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Theorem inxpss 39515
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
StepHypRef Expression
1 brinxp2 5737 . . . . 5 (𝑥(𝑅 ∩ (𝐴 × 𝐵))𝑦 ↔ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑥𝑅𝑦))
21imbi1i 352 . . . 4 ((𝑥(𝑅 ∩ (𝐴 × 𝐵))𝑦 → 𝑥𝑆𝑦) ↔ (((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑥𝑅𝑦) → 𝑥𝑆𝑦))
3 impexp 456 . . . 4 ((((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝑥𝑅𝑦) → 𝑥𝑆𝑦) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → (𝑥𝑅𝑦 → 𝑥𝑆𝑦)))
42, 3bitri 278 . . 3 ((𝑥(𝑅 ∩ (𝐴 × 𝐵))𝑦 → 𝑥𝑆𝑦) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → (𝑥𝑅𝑦 → 𝑥𝑆𝑦)))
542albii 1853 . 2 (∀𝑥∀𝑦(𝑥(𝑅 ∩ (𝐴 × 𝐵))𝑦 → 𝑥𝑆𝑦) ↔ ∀𝑥∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → (𝑥𝑅𝑦 → 𝑥𝑆𝑦)))
6 relinxp 5801 . . 3 Rel (𝑅 ∩ (𝐴 × 𝐵))
7 ssrel3 5771 . . 3 (Rel (𝑅 ∩ (𝐴 × 𝐵)) → ((𝑅 ∩ (𝐴 × 𝐵)) ⊆ 𝑆 ↔ ∀𝑥∀𝑦(𝑥(𝑅 ∩ (𝐴 × 𝐵))𝑦 → 𝑥𝑆𝑦)))
86, 7ax-mp 5 . 2 ((𝑅 ∩ (𝐴 × 𝐵)) ⊆ 𝑆 ↔ ∀𝑥∀𝑦(𝑥(𝑅 ∩ (𝐴 × 𝐵))𝑦 → 𝑥𝑆𝑦))
9 r2al 3202 . 2 (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝑥𝑅𝑦 → 𝑥𝑆𝑦) ↔ ∀𝑥∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → (𝑥𝑅𝑦 → 𝑥𝑆𝑦)))
105, 8, 93bitr4i 306 1 ((𝑅 ∩ (𝐴 × 𝐵)) ⊆ 𝑆 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝑥𝑅𝑦 → 𝑥𝑆𝑦))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568   ∈ wcel 2147  ∀wral 3080   ⊆ wss 3900   ∩ cin 4092   class class class wbr 5108   × cxp 5656  Rel wrel 5663
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 5255  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 2102  df-clab 2742  df-cleq 2755  df-clel 2839  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 4841  df-br 5109  df-opab 5174  df-xp 5664  df-rel 5665
This theorem is used by:  idinxpss  39516
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