (𝐴 × 𝐵) = 𝐵)
39	38	ad2antrl 740	. . . . 5 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → ran (𝐴 × 𝐵) = 𝐵)
40	37, 39	eqtrd 2804	. . . 4 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → ran 𝑅 = 𝐵)
41	36, 40	xpeq12d 5693	. . 3 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → (dom 𝑅 × ran 𝑅) = (𝐴 × 𝐵))
42	32, 41	eqtr4d 2807	. 2 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → 𝑅 = (dom 𝑅 × ran 𝑅))
43	17, 31, 42	pm2.61dda 826	1 ⊢ (𝑅 = (𝐴 × 𝐵) → 𝑅 = (dom 𝑅 × ran 𝑅))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 400   = wceq 1567   ≠ wne 2964  ∅c0 4294   × cxp 5660  dom cdm 5662  ran crn 5663

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-11 2198  ax-ext 2741  ax-sep 5261  ax-pr 5405

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1103  df-tru 1570  df-fal 1580  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-ne 2965  df-ral 3086  df-rex 3096  df-rab 3424  df-v 3465  df-dif 3916  df-un 3918  df-in 3920  df-ss 3930  df-nul 4295  df-if 4493  df-sn 4595  df-pr 4597  df-op 4601  df-br 5114  df-opab 5178  df-xp 5668  df-rel 5669  df-cnv 5670  df-dm 5672  df-rn 5673

This theorem is referenced by: (None)