Theorem inxpss 38851

Description: Two ways to say that an intersection with a Cartesian product is a subclass. (Contributed by Peter Mazsa, 16-Jul-2019.)

Assertion

Ref	Expression
inxpss	⊢ ((𝑅 ∩ (𝐴 × 𝐵)) ⊆ 𝑆 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝑥𝑅𝑦 → 𝑥𝑆𝑦))

Distinct variable groups:   𝑥,𝐴,𝑦   𝑥,𝐵,𝑦   𝑥,𝑅,𝑦   𝑥,𝑆,𝑦

Proof of Theorem inxpss

Step	Hyp	Ref	Expression
1		brinxp2 5737	. . . . 5 ⊢ (𝑥(𝑅 ∩ (𝐴 × 𝐵))𝑦 ↔ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑥𝑅𝑦))
2	1	imbi1i 352	. . . 4 ⊢ ((𝑥(𝑅 ∩ (𝐴 × 𝐵))𝑦 → 𝑥𝑆𝑦) ↔ (((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑥𝑅𝑦) → 𝑥𝑆𝑦))
3		impexp 455	. . . 4 ⊢ ((((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑥𝑅𝑦) → 𝑥𝑆𝑦) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → (𝑥𝑅𝑦 → 𝑥𝑆𝑦)))
4	2, 3	bitri 278	. . 3 ⊢ ((𝑥(𝑅 ∩ (𝐴 × 𝐵))𝑦 → 𝑥𝑆𝑦) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → (𝑥𝑅𝑦 → 𝑥𝑆𝑦)))
5	4	2albii 1847	. 2 ⊢ (∀𝑥∀𝑦(𝑥(𝑅 ∩ (𝐴 × 𝐵))𝑦 → 𝑥𝑆𝑦) ↔ ∀𝑥∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → (𝑥𝑅𝑦 → 𝑥𝑆𝑦)))
6		relinxp 5799	. . 3 ⊢ Rel (𝑅 ∩ (𝐴 × 𝐵))
7		ssrel3 5770	. . 3 ⊢ (Rel (𝑅 ∩ (𝐴 × 𝐵)) → ((𝑅 ∩ (𝐴 × 𝐵)) ⊆ 𝑆 ↔ ∀𝑥∀𝑦(𝑥(𝑅 ∩ (𝐴 × 𝐵))𝑦 → 𝑥𝑆𝑦)))
8	6, 7	ax-mp 5	. 2 ⊢ ((𝑅 ∩ (𝐴 × 𝐵)) ⊆ 𝑆 ↔ ∀𝑥∀𝑦(𝑥(𝑅 ∩ (𝐴 × 𝐵))𝑦 → 𝑥𝑆𝑦))
9		r2al 3207	. 2 ⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝑥𝑅𝑦 → 𝑥𝑆𝑦) ↔ ∀𝑥∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → (𝑥𝑅𝑦 → 𝑥𝑆𝑦)))
10	5, 8, 9	3bitr4i 306	1 ⊢ ((𝑅 ∩ (𝐴 × 𝐵)) ⊆ 𝑆 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝑥𝑅𝑦 → 𝑥𝑆𝑦))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 400  ∀wal 1565   ∈ wcel 2149  ∀wral 3085   ∩ cin 3912   ⊆ wss 3913   class class class wbr 5110   × cxp 5657  Rel wrel 5664

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-ext 2741  ax-sep 5258  ax-pr 5402

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-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 4490  df-sn 4592  df-pr 4594  df-op 4598  df-br 5111  df-opab 5175  df-xp 5665  df-rel 5666

This theorem is referenced by:  idinxpss  38852