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Theorem bj-xpcossxp 37814
Description: The composition of two Cartesian products is included in the expected Cartesian product. There is equality if (𝐵𝐶) ≠ ∅, see xpcogend 15013. (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 5712 . . . . . . 7 (𝑥(𝐴 × 𝐵)𝑦 ↔ (𝑥𝐴𝑦𝐵))
2 brxp 5712 . . . . . . 7 (𝑦(𝐶 × 𝐷)𝑡 ↔ (𝑦𝐶𝑡𝐷))
31, 2anbi12i 639 . . . . . 6 ((𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡) ↔ ((𝑥𝐴𝑦𝐵) ∧ (𝑦𝐶𝑡𝐷)))
4 an43 670 . . . . . 6 (((𝑥𝐴𝑦𝐵) ∧ (𝑦𝐶𝑡𝐷)) ↔ ((𝑥𝐴𝑡𝐷) ∧ (𝑦𝐵𝑦𝐶)))
53, 4bitri 278 . . . . 5 ((𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡) ↔ ((𝑥𝐴𝑡𝐷) ∧ (𝑦𝐵𝑦𝐶)))
65exbii 1878 . . . 4 (∃𝑦(𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡) ↔ ∃𝑦((𝑥𝐴𝑡𝐷) ∧ (𝑦𝐵𝑦𝐶)))
7 19.42v 1983 . . . . 5 (∃𝑦((𝑥𝐴𝑡𝐷) ∧ (𝑦𝐵𝑦𝐶)) ↔ ((𝑥𝐴𝑡𝐷) ∧ ∃𝑦(𝑦𝐵𝑦𝐶)))
87simplbi 501 . . . 4 (∃𝑦((𝑥𝐴𝑡𝐷) ∧ (𝑦𝐵𝑦𝐶)) → (𝑥𝐴𝑡𝐷))
96, 8sylbi 220 . . 3 (∃𝑦(𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡) → (𝑥𝐴𝑡𝐷))
109ssopab2i 5537 . 2 {⟨𝑥, 𝑡⟩ ∣ ∃𝑦(𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡)} ⊆ {⟨𝑥, 𝑡⟩ ∣ (𝑥𝐴𝑡𝐷)}
11 df-co 5672 . 2 ((𝐶 × 𝐷) ∘ (𝐴 × 𝐵)) = {⟨𝑥, 𝑡⟩ ∣ ∃𝑦(𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡)}
12 df-xp 5669 . 2 (𝐴 × 𝐷) = {⟨𝑥, 𝑡⟩ ∣ (𝑥𝐴𝑡𝐷)}
1310, 11, 123sstr4i 3989 1 ((𝐶 × 𝐷) ∘ (𝐴 × 𝐵)) ⊆ (𝐴 × 𝐷)
Colors of variables: wff setvar class
Syntax hints:  wa 400  wex 1809  wcel 2143  wss 3906   class class class wbr 5110  {copab 5174   × cxp 5661  ccom 5667
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735  ax-sep 5258  ax-pr 5406
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-ral 3080  df-rex 3090  df-rab 3417  df-v 3457  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4288  df-if 4489  df-sn 4591  df-pr 4593  df-op 4597  df-br 5111  df-opab 5175  df-xp 5669  df-co 5672
This theorem is referenced by:  bj-imdirco  37815
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