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Theorem disjxp1 41698
 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 animorrl 978 . . . 4 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦 = 𝑧) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
2 csbxp 5614 . . . . . . 7 𝑦 / 𝑥(𝐵 × 𝐶) = (𝑦 / 𝑥𝐵 × 𝑦 / 𝑥𝐶)
3 csbxp 5614 . . . . . . 7 𝑧 / 𝑥(𝐵 × 𝐶) = (𝑧 / 𝑥𝐵 × 𝑧 / 𝑥𝐶)
42, 3ineq12i 4137 . . . . . 6 (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ((𝑦 / 𝑥𝐵 × 𝑦 / 𝑥𝐶) ∩ (𝑧 / 𝑥𝐵 × 𝑧 / 𝑥𝐶))
5 simpll 766 . . . . . . . . 9 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → 𝜑)
6 simplrl 776 . . . . . . . . 9 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → 𝑦𝐴)
7 simplrr 777 . . . . . . . . 9 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → 𝑧𝐴)
85, 6, 7jca31 518 . . . . . . . 8 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → ((𝜑𝑦𝐴) ∧ 𝑧𝐴))
9 simpr 488 . . . . . . . . 9 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → 𝑦𝑧)
109neneqd 2992 . . . . . . . 8 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → ¬ 𝑦 = 𝑧)
11 disjxp1.1 . . . . . . . . . . . 12 (𝜑Disj 𝑥𝐴 𝐵)
12 disjors 5011 . . . . . . . . . . . 12 (Disj 𝑥𝐴 𝐵 ↔ ∀𝑦𝐴𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
1311, 12sylib 221 . . . . . . . . . . 11 (𝜑 → ∀𝑦𝐴𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
1413r19.21bi 3173 . . . . . . . . . 10 ((𝜑𝑦𝐴) → ∀𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
1514r19.21bi 3173 . . . . . . . . 9 (((𝜑𝑦𝐴) ∧ 𝑧𝐴) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
1615ord 861 . . . . . . . 8 (((𝜑𝑦𝐴) ∧ 𝑧𝐴) → (¬ 𝑦 = 𝑧 → (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
178, 10, 16sylc 65 . . . . . . 7 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅)
18 xpdisj1 5985 . . . . . . 7 ((𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅ → ((𝑦 / 𝑥𝐵 × 𝑦 / 𝑥𝐶) ∩ (𝑧 / 𝑥𝐵 × 𝑧 / 𝑥𝐶)) = ∅)
1917, 18syl 17 . . . . . 6 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → ((𝑦 / 𝑥𝐵 × 𝑦 / 𝑥𝐶) ∩ (𝑧 / 𝑥𝐵 × 𝑧 / 𝑥𝐶)) = ∅)
204, 19syl5eq 2845 . . . . 5 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅)
2120olcd 871 . . . 4 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
221, 21pm2.61dane 3074 . . 3 ((𝜑 ∧ (𝑦𝐴𝑧𝐴)) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
2322ralrimivva 3156 . 2 (𝜑 → ∀𝑦𝐴𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
24 disjors 5011 . 2 (Disj 𝑥𝐴 (𝐵 × 𝐶) ↔ ∀𝑦𝐴𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
2523, 24sylibr 237 1 (𝜑Disj 𝑥𝐴 (𝐵 × 𝐶))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 399   ∨ wo 844   = wceq 1538   ∈ wcel 2111   ≠ wne 2987  ∀wral 3106  ⦋csb 3828   ∩ cin 3880  ∅c0 4243  Disj wdisj 4995   × cxp 5517 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-10 2142  ax-11 2158  ax-12 2175  ax-ext 2770  ax-sep 5167  ax-nul 5174  ax-pr 5295 This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-3an 1086  df-tru 1541  df-fal 1551  df-ex 1782  df-nf 1786  df-sb 2070  df-mo 2598  df-clab 2777  df-cleq 2791  df-clel 2870  df-nfc 2938  df-ne 2988  df-ral 3111  df-rmo 3114  df-rab 3115  df-v 3443  df-sbc 3721  df-csb 3829  df-dif 3884  df-un 3886  df-in 3888  df-ss 3898  df-nul 4244  df-if 4426  df-sn 4526  df-pr 4528  df-op 4532  df-disj 4996  df-opab 5093  df-xp 5525  df-rel 5526 This theorem is referenced by:  disjsnxp  41699
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