Theorem dmrnxp 49768

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 488	. . . 4 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → 𝑅 = (𝐴 × 𝐵))
2		nne 2959	. . . . . . 7 ⊢ (¬ 𝐴 ≠ ∅ ↔ 𝐴 = ∅)
3	2	bilani 510	. . . . . 6 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → 𝐴 = ∅)
4	3	xpeq1d 5684	. . . . 5 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → (𝐴 × 𝐵) = (∅ × 𝐵))
5		0xp 5754	. . . . 5 ⊢ (∅ × 𝐵) = ∅
6	4, 5	eqtrdi 2811	. . . 4 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → (𝐴 × 𝐵) = ∅)
7	1, 6	eqtrd 2795	. . 3 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → 𝑅 = ∅)
8	7	dmeqd 5889	. . . . . 6 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → dom 𝑅 = dom ∅)
9		dm0 5904	. . . . . 6 ⊢ dom ∅ = ∅
10	8, 9	eqtrdi 2811	. . . . 5 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → dom 𝑅 = ∅)
11	7	rneqd 5922	. . . . . 6 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → ran 𝑅 = ran ∅)
12		rn0 5910	. . . . . 6 ⊢ ran ∅ = ∅
13	11, 12	eqtrdi 2811	. . . . 5 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → ran 𝑅 = ∅)
14	10, 13	xpeq12d 5686	. . . 4 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → (dom 𝑅 × ran 𝑅) = (∅ × ∅))
15		0xp 5754	. . . 4 ⊢ (∅ × ∅) = ∅
16	14, 15	eqtrdi 2811	. . 3 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → (dom 𝑅 × ran 𝑅) = ∅)
17	7, 16	eqtr4d 2798	. 2 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐴 ≠ ∅) → 𝑅 = (dom 𝑅 × ran 𝑅))
18		simpl 488	. . . 4 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → 𝑅 = (𝐴 × 𝐵))
19		nne 2959	. . . . . . 7 ⊢ (¬ 𝐵 ≠ ∅ ↔ 𝐵 = ∅)
20	19	bilani 510	. . . . . 6 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → 𝐵 = ∅)
21	20	xpeq2d 5685	. . . . 5 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → (𝐴 × 𝐵) = (𝐴 × ∅))
22		xp0 5755	. . . . 5 ⊢ (𝐴 × ∅) = ∅
23	21, 22	eqtrdi 2811	. . . 4 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → (𝐴 × 𝐵) = ∅)
24	18, 23	eqtrd 2795	. . 3 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → 𝑅 = ∅)
25	24	dmeqd 5889	. . . . . 6 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → dom 𝑅 = dom ∅)
26	25, 9	eqtrdi 2811	. . . . 5 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → dom 𝑅 = ∅)
27	24	rneqd 5922	. . . . . 6 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → ran 𝑅 = ran ∅)
28	27, 12	eqtrdi 2811	. . . . 5 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → ran 𝑅 = ∅)
29	26, 28	xpeq12d 5686	. . . 4 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → (dom 𝑅 × ran 𝑅) = (∅ × ∅))
30	29, 15	eqtrdi 2811	. . 3 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → (dom 𝑅 × ran 𝑅) = ∅)
31	24, 30	eqtr4d 2798	. 2 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ ¬ 𝐵 ≠ ∅) → 𝑅 = (dom 𝑅 × ran 𝑅))
32		simpl 488	. . 3 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → 𝑅 = (𝐴 × 𝐵))
33	32	dmeqd 5889	. . . . 5 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → dom 𝑅 = dom (𝐴 × 𝐵))
34		dmxp 5913	. . . . . 6 ⊢ (𝐵 ≠ ∅ → dom (𝐴 × 𝐵) = 𝐴)
35	34	ad2antll 742	. . . . 5 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → dom (𝐴 × 𝐵) = 𝐴)
36	33, 35	eqtrd 2795	. . . 4 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → dom 𝑅 = 𝐴)
37	32	rneqd 5922	. . . . 5 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → ran 𝑅 = ran (𝐴 × 𝐵))
38		rnxp 6163	. . . . . 6 ⊢ (𝐴 ≠ ∅ → ran (𝐴 × 𝐵) = 𝐵)
39	38	ad2antrl 741	. . . . 5 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → ran (𝐴 × 𝐵) = 𝐵)
40	37, 39	eqtrd 2795	. . . 4 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → ran 𝑅 = 𝐵)
41	36, 40	xpeq12d 5686	. . 3 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → (dom 𝑅 × ran 𝑅) = (𝐴 × 𝐵))
42	32, 41	eqtr4d 2798	. 2 ⊢ ((𝑅 = (𝐴 × 𝐵) ∧ (𝐴 ≠ ∅ ∧ 𝐵 ≠ ∅)) → 𝑅 = (dom 𝑅 × ran 𝑅))
43	17, 31, 42	pm2.61dda 827	1 ⊢ (𝑅 = (𝐴 × 𝐵) → 𝑅 = (dom 𝑅 × ran 𝑅))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   = wceq 1570   ≠ wne 2955  ∅c0 4279   × cxp 5653  dom cdm 5655  ran crn 5656

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732  ax-sep 5251  ax-pr 5398

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-ne 2956  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-br 5104  df-opab 5168  df-xp 5661  df-rel 5662  df-cnv 5663  df-dm 5665  df-rn 5666

This theorem is used by: (None)