Theorem disji2 5087

Description: Property of a disjoint collection: if 𝐵(𝑋) = 𝐶 and 𝐵(𝑌) = 𝐷, and 𝑋 ≠ 𝑌, then 𝐶 and 𝐷 are disjoint. (Contributed by Mario Carneiro, 14-Nov-2016.)

Hypotheses

Ref	Expression
disji.1	⊢ (𝑥 = 𝑋 → 𝐵 = 𝐶)
disji.2	⊢ (𝑥 = 𝑌 → 𝐵 = 𝐷)

Assertion

Ref	Expression
disji2	⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ 𝑋 ≠ 𝑌) → (𝐶 ∩ 𝐷) = ∅)

Distinct variable groups:   𝑥,𝐴   𝑥,𝐶   𝑥,𝐷   𝑥,𝑋   𝑥,𝑌

Allowed substitution hint:   𝐵(𝑥)

Proof of Theorem disji2

Dummy variables 𝑦 𝑧 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		df-ne 2957	. . 3 ⊢ (𝑋 ≠ 𝑌 ↔ ¬ 𝑋 = 𝑌)
2		disjors 5086	. . . . . 6 ⊢ (Disj 𝑥 ∈ 𝐴 𝐵 ↔ ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
3		eqeq1 2765	. . . . . . . 8 ⊢ (𝑦 = 𝑋 → (𝑦 = 𝑧 ↔ 𝑋 = 𝑧))
4		nfcv 2923	. . . . . . . . . . 11 ⊢ Ⅎ𝑥𝑋
5		nfcv 2923	. . . . . . . . . . 11 ⊢ Ⅎ𝑥𝐶
6		disji.1	. . . . . . . . . . 11 ⊢ (𝑥 = 𝑋 → 𝐵 = 𝐶)
7	4, 5, 6	csbhypf 3875	. . . . . . . . . 10 ⊢ (𝑦 = 𝑋 → ⦋𝑦 / 𝑥⦌𝐵 = 𝐶)
8	7	ineq1d 4165	. . . . . . . . 9 ⊢ (𝑦 = 𝑋 → (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = (𝐶 ∩ ⦋𝑧 / 𝑥⦌𝐵))
9	8	eqeq1d 2763	. . . . . . . 8 ⊢ (𝑦 = 𝑋 → ((⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅ ↔ (𝐶 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
10	3, 9	orbi12d 932	. . . . . . 7 ⊢ (𝑦 = 𝑋 → ((𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅) ↔ (𝑋 = 𝑧 ∨ (𝐶 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅)))
11		eqeq2 2773	. . . . . . . 8 ⊢ (𝑧 = 𝑌 → (𝑋 = 𝑧 ↔ 𝑋 = 𝑌))
12		nfcv 2923	. . . . . . . . . . 11 ⊢ Ⅎ𝑥𝑌
13		nfcv 2923	. . . . . . . . . . 11 ⊢ Ⅎ𝑥𝐷
14		disji.2	. . . . . . . . . . 11 ⊢ (𝑥 = 𝑌 → 𝐵 = 𝐷)
15	12, 13, 14	csbhypf 3875	. . . . . . . . . 10 ⊢ (𝑧 = 𝑌 → ⦋𝑧 / 𝑥⦌𝐵 = 𝐷)
16	15	ineq2d 4166	. . . . . . . . 9 ⊢ (𝑧 = 𝑌 → (𝐶 ∩ ⦋𝑧 / 𝑥⦌𝐵) = (𝐶 ∩ 𝐷))
17	16	eqeq1d 2763	. . . . . . . 8 ⊢ (𝑧 = 𝑌 → ((𝐶 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅ ↔ (𝐶 ∩ 𝐷) = ∅))
18	11, 17	orbi12d 932	. . . . . . 7 ⊢ (𝑧 = 𝑌 → ((𝑋 = 𝑧 ∨ (𝐶 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅) ↔ (𝑋 = 𝑌 ∨ (𝐶 ∩ 𝐷) = ∅)))
19	10, 18	rspc2v 3587	. . . . . 6 ⊢ ((𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) → (∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅) → (𝑋 = 𝑌 ∨ (𝐶 ∩ 𝐷) = ∅)))
20	2, 19	biimtrid 245	. . . . 5 ⊢ ((𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) → (Disj 𝑥 ∈ 𝐴 𝐵 → (𝑋 = 𝑌 ∨ (𝐶 ∩ 𝐷) = ∅)))
21	20	impcom 413	. . . 4 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (𝑋 = 𝑌 ∨ (𝐶 ∩ 𝐷) = ∅))
22	21	ord 878	. . 3 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (¬ 𝑋 = 𝑌 → (𝐶 ∩ 𝐷) = ∅))
23	1, 22	biimtrid 245	. 2 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (𝑋 ≠ 𝑌 → (𝐶 ∩ 𝐷) = ∅))
24	23	3impia 1135	1 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ 𝑋 ≠ 𝑌) → (𝐶 ∩ 𝐷) = ∅)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   ∨ wo 861   ∧ w3a 1103   = wceq 1570   ∈ wcel 2145   ≠ wne 2956  ∀wral 3077  ⦋csb 3847   ∩ cin 3898  ∅c0 4279  Disj wdisj 5070

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

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-rmo 3366  df-rab 3414  df-v 3453  df-sbc 3740  df-csb 3848  df-dif 3902  df-in 3906  df-nul 4280  df-disj 5071

This theorem is used by:  disji  5088  disjxiun  5100  voliunlem1  25871  symgcntz  33646  tocyccntz  33705  disjf1  46197