Theorem disjxp1 42506

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 977	. . . 4 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 = 𝑧) → (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
2		csbxp 5676	. . . . . . 7 ⊢ ⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) = (⦋𝑦 / 𝑥⦌𝐵 × ⦋𝑦 / 𝑥⦌𝐶)
3		csbxp 5676	. . . . . . 7 ⊢ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶) = (⦋𝑧 / 𝑥⦌𝐵 × ⦋𝑧 / 𝑥⦌𝐶)
4	2, 3	ineq12i 4141	. . . . . 6 ⊢ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ((⦋𝑦 / 𝑥⦌𝐵 × ⦋𝑦 / 𝑥⦌𝐶) ∩ (⦋𝑧 / 𝑥⦌𝐵 × ⦋𝑧 / 𝑥⦌𝐶))
5		simpll 763	. . . . . . . . 9 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → 𝜑)
6		simplrl 773	. . . . . . . . 9 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → 𝑦 ∈ 𝐴)
7		simplrr 774	. . . . . . . . 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 2947	. . . . . . . 8 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → ¬ 𝑦 = 𝑧)
11		disjxp1.1	. . . . . . . . . . . 12 ⊢ (𝜑 → Disj 𝑥 ∈ 𝐴 𝐵)
12		disjors 5051	. . . . . . . . . . . 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 3132	. . . . . . . . . 10 ⊢ ((𝜑 ∧ 𝑦 ∈ 𝐴) → ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
15	14	r19.21bi 3132	. . . . . . . . 9 ⊢ (((𝜑 ∧ 𝑦 ∈ 𝐴) ∧ 𝑧 ∈ 𝐴) → (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
16	15	ord 860	. . . . . . . 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 6053	. . . . . . 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 2791	. . . . 5 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅)
21	20	olcd 870	. . . 4 ⊢ (((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) ∧ 𝑦 ≠ 𝑧) → (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
22	1, 21	pm2.61dane 3031	. . 3 ⊢ ((𝜑 ∧ (𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴)) → (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
23	22	ralrimivva 3114	. 2 ⊢ (𝜑 → ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌(𝐵 × 𝐶) ∩ ⦋𝑧 / 𝑥⦌(𝐵 × 𝐶)) = ∅))
24		disjors 5051	. 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 395   ∨ wo 843   = wceq 1539   ∈ wcel 2108   ≠ wne 2942  ∀wral 3063  ⦋csb 3828   ∩ cin 3882  ∅c0 4253  Disj wdisj 5035   × cxp 5578

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-10 2139  ax-11 2156  ax-12 2173  ax-ext 2709  ax-sep 5218  ax-nul 5225  ax-pr 5347

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-3an 1087  df-tru 1542  df-fal 1552  df-ex 1784  df-nf 1788  df-sb 2069  df-mo 2540  df-clab 2716  df-cleq 2730  df-clel 2817  df-nfc 2888  df-ne 2943  df-ral 3068  df-rmo 3071  df-rab 3072  df-v 3424  df-sbc 3712  df-csb 3829  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4254  df-if 4457  df-sn 4559  df-pr 4561  df-op 4565  df-disj 5036  df-opab 5133  df-xp 5586  df-rel 5587

This theorem is referenced by:  disjsnxp  42507