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Theorem bj-xpcossxp 34869
Description: The composition of two Cartesian products is included in the expected Cartesian product. There is equality if (𝐵𝐶) ≠ ∅, see xpcogend 14366. (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 5563 . . . . . . 7 (𝑥(𝐴 × 𝐵)𝑦 ↔ (𝑥𝐴𝑦𝐵))
2 brxp 5563 . . . . . . 7 (𝑦(𝐶 × 𝐷)𝑡 ↔ (𝑦𝐶𝑡𝐷))
31, 2anbi12i 630 . . . . . 6 ((𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡) ↔ ((𝑥𝐴𝑦𝐵) ∧ (𝑦𝐶𝑡𝐷)))
4 an43 658 . . . . . 6 (((𝑥𝐴𝑦𝐵) ∧ (𝑦𝐶𝑡𝐷)) ↔ ((𝑥𝐴𝑡𝐷) ∧ (𝑦𝐵𝑦𝐶)))
53, 4bitri 278 . . . . 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 502 . . . 4 (∃𝑦((𝑥𝐴𝑡𝐷) ∧ (𝑦𝐵𝑦𝐶)) → (𝑥𝐴𝑡𝐷))
96, 8sylbi 220 . . 3 (∃𝑦(𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡) → (𝑥𝐴𝑡𝐷))
109ssopab2i 5400 . 2 {⟨𝑥, 𝑡⟩ ∣ ∃𝑦(𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡)} ⊆ {⟨𝑥, 𝑡⟩ ∣ (𝑥𝐴𝑡𝐷)}
11 df-co 5526 . 2 ((𝐶 × 𝐷) ∘ (𝐴 × 𝐵)) = {⟨𝑥, 𝑡⟩ ∣ ∃𝑦(𝑥(𝐴 × 𝐵)𝑦𝑦(𝐶 × 𝐷)𝑡)}
12 df-xp 5523 . 2 (𝐴 × 𝐷) = {⟨𝑥, 𝑡⟩ ∣ (𝑥𝐴𝑡𝐷)}
1310, 11, 123sstr4i 3931 1 ((𝐶 × 𝐷) ∘ (𝐴 × 𝐵)) ⊆ (𝐴 × 𝐷)
Colors of variables: wff setvar class
Syntax hints:  wa 400  wex 1782  wcel 2112  wss 3854   class class class wbr 5025  {copab 5087   × cxp 5515  ccom 5521
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1912  ax-6 1971  ax-7 2016  ax-8 2114  ax-9 2122  ax-ext 2730  ax-sep 5162  ax-nul 5169  ax-pr 5291
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 846  df-3an 1087  df-tru 1542  df-ex 1783  df-sb 2071  df-clab 2737  df-cleq 2751  df-clel 2831  df-ral 3073  df-rex 3074  df-v 3409  df-dif 3857  df-un 3859  df-in 3861  df-ss 3871  df-nul 4222  df-if 4414  df-sn 4516  df-pr 4518  df-op 4522  df-br 5026  df-opab 5088  df-xp 5523  df-co 5526
This theorem is referenced by:  bj-imdirco  34870
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