Theorem disjxp1 43267

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 979	. . . 4 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 = 𝑧) → (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
2		csbxp 5731	. . . . . . 7 ⊢ ⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) = (⦋𝑦 / 𝑥⦌𝐵 × ⦋𝑦 / 𝑥⦌𝐶)
3		csbxp 5731	. . . . . . 7 ⊢ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶) = (⦋𝑧 / 𝑥⦌𝐵 × ⦋𝑧 / 𝑥⦌𝐶)
4	2, 3	ineq12i 4170	. . . . . 6 ⊢ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ((⦋𝑦 / 𝑥⦌𝐵 × ⦋𝑦 / 𝑥⦌𝐶) ∩ (⦋𝑧 / 𝑥⦌𝐵 × ⦋𝑧 / 𝑥⦌𝐶))
5		simpll 765	. . . . . . . . 9 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → 𝜑)
6		simplrl 775	. . . . . . . . 9 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → 𝑦 ∈ 𝐴)
7		simplrr 776	. . . . . . . . 9 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → 𝑧 ∈ 𝐴)
8	5, 6, 7	jca31 515	. . . . . . . 8 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → ((𝜑 ∧ 𝑦 ∈ 𝐴) ∧ 𝑧 ∈ 𝐴))
9		simpr 485	. . . . . . . . 9 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → 𝑦 ≠ 𝑧)
10	9	neneqd 2948	. . . . . . . 8 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → ¬ 𝑦 = 𝑧)
11		disjxp1.1	. . . . . . . . . . . 12 ⊢ (𝜑 → Disj 𝑥 ∈ 𝐴 𝐵)
12		disjors 5086	. . . . . . . . . . . 12 ⊢ (Disj 𝑥 ∈ 𝐴 𝐵 ↔ ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
13	11, 12	sylib 217	. . . . . . . . . . 11 ⊢ (𝜑 → ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
14	13	r19.21bi 3234	. . . . . . . . . 10 ⊢ ((𝜑 ∧ 𝑦 ∈ 𝐴) → ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
15	14	r19.21bi 3234	. . . . . . . . 9 ⊢ (((𝜑 ∧ 𝑦 ∈ 𝐴) ∧ 𝑧 ∈ 𝐴) → (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
16	15	ord 862	. . . . . . . 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 6113	. . . . . . 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 2788	. . . . 5 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅)
21	20	olcd 872	. . . 4 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
22	1, 21	pm2.61dane 3032	. . 3 ⊢ ((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) → (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
23	22	ralrimivva 3197	. 2 ⊢ (𝜑 → ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
24		disjors 5086	. 2 ⊢ (Disj 𝑥 ∈ 𝐴 (𝐵 × 𝐶) ↔ ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
25	23, 24	sylibr 233	1 ⊢ (𝜑 → Disj 𝑥 ∈ 𝐴 (𝐵 × 𝐶))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 396   ∨ wo 845   = wceq 1541   ∈ wcel 2106   ≠ wne 2943  ∀wral 3064  ⦋csb 3855   ∩ cin 3909  ∅c0 4282  Disj wdisj 5070   × cxp 5631

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 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-10 2137  ax-11 2154  ax-12 2171  ax-ext 2707  ax-sep 5256  ax-nul 5263  ax-pr 5384

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 846  df-3an 1089  df-tru 1544  df-fal 1554  df-ex 1782  df-nf 1786  df-sb 2068  df-mo 2538  df-clab 2714  df-cleq 2728  df-clel 2814  df-nfc 2889  df-ne 2944  df-ral 3065  df-rmo 3353  df-rab 3408  df-v 3447  df-sbc 3740  df-csb 3856  df-dif 3913  df-un 3915  df-in 3917  df-ss 3927  df-nul 4283  df-if 4487  df-sn 4587  df-pr 4589  df-op 4593  df-disj 5071  df-opab 5168  df-xp 5639  df-rel 5640

This theorem is referenced by:  disjsnxp  43268