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Theorem disjxp1 39755
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 399 . . . . 5 (𝑦 = 𝑧 → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
21adantl 473 . . . 4 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦 = 𝑧) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
3 simpl 474 . . . . 5 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ ¬ 𝑦 = 𝑧) → (𝜑 ∧ (𝑦𝐴𝑧𝐴)))
4 neqne 2940 . . . . . 6 𝑦 = 𝑧𝑦𝑧)
54adantl 473 . . . . 5 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ ¬ 𝑦 = 𝑧) → 𝑦𝑧)
6 csbxp 5357 . . . . . . . . 9 𝑦 / 𝑥(𝐵 × 𝐶) = (𝑦 / 𝑥𝐵 × 𝑦 / 𝑥𝐶)
7 csbxp 5357 . . . . . . . . 9 𝑧 / 𝑥(𝐵 × 𝐶) = (𝑧 / 𝑥𝐵 × 𝑧 / 𝑥𝐶)
86, 7ineq12i 3955 . . . . . . . 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 807 . . . . . . . . . 10 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → 𝜑)
11 simplrl 819 . . . . . . . . . 10 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → 𝑦𝐴)
12 simplrr 820 . . . . . . . . . 10 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → 𝑧𝐴)
1310, 11, 12jca31 558 . . . . . . . . 9 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → ((𝜑𝑦𝐴) ∧ 𝑧𝐴))
14 simpr 479 . . . . . . . . . 10 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → 𝑦𝑧)
1514neneqd 2937 . . . . . . . . 9 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → ¬ 𝑦 = 𝑧)
16 disjxp1.1 . . . . . . . . . . . . 13 (𝜑Disj 𝑥𝐴 𝐵)
17 disjors 4787 . . . . . . . . . . . . 13 (Disj 𝑥𝐴 𝐵 ↔ ∀𝑦𝐴𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
1816, 17sylib 208 . . . . . . . . . . . 12 (𝜑 → ∀𝑦𝐴𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
1918r19.21bi 3070 . . . . . . . . . . 11 ((𝜑𝑦𝐴) → ∀𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
2019r19.21bi 3070 . . . . . . . . . 10 (((𝜑𝑦𝐴) ∧ 𝑧𝐴) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
2120ord 391 . . . . . . . . 9 (((𝜑𝑦𝐴) ∧ 𝑧𝐴) → (¬ 𝑦 = 𝑧 → (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
2213, 15, 21sylc 65 . . . . . . . 8 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅)
23 xpdisj1 5713 . . . . . . . 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 2794 . . . . . 6 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅)
26 olc 398 . . . . . 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 696 . . . 4 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ ¬ 𝑦 = 𝑧) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
292, 28pm2.61dan 867 . . 3 ((𝜑 ∧ (𝑦𝐴𝑧𝐴)) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
3029ralrimivva 3109 . 2 (𝜑 → ∀𝑦𝐴𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
31 disjors 4787 . 2 (Disj 𝑥𝐴 (𝐵 × 𝐶) ↔ ∀𝑦𝐴𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
3230, 31sylibr 224 1 (𝜑Disj 𝑥𝐴 (𝐵 × 𝐶))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wo 382  wa 383   = wceq 1632  wcel 2139  wne 2932  wral 3050  csb 3674  cin 3714  c0 4058  Disj wdisj 4772   × cxp 5264
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1871  ax-4 1886  ax-5 1988  ax-6 2054  ax-7 2090  ax-9 2148  ax-10 2168  ax-11 2183  ax-12 2196  ax-13 2391  ax-ext 2740  ax-sep 4933  ax-nul 4941  ax-pr 5055
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-3an 1074  df-tru 1635  df-fal 1638  df-ex 1854  df-nf 1859  df-sb 2047  df-eu 2611  df-mo 2612  df-clab 2747  df-cleq 2753  df-clel 2756  df-nfc 2891  df-ne 2933  df-ral 3055  df-rex 3056  df-reu 3057  df-rmo 3058  df-rab 3059  df-v 3342  df-sbc 3577  df-csb 3675  df-dif 3718  df-un 3720  df-in 3722  df-ss 3729  df-nul 4059  df-if 4231  df-sn 4322  df-pr 4324  df-op 4328  df-disj 4773  df-opab 4865  df-xp 5272  df-rel 5273
This theorem is referenced by:  disjsnxp  39756
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