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Theorem bj-xpcossxp 36591
Description: The composition of two Cartesian products is included in the expected Cartesian product. There is equality if (𝐵𝐶) ≠ ∅, see xpcogend 14939. (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 5721 . . . . . . 7 (𝑥(𝐴 × 𝐵)𝑦 ↔ (𝑥𝐴𝑦𝐵))
2 brxp 5721 . . . . . . 7 (𝑦(𝐶 × 𝐷)𝑡 ↔ (𝑦𝐶𝑡𝐷))
31, 2anbi12i 626 . . . . . 6 ((𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡) ↔ ((𝑥𝐴𝑦𝐵) ∧ (𝑦𝐶𝑡𝐷)))
4 an43 657 . . . . . 6 (((𝑥𝐴𝑦𝐵) ∧ (𝑦𝐶𝑡𝐷)) ↔ ((𝑥𝐴𝑡𝐷) ∧ (𝑦𝐵𝑦𝐶)))
53, 4bitri 275 . . . . 5 ((𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡) ↔ ((𝑥𝐴𝑡𝐷) ∧ (𝑦𝐵𝑦𝐶)))
65exbii 1843 . . . 4 (∃𝑦(𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡) ↔ ∃𝑦((𝑥𝐴𝑡𝐷) ∧ (𝑦𝐵𝑦𝐶)))
7 19.42v 1950 . . . . 5 (∃𝑦((𝑥𝐴𝑡𝐷) ∧ (𝑦𝐵𝑦𝐶)) ↔ ((𝑥𝐴𝑡𝐷) ∧ ∃𝑦(𝑦𝐵𝑦𝐶)))
87simplbi 497 . . . 4 (∃𝑦((𝑥𝐴𝑡𝐷) ∧ (𝑦𝐵𝑦𝐶)) → (𝑥𝐴𝑡𝐷))
96, 8sylbi 216 . . 3 (∃𝑦(𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡) → (𝑥𝐴𝑡𝐷))
109ssopab2i 5546 . 2 {⟨𝑥, 𝑡⟩ ∣ ∃𝑦(𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡)} ⊆ {⟨𝑥, 𝑡⟩ ∣ (𝑥𝐴𝑡𝐷)}
11 df-co 5681 . 2 ((𝐶 × 𝐷) ∘ (𝐴 × 𝐵)) = {⟨𝑥, 𝑡⟩ ∣ ∃𝑦(𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡)}
12 df-xp 5678 . 2 (𝐴 × 𝐷) = {⟨𝑥, 𝑡⟩ ∣ (𝑥𝐴𝑡𝐷)}
1310, 11, 123sstr4i 4021 1 ((𝐶 × 𝐷) ∘ (𝐴 × 𝐵)) ⊆ (𝐴 × 𝐷)
Colors of variables: wff setvar class
Syntax hints:  wa 395  wex 1774  wcel 2099  wss 3944   class class class wbr 5142  {copab 5204   × cxp 5670  ccom 5676
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-ext 2698  ax-sep 5293  ax-nul 5300  ax-pr 5423
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-3an 1087  df-tru 1537  df-fal 1547  df-ex 1775  df-sb 2061  df-clab 2705  df-cleq 2719  df-clel 2805  df-ral 3057  df-rex 3066  df-rab 3428  df-v 3471  df-dif 3947  df-un 3949  df-in 3951  df-ss 3961  df-nul 4319  df-if 4525  df-sn 4625  df-pr 4627  df-op 4631  df-br 5143  df-opab 5205  df-xp 5678  df-co 5681
This theorem is referenced by:  bj-imdirco  36592
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