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Theorem disjxp1 40050
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 898 . . . . 5 (𝑦 = 𝑧 → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
21adantl 475 . . . 4 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦 = 𝑧) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
3 simpl 476 . . . . 5 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ ¬ 𝑦 = 𝑧) → (𝜑 ∧ (𝑦𝐴𝑧𝐴)))
4 neqne 3007 . . . . . 6 𝑦 = 𝑧𝑦𝑧)
54adantl 475 . . . . 5 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ ¬ 𝑦 = 𝑧) → 𝑦𝑧)
6 csbxp 5439 . . . . . . . . 9 𝑦 / 𝑥(𝐵 × 𝐶) = (𝑦 / 𝑥𝐵 × 𝑦 / 𝑥𝐶)
7 csbxp 5439 . . . . . . . . 9 𝑧 / 𝑥(𝐵 × 𝐶) = (𝑧 / 𝑥𝐵 × 𝑧 / 𝑥𝐶)
86, 7ineq12i 4041 . . . . . . . 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 783 . . . . . . . . . 10 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → 𝜑)
11 simplrl 795 . . . . . . . . . 10 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → 𝑦𝐴)
12 simplrr 796 . . . . . . . . . 10 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → 𝑧𝐴)
1310, 11, 12jca31 510 . . . . . . . . 9 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → ((𝜑𝑦𝐴) ∧ 𝑧𝐴))
14 simpr 479 . . . . . . . . . 10 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → 𝑦𝑧)
1514neneqd 3004 . . . . . . . . 9 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → ¬ 𝑦 = 𝑧)
16 disjxp1.1 . . . . . . . . . . . . 13 (𝜑Disj 𝑥𝐴 𝐵)
17 disjors 4858 . . . . . . . . . . . . 13 (Disj 𝑥𝐴 𝐵 ↔ ∀𝑦𝐴𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
1816, 17sylib 210 . . . . . . . . . . . 12 (𝜑 → ∀𝑦𝐴𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
1918r19.21bi 3141 . . . . . . . . . . 11 ((𝜑𝑦𝐴) → ∀𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
2019r19.21bi 3141 . . . . . . . . . 10 (((𝜑𝑦𝐴) ∧ 𝑧𝐴) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
2120ord 895 . . . . . . . . 9 (((𝜑𝑦𝐴) ∧ 𝑧𝐴) → (¬ 𝑦 = 𝑧 → (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
2213, 15, 21sylc 65 . . . . . . . 8 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅)
23 xpdisj1 5800 . . . . . . . 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 2861 . . . . . 6 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅)
26 olc 899 . . . . . 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 579 . . . 4 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ ¬ 𝑦 = 𝑧) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
292, 28pm2.61dan 847 . . 3 ((𝜑 ∧ (𝑦𝐴𝑧𝐴)) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
3029ralrimivva 3180 . 2 (𝜑 → ∀𝑦𝐴𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
31 disjors 4858 . 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 386  wo 878   = wceq 1656  wcel 2164  wne 2999  wral 3117  csb 3757  cin 3797  c0 4146  Disj wdisj 4843   × cxp 5344
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1894  ax-4 1908  ax-5 2009  ax-6 2075  ax-7 2112  ax-9 2173  ax-10 2192  ax-11 2207  ax-12 2220  ax-13 2389  ax-ext 2803  ax-sep 5007  ax-nul 5015  ax-pr 5129
This theorem depends on definitions:  df-bi 199  df-an 387  df-or 879  df-3an 1113  df-tru 1660  df-fal 1670  df-ex 1879  df-nf 1883  df-sb 2068  df-mo 2605  df-eu 2640  df-clab 2812  df-cleq 2818  df-clel 2821  df-nfc 2958  df-ne 3000  df-ral 3122  df-rex 3123  df-reu 3124  df-rmo 3125  df-rab 3126  df-v 3416  df-sbc 3663  df-csb 3758  df-dif 3801  df-un 3803  df-in 3805  df-ss 3812  df-nul 4147  df-if 4309  df-sn 4400  df-pr 4402  df-op 4406  df-disj 4844  df-opab 4938  df-xp 5352  df-rel 5353
This theorem is referenced by:  disjsnxp  40051
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