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Theorem disjxp1 39238
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 400 . . . . 5 (𝑦 = 𝑧 → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
21adantl 482 . . . 4 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦 = 𝑧) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
3 simpl 473 . . . . 5 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ ¬ 𝑦 = 𝑧) → (𝜑 ∧ (𝑦𝐴𝑧𝐴)))
4 neqne 2802 . . . . . 6 𝑦 = 𝑧𝑦𝑧)
54adantl 482 . . . . 5 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ ¬ 𝑦 = 𝑧) → 𝑦𝑧)
6 csbxp 5200 . . . . . . . . 9 𝑦 / 𝑥(𝐵 × 𝐶) = (𝑦 / 𝑥𝐵 × 𝑦 / 𝑥𝐶)
7 csbxp 5200 . . . . . . . . 9 𝑧 / 𝑥(𝐵 × 𝐶) = (𝑧 / 𝑥𝐵 × 𝑧 / 𝑥𝐶)
86, 7ineq12i 3812 . . . . . . . 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 790 . . . . . . . . . 10 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → 𝜑)
11 simplrl 800 . . . . . . . . . 10 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → 𝑦𝐴)
12 simplrr 801 . . . . . . . . . 10 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → 𝑧𝐴)
1310, 11, 12jca31 557 . . . . . . . . 9 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → ((𝜑𝑦𝐴) ∧ 𝑧𝐴))
14 simpr 477 . . . . . . . . . 10 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → 𝑦𝑧)
1514neneqd 2799 . . . . . . . . 9 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → ¬ 𝑦 = 𝑧)
16 disjxp1.1 . . . . . . . . . . . . 13 (𝜑Disj 𝑥𝐴 𝐵)
17 disjors 4635 . . . . . . . . . . . . 13 (Disj 𝑥𝐴 𝐵 ↔ ∀𝑦𝐴𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
1816, 17sylib 208 . . . . . . . . . . . 12 (𝜑 → ∀𝑦𝐴𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
1918r19.21bi 2932 . . . . . . . . . . 11 ((𝜑𝑦𝐴) → ∀𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
2019r19.21bi 2932 . . . . . . . . . 10 (((𝜑𝑦𝐴) ∧ 𝑧𝐴) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
2120ord 392 . . . . . . . . 9 (((𝜑𝑦𝐴) ∧ 𝑧𝐴) → (¬ 𝑦 = 𝑧 → (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅))
2213, 15, 21sylc 65 . . . . . . . 8 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → (𝑦 / 𝑥𝐵𝑧 / 𝑥𝐵) = ∅)
23 xpdisj1 5555 . . . . . . . 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 2656 . . . . . 6 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ 𝑦𝑧) → (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅)
26 olc 399 . . . . . 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 693 . . . 4 (((𝜑 ∧ (𝑦𝐴𝑧𝐴)) ∧ ¬ 𝑦 = 𝑧) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
292, 28pm2.61dan 832 . . 3 ((𝜑 ∧ (𝑦𝐴𝑧𝐴)) → (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
3029ralrimivva 2971 . 2 (𝜑 → ∀𝑦𝐴𝑧𝐴 (𝑦 = 𝑧 ∨ (𝑦 / 𝑥(𝐵 × 𝐶) ∩ 𝑧 / 𝑥(𝐵 × 𝐶)) = ∅))
31 disjors 4635 . 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 383  wa 384   = wceq 1483  wcel 1990  wne 2794  wral 2912  csb 3533  cin 3573  c0 3915  Disj wdisj 4620   × cxp 5112
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602  ax-sep 4781  ax-nul 4789  ax-pr 4906
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1039  df-tru 1486  df-fal 1489  df-ex 1705  df-nf 1710  df-sb 1881  df-eu 2474  df-mo 2475  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-ne 2795  df-ral 2917  df-rex 2918  df-reu 2919  df-rmo 2920  df-rab 2921  df-v 3202  df-sbc 3436  df-csb 3534  df-dif 3577  df-un 3579  df-in 3581  df-ss 3588  df-nul 3916  df-if 4087  df-sn 4178  df-pr 4180  df-op 4184  df-disj 4621  df-opab 4713  df-xp 5120  df-rel 5121
This theorem is referenced by:  disjsnxp  39239
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