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Theorem disjxp1 39997
Description: The sets of a cartesian product are disjoint if the sets in the first argument are disjoint. (Contributed by Glauco Siliprandi, 11-Oct-2020.)
Hypothesis
Ref Expression
disjxp1.1 (𝜑Disj 𝑥𝐴 𝐵)
Assertion
Ref Expression
disjxp1 (𝜑Disj 𝑥𝐴 (𝐵 × 𝐶))
Distinct variable group:   𝑥,𝐴
Allowed substitution hints:   𝜑(𝑥)   𝐵(𝑥)   𝐶(𝑥)

Proof of Theorem disjxp1
Dummy variables 𝑦 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 orc 894 . . . . 5 (𝑦 = 𝑧 → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
21adantl 474 . . . 4 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦 = 𝑧) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
3 simpl 475 . . . . 5 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ ¬ 𝑦 = 𝑧) → (𝜑 ∧ (𝑦𝐴𝑧𝐴)))
4 neqne 2979 . . . . . 6 𝑦 = 𝑧𝑦𝑧)
54adantl 474 . . . . 5 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ ¬ 𝑦 = 𝑧) → 𝑦𝑧)
6 csbxp 5405 . . . . . . . . 9 𝑦 / 𝑥(𝐵 × 𝐶) = (𝑦 / 𝑥𝐵 × 𝑦 / 𝑥𝐶)
7 csbxp 5405 . . . . . . . . 9 𝑧 / 𝑥(𝐵 × 𝐶) = (𝑧 / 𝑥𝐵 × 𝑧 / 𝑥𝐶)
86, 7ineq12i 4010 . . . . . . . 8 (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ((𝑦 / 𝑥𝐵 × 𝑦 / 𝑥𝐶) ∩ (𝑧 / 𝑥𝐵 × 𝑧 / 𝑥𝐶))
98a1i 11 . . . . . . 7 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ((𝑦 / 𝑥𝐵 × 𝑦 / 𝑥𝐶) ∩ (𝑧 / 𝑥𝐵 × 𝑧 / 𝑥𝐶)))
10 simpll 784 . . . . . . . . . 10 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → 𝜑)
11 simplrl 796 . . . . . . . . . 10 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → 𝑦𝐴)
12 simplrr 797 . . . . . . . . . 10 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → 𝑧𝐴)
1310, 11, 12jca31 511 . . . . . . . . 9 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → ((𝜑𝑦𝐴) ∧ 𝑧𝐴))
14 simpr 478 . . . . . . . . . 10 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → 𝑦𝑧)
1514neneqd 2976 . . . . . . . . 9 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → ¬ 𝑦 = 𝑧)
16 disjxp1.1 . . . . . . . . . . . . 13 (𝜑Disj 𝑥𝐴 𝐵)
17 disjors 4826 . . . . . . . . . . . . 13 (Disj 𝑥𝐴 𝐵 ↔ ∀𝑦𝐴𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
1816, 17sylib 210 . . . . . . . . . . . 12 (𝜑 → ∀𝑦𝐴𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
1918r19.21bi 3113 . . . . . . . . . . 11 ((𝜑𝑦𝐴) → ∀𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
2019r19.21bi 3113 . . . . . . . . . 10 (((𝜑𝑦𝐴) ∧ 𝑧𝐴) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
2120ord 891 . . . . . . . . 9 (((𝜑𝑦𝐴) ∧ 𝑧𝐴) → (¬ 𝑦 = 𝑧 → (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
2213, 15, 21sylc 65 . . . . . . . 8 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅)
23 xpdisj1 5772 . . . . . . . 8 ((𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅ → ((𝑦 / 𝑥𝐵 × 𝑦 / 𝑥𝐶) ∩ (𝑧 / 𝑥𝐵 × 𝑧 / 𝑥𝐶)) = ∅)
2422, 23syl 17 . . . . . . 7 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → ((𝑦 / 𝑥𝐵 × 𝑦 / 𝑥𝐶) ∩ (𝑧 / 𝑥𝐵 × 𝑧 / 𝑥𝐶)) = ∅)
259, 24eqtrd 2833 . . . . . 6 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅)
26 olc 895 . . . . . 6 ((𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅ → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
2725, 26syl 17 . . . . 5 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
283, 5, 27syl2anc 580 . . . 4 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ ¬ 𝑦 = 𝑧) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
292, 28pm2.61dan 848 . . 3 ((𝜑 ∧ (𝑦𝐴𝑧𝐴)) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
3029ralrimivva 3152 . 2 (𝜑 → ∀𝑦𝐴𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
31 disjors 4826 . 2 (Disj 𝑥𝐴 (𝐵 × 𝐶) ↔ ∀𝑦𝐴𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
3230, 31sylibr 226 1 (𝜑Disj 𝑥𝐴 (𝐵 × 𝐶))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wa 385  wo 874   = wceq 1653  wcel 2157  wne 2971  wral 3089  csb 3728  cin 3768  c0 4115  Disj wdisj 4811   × cxp 5310
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905  ax-5 2006  ax-6 2072  ax-7 2107  ax-9 2166  ax-10 2185  ax-11 2200  ax-12 2213  ax-13 2377  ax-ext 2777  ax-sep 4975  ax-nul 4983  ax-pr 5097
This theorem depends on definitions:  df-bi 199  df-an 386  df-or 875  df-3an 1110  df-tru 1657  df-fal 1667  df-ex 1876  df-nf 1880  df-sb 2065  df-mo 2591  df-eu 2609  df-clab 2786  df-cleq 2792  df-clel 2795  df-nfc 2930  df-ne 2972  df-ral 3094  df-rex 3095  df-reu 3096  df-rmo 3097  df-rab 3098  df-v 3387  df-sbc 3634  df-csb 3729  df-dif 3772  df-un 3774  df-in 3776  df-ss 3783  df-nul 4116  df-if 4278  df-sn 4369  df-pr 4371  df-op 4375  df-disj 4812  df-opab 4906  df-xp 5318  df-rel 5319
This theorem is referenced by:  disjsnxp  39998
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