Theorem intxpd 50146

Description: Intersection of Cartesian products is the Cartesian product of intersection of domains and ranges. See also inxp 5814 and iinxp 50145. (Contributed by Zhi Wang, 30-Oct-2025.)

Hypotheses

Ref	Expression
intxpd.1	⊢ (𝜑 → 𝐴 ≠ ∅)
intxpd.2	⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑥 = (dom 𝑥 × ran 𝑥))
intxpd.3	⊢ 𝑋 = ∩ 𝑥 ∈ 𝐴 dom 𝑥
intxpd.4	⊢ 𝑌 = ∩ 𝑥 ∈ 𝐴 ran 𝑥

Assertion

Ref	Expression
intxpd	⊢ (𝜑 → ∩ 𝐴 = (𝑋 × 𝑌))

Distinct variable groups:   𝑥,𝐴   𝜑,𝑥

Allowed substitution hints:   𝑋(𝑥)   𝑌(𝑥)

Proof of Theorem intxpd

Step	Hyp	Ref	Expression
1		intiin 5021	. . . 4 ⊢ ∩ 𝐴 = ∩ 𝑥 ∈ 𝐴 𝑥
2		intxpd.2	. . . . 5 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑥 = (dom 𝑥 × ran 𝑥))
3	2	iineq2dv 4980	. . . 4 ⊢ (𝜑 → ∩ 𝑥 ∈ 𝐴 𝑥 = ∩ 𝑥 ∈ 𝐴 (dom 𝑥 × ran 𝑥))
4	1, 3	eqtrid 2810	. . 3 ⊢ (𝜑 → ∩ 𝐴 = ∩ 𝑥 ∈ 𝐴 (dom 𝑥 × ran 𝑥))
5		intxpd.1	. . . 4 ⊢ (𝜑 → 𝐴 ≠ ∅)
6		iinxp 50145	. . . 4 ⊢ (𝐴 ≠ ∅ → ∩ 𝑥 ∈ 𝐴 (dom 𝑥 × ran 𝑥) = (∩ 𝑥 ∈ 𝐴 dom 𝑥 × ∩ 𝑥 ∈ 𝐴 ran 𝑥))
7	5, 6	syl 18	. . 3 ⊢ (𝜑 → ∩ 𝑥 ∈ 𝐴 (dom 𝑥 × ran 𝑥) = (∩ 𝑥 ∈ 𝐴 dom 𝑥 × ∩ 𝑥 ∈ 𝐴 ran 𝑥))
8	4, 7	eqtrd 2798	. 2 ⊢ (𝜑 → ∩ 𝐴 = (∩ 𝑥 ∈ 𝐴 dom 𝑥 × ∩ 𝑥 ∈ 𝐴 ran 𝑥))
9		intxpd.3	. . 3 ⊢ 𝑋 = ∩ 𝑥 ∈ 𝐴 dom 𝑥
10		intxpd.4	. . 3 ⊢ 𝑌 = ∩ 𝑥 ∈ 𝐴 ran 𝑥
11	9, 10	xpeq12i 5683	. 2 ⊢ (𝑋 × 𝑌) = (∩ 𝑥 ∈ 𝐴 dom 𝑥 × ∩ 𝑥 ∈ 𝐴 ran 𝑥)
12	8, 11	eqtr4di 2816	1 ⊢ (𝜑 → ∩ 𝐴 = (𝑋 × 𝑌))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570   ∈ wcel 2147   ≠ wne 2959  ∅c0 4281  ∩ cint 4910  ∩ ciin 4955   × 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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735  ax-sep 5252  ax-pr 5395

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 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ne 2960  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-int 4911  df-iin 4957  df-opab 5171  df-xp 5661  df-rel 5662

This theorem is used by: (None)