Theorem disjxp1 45070

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	⊢ (𝜑 → Disj 𝑥 ∈ 𝐴 𝐵)

Assertion

Ref	Expression
disjxp1	⊢ (𝜑 → Disj 𝑥 ∈ 𝐴 (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.

Step	Hyp	Ref	Expression
1		animorrl 982	. . . 4 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 = 𝑧) → (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
2		csbxp 5741	. . . . . . 7 ⊢ ⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) = (⦋𝑦 / 𝑥⦌𝐵 × ⦋𝑦 / 𝑥⦌𝐶)
3		csbxp 5741	. . . . . . 7 ⊢ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶) = (⦋𝑧 / 𝑥⦌𝐵 × ⦋𝑧 / 𝑥⦌𝐶)
4	2, 3	ineq12i 4184	. . . . . 6 ⊢ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ((⦋𝑦 / 𝑥⦌𝐵 × ⦋𝑦 / 𝑥⦌𝐶) ∩ (⦋𝑧 / 𝑥⦌𝐵 × ⦋𝑧 / 𝑥⦌𝐶))
5		simpll 766	. . . . . . . . 9 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → 𝜑)
6		simplrl 776	. . . . . . . . 9 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → 𝑦 ∈ 𝐴)
7		simplrr 777	. . . . . . . . 9 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → 𝑧 ∈ 𝐴)
8	5, 6, 7	jca31 514	. . . . . . . 8 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → ((𝜑 ∧ 𝑦 ∈ 𝐴) ∧ 𝑧 ∈ 𝐴))
9		simpr 484	. . . . . . . . 9 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → 𝑦 ≠ 𝑧)
10	9	neneqd 2931	. . . . . . . 8 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → ¬ 𝑦 = 𝑧)
11		disjxp1.1	. . . . . . . . . . . 12 ⊢ (𝜑 → Disj 𝑥 ∈ 𝐴 𝐵)
12		disjors 5093	. . . . . . . . . . . 12 ⊢ (Disj 𝑥 ∈ 𝐴 𝐵 ↔ ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
13	11, 12	sylib 218	. . . . . . . . . . 11 ⊢ (𝜑 → ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
14	13	r19.21bi 3230	. . . . . . . . . 10 ⊢ ((𝜑 ∧ 𝑦 ∈ 𝐴) → ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
15	14	r19.21bi 3230	. . . . . . . . 9 ⊢ (((𝜑 ∧ 𝑦 ∈ 𝐴) ∧ 𝑧 ∈ 𝐴) → (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
16	15	ord 864	. . . . . . . 8 ⊢ (((𝜑 ∧ 𝑦 ∈ 𝐴) ∧ 𝑧 ∈ 𝐴) → (¬ 𝑦 = 𝑧 → (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
17	8, 10, 16	sylc 65	. . . . . . 7 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅)
18		xpdisj1 6137	. . . . . . 7 ⊢ ((⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅ → ((⦋𝑦 / 𝑥⦌𝐵 × ⦋𝑦 / 𝑥⦌𝐶) ∩ (⦋𝑧 / 𝑥⦌𝐵 × ⦋𝑧 / 𝑥⦌𝐶)) = ∅)
19	17, 18	syl 17	. . . . . 6 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → ((⦋𝑦 / 𝑥⦌𝐵 × ⦋𝑦 / 𝑥⦌𝐶) ∩ (⦋𝑧 / 𝑥⦌𝐵 × ⦋𝑧 / 𝑥⦌𝐶)) = ∅)
20	4, 19	eqtrid 2777	. . . . 5 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅)
21	20	olcd 874	. . . 4 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
22	1, 21	pm2.61dane 3013	. . 3 ⊢ ((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) → (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
23	22	ralrimivva 3181	. 2 ⊢ (𝜑 → ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
24		disjors 5093	. 2 ⊢ (Disj 𝑥 ∈ 𝐴 (𝐵 × 𝐶) ↔ ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
25	23, 24	sylibr 234	1 ⊢ (𝜑 → Disj 𝑥 ∈ 𝐴 (𝐵 × 𝐶))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 395   ∨ wo 847   = wceq 1540   ∈ wcel 2109   ≠ wne 2926  ∀wral 3045  ⦋csb 3865   ∩ cin 3916  ∅c0 4299  Disj wdisj 5077   × cxp 5639

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-10 2142  ax-11 2158  ax-12 2178  ax-ext 2702  ax-sep 5254  ax-nul 5264  ax-pr 5390

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2066  df-mo 2534  df-clab 2709  df-cleq 2722  df-clel 2804  df-nfc 2879  df-ne 2927  df-ral 3046  df-rex 3055  df-rmo 3356  df-rab 3409  df-v 3452  df-sbc 3757  df-csb 3866  df-dif 3920  df-un 3922  df-in 3924  df-ss 3934  df-nul 4300  df-if 4492  df-sn 4593  df-pr 4595  df-op 4599  df-disj 5078  df-opab 5173  df-xp 5647  df-rel 5648

This theorem is referenced by:  disjsnxp  45071