Theorem disjxp1 46029

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 996	. . . 4 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 = 𝑧) → (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
2		csbxp 5752	. . . . . . 7 ⊢ ⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) = (⦋𝑦 / 𝑥⦌𝐵 × ⦋𝑦 / 𝑥⦌𝐶)
3		csbxp 5752	. . . . . . 7 ⊢ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶) = (⦋𝑧 / 𝑥⦌𝐵 × ⦋𝑧 / 𝑥⦌𝐶)
4	2, 3	ineq12i 4164	. . . . . 6 ⊢ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ((⦋𝑦 / 𝑥⦌𝐵 × ⦋𝑦 / 𝑥⦌𝐶) ∩ (⦋𝑧 / 𝑥⦌𝐵 × ⦋𝑧 / 𝑥⦌𝐶))
5		simpll 779	. . . . . . . . 9 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → 𝜑)
6		simplrl 789	. . . . . . . . 9 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → 𝑦 ∈ 𝐴)
7		simplrr 790	. . . . . . . . 9 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → 𝑧 ∈ 𝐴)
8	5, 6, 7	jca31 524	. . . . . . . 8 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → ((𝜑 ∧ 𝑦 ∈ 𝐴) ∧ 𝑧 ∈ 𝐴))
9		simpr 490	. . . . . . . . 9 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → 𝑦 ≠ 𝑧)
10	9	neneqd 2961	. . . . . . . 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 221	. . . . . . . . . . 11 ⊢ (𝜑 → ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
14	13	r19.21bi 3255	. . . . . . . . . 10 ⊢ ((𝜑 ∧ 𝑦 ∈ 𝐴) → ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
15	14	r19.21bi 3255	. . . . . . . . 9 ⊢ (((𝜑 ∧ 𝑦 ∈ 𝐴) ∧ 𝑧 ∈ 𝐴) → (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
16	15	ord 878	. . . . . . . 8 ⊢ (((𝜑 ∧ 𝑦 ∈ 𝐴) ∧ 𝑧 ∈ 𝐴) → (¬ 𝑦 = 𝑧 → (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
17	8, 10, 16	sylc 66	. . . . . . 7 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅)
18		xpdisj1 6151	. . . . . . 7 ⊢ ((⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅ → ((⦋𝑦 / 𝑥⦌𝐵 × ⦋𝑦 / 𝑥⦌𝐶) ∩ (⦋𝑧 / 𝑥⦌𝐵 × ⦋𝑧 / 𝑥⦌𝐶)) = ∅)
19	17, 18	syl 18	. . . . . 6 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → ((⦋𝑦 / 𝑥⦌𝐵 × ⦋𝑦 / 𝑥⦌𝐶) ∩ (⦋𝑧 / 𝑥⦌𝐵 × ⦋𝑧 / 𝑥⦌𝐶)) = ∅)
20	4, 19	eqtrid 2808	. . . . 5 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅)
21	20	olcd 888	. . . 4 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
22	1, 21	pm2.61dane 3043	. . 3 ⊢ ((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) → (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
23	22	ralrimivva 3206	. 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 237	1 ⊢ (𝜑 → Disj 𝑥 ∈ 𝐴 (𝐵 × 𝐶))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   ∨ wo 861   = wceq 1570   ∈ wcel 2145   ≠ wne 2956  ∀wral 3077  ⦋csb 3847   ∩ cin 3898  ∅c0 4279  Disj wdisj 5070   × cxp 5649

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2733  ax-sep 5249  ax-pr 5391

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-mo 2565  df-clab 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-ne 2957  df-ral 3078  df-rex 3088  df-rmo 3366  df-rab 3414  df-v 3453  df-sbc 3740  df-csb 3848  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-disj 5071  df-opab 5168  df-xp 5657  df-rel 5658

This theorem is used by:  disjsnxp  46030