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Theorem bj-xpcossxp 37519
Description: The composition of two Cartesian products is included in the expected Cartesian product. There is equality if (𝐵𝐶) ≠ ∅, see xpcogend 14927. (Contributed by BJ, 22-May-2024.)
Assertion
Ref Expression
bj-xpcossxp ((𝐶 × 𝐷) ∘ (𝐴 × 𝐵)) ⊆ (𝐴 × 𝐷)

Proof of Theorem bj-xpcossxp
Dummy variables 𝑥 𝑦 𝑡 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 brxp 5673 . . . . . . 7 (𝑥(𝐴 × 𝐵)𝑦 ↔ (𝑥𝐴𝑦𝐵))
2 brxp 5673 . . . . . . 7 (𝑦(𝐶 × 𝐷)𝑡 ↔ (𝑦𝐶𝑡𝐷))
31, 2anbi12i 629 . . . . . 6 ((𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡) ↔ ((𝑥𝐴𝑦𝐵) ∧ (𝑦𝐶𝑡𝐷)))
4 an43 659 . . . . . 6 (((𝑥𝐴𝑦𝐵) ∧ (𝑦𝐶𝑡𝐷)) ↔ ((𝑥𝐴𝑡𝐷) ∧ (𝑦𝐵𝑦𝐶)))
53, 4bitri 275 . . . . 5 ((𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡) ↔ ((𝑥𝐴𝑡𝐷) ∧ (𝑦𝐵𝑦𝐶)))
65exbii 1850 . . . 4 (∃𝑦(𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡) ↔ ∃𝑦((𝑥𝐴𝑡𝐷) ∧ (𝑦𝐵𝑦𝐶)))
7 19.42v 1955 . . . . 5 (∃𝑦((𝑥𝐴𝑡𝐷) ∧ (𝑦𝐵𝑦𝐶)) ↔ ((𝑥𝐴𝑡𝐷) ∧ ∃𝑦(𝑦𝐵𝑦𝐶)))
87simplbi 496 . . . 4 (∃𝑦((𝑥𝐴𝑡𝐷) ∧ (𝑦𝐵𝑦𝐶)) → (𝑥𝐴𝑡𝐷))
96, 8sylbi 217 . . 3 (∃𝑦(𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡) → (𝑥𝐴𝑡𝐷))
109ssopab2i 5498 . 2 {⟨𝑥, 𝑡⟩ ∣ ∃𝑦(𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡)} ⊆ {⟨𝑥, 𝑡⟩ ∣ (𝑥𝐴𝑡𝐷)}
11 df-co 5633 . 2 ((𝐶 × 𝐷) ∘ (𝐴 × 𝐵)) = {⟨𝑥, 𝑡⟩ ∣ ∃𝑦(𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡)}
12 df-xp 5630 . 2 (𝐴 × 𝐷) = {⟨𝑥, 𝑡⟩ ∣ (𝑥𝐴𝑡𝐷)}
1310, 11, 123sstr4i 3974 1 ((𝐶 × 𝐷) ∘ (𝐴 × 𝐵)) ⊆ (𝐴 × 𝐷)
Colors of variables: wff setvar class
Syntax hints:  wa 395  wex 1781  wcel 2114  wss 3890   class class class wbr 5086  {copab 5148   × cxp 5622  ccom 5628
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 5231  ax-pr 5370
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 3391  df-v 3432  df-dif 3893  df-un 3895  df-in 3897  df-ss 3907  df-nul 4275  df-if 4468  df-sn 4569  df-pr 4571  df-op 4575  df-br 5087  df-opab 5149  df-xp 5630  df-co 5633
This theorem is referenced by:  bj-imdirco  37520
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