Theorem inxpss 38568

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 5710	. . . . 5 ⊢ (𝑥(𝑅 ∩ (𝐴 × 𝐵))𝑦 ↔ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑥𝑅𝑦))
2	1	imbi1i 349	. . . 4 ⊢ ((𝑥(𝑅 ∩ (𝐴 × 𝐵))𝑦 → 𝑥𝑆𝑦) ↔ (((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑥𝑅𝑦) → 𝑥𝑆𝑦))
3		impexp 450	. . . 4 ⊢ ((((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑥𝑅𝑦) → 𝑥𝑆𝑦) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → (𝑥𝑅𝑦 → 𝑥𝑆𝑦)))
4	2, 3	bitri 275	. . 3 ⊢ ((𝑥(𝑅 ∩ (𝐴 × 𝐵))𝑦 → 𝑥𝑆𝑦) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → (𝑥𝑅𝑦 → 𝑥𝑆𝑦)))
5	4	2albii 1822	. 2 ⊢ (∀𝑥∀𝑦(𝑥(𝑅 ∩ (𝐴 × 𝐵))𝑦 → 𝑥𝑆𝑦) ↔ ∀𝑥∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → (𝑥𝑅𝑦 → 𝑥𝑆𝑦)))
6		relinxp 5771	. . 3 ⊢ Rel (𝑅 ∩ (𝐴 × 𝐵))
7		ssrel3 5743	. . 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 3174	. 2 ⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝑥𝑅𝑦 → 𝑥𝑆𝑦) ↔ ∀𝑥∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → (𝑥𝑅𝑦 → 𝑥𝑆𝑦)))
10	5, 8, 9	3bitr4i 303	1 ⊢ ((𝑅 ∩ (𝐴 × 𝐵)) ⊆ 𝑆 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝑥𝑅𝑦 → 𝑥𝑆𝑦))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395  ∀wal 1540   ∈ wcel 2114  ∀wral 3052   ∩ cin 3902   ⊆ wss 3903   class class class wbr 5100   × cxp 5630  Rel wrel 5637

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709  ax-sep 5243  ax-pr 5379

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ral 3053  df-rex 3063  df-rab 3402  df-v 3444  df-dif 3906  df-un 3908  df-in 3910  df-ss 3920  df-nul 4288  df-if 4482  df-sn 4583  df-pr 4585  df-op 4589  df-br 5101  df-opab 5163  df-xp 5638  df-rel 5639

This theorem is referenced by:  idinxpss  38569