Theorem disjxp1 42231

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 981	. . . 4 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 = 𝑧) → (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
2		csbxp 5632	. . . . . . 7 ⊢ ⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) = (⦋𝑦 / 𝑥⦌𝐵 × ⦋𝑦 / 𝑥⦌𝐶)
3		csbxp 5632	. . . . . . 7 ⊢ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶) = (⦋𝑧 / 𝑥⦌𝐵 × ⦋𝑧 / 𝑥⦌𝐶)
4	2, 3	ineq12i 4111	. . . . . 6 ⊢ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ((⦋𝑦 / 𝑥⦌𝐵 × ⦋𝑦 / 𝑥⦌𝐶) ∩ (⦋𝑧 / 𝑥⦌𝐵 × ⦋𝑧 / 𝑥⦌𝐶))
5		simpll 767	. . . . . . . . 9 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → 𝜑)
6		simplrl 777	. . . . . . . . 9 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → 𝑦 ∈ 𝐴)
7		simplrr 778	. . . . . . . . 9 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → 𝑧 ∈ 𝐴)
8	5, 6, 7	jca31 518	. . . . . . . 8 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → ((𝜑 ∧ 𝑦 ∈ 𝐴) ∧ 𝑧 ∈ 𝐴))
9		simpr 488	. . . . . . . . 9 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → 𝑦 ≠ 𝑧)
10	9	neneqd 2937	. . . . . . . 8 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → ¬ 𝑦 = 𝑧)
11		disjxp1.1	. . . . . . . . . . . 12 ⊢ (𝜑 → Disj 𝑥 ∈ 𝐴 𝐵)
12		disjors 5020	. . . . . . . . . . . 12 ⊢ (Disj 𝑥 ∈ 𝐴 𝐵 ↔ ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
13	11, 12	sylib 221	. . . . . . . . . . 11 ⊢ (𝜑 → ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
14	13	r19.21bi 3120	. . . . . . . . . 10 ⊢ ((𝜑 ∧ 𝑦 ∈ 𝐴) → ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
15	14	r19.21bi 3120	. . . . . . . . 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 6004	. . . . . . 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	syl5eq 2783	. . . . 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 3019	. . 3 ⊢ ((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) → (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
23	22	ralrimivva 3102	. 2 ⊢ (𝜑 → ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
24		disjors 5020	. 2 ⊢ (Disj 𝑥 ∈ 𝐴 (𝐵 × 𝐶) ↔ ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
25	23, 24	sylibr 237	1 ⊢ (𝜑 → Disj 𝑥 ∈ 𝐴 (𝐵 × 𝐶))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 399   ∨ wo 847   = wceq 1543   ∈ wcel 2112   ≠ wne 2932  ∀wral 3051  ⦋csb 3798   ∩ cin 3852  ∅c0 4223  Disj wdisj 5004   × cxp 5534

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2018  ax-8 2114  ax-9 2122  ax-10 2143  ax-11 2160  ax-12 2177  ax-ext 2708  ax-sep 5177  ax-nul 5184  ax-pr 5307

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 848  df-3an 1091  df-tru 1546  df-fal 1556  df-ex 1788  df-nf 1792  df-sb 2073  df-mo 2539  df-clab 2715  df-cleq 2728  df-clel 2809  df-nfc 2879  df-ne 2933  df-ral 3056  df-rmo 3059  df-rab 3060  df-v 3400  df-sbc 3684  df-csb 3799  df-dif 3856  df-un 3858  df-in 3860  df-ss 3870  df-nul 4224  df-if 4426  df-sn 4528  df-pr 4530  df-op 4534  df-disj 5005  df-opab 5102  df-xp 5542  df-rel 5543

This theorem is referenced by:  disjsnxp  42232