Theorem disji2f 33220

Description: Property of a disjoint collection: if 𝐵(𝑥) = 𝐶 and 𝐵(𝑌) = 𝐷, and 𝑥 ≠ 𝑌, then 𝐵 and 𝐶 are disjoint. (Contributed by Thierry Arnoux, 30-Dec-2016.)

Hypotheses

Ref	Expression
disjif.1	⊢ Ⅎ𝑥𝐶
disjif.2	⊢ (𝑥 = 𝑌 → 𝐵 = 𝐶)

Assertion

Ref	Expression
disji2f	⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ 𝑥 ≠ 𝑌) → (𝐵 ∩ 𝐶) = ∅)

Distinct variable groups:   𝑥,𝐴   𝑥,𝑌

Allowed substitution hints:   𝐵(𝑥)   𝐶(𝑥)

Proof of Theorem disji2f

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

Step	Hyp	Ref	Expression
1		df-ne 2956	. . 3 ⊢ (𝑥 ≠ 𝑌 ↔ ¬ 𝑥 = 𝑌)
2		disjors 5085	. . . . . 6 ⊢ (Disj 𝑥 ∈ 𝐴 𝐵 ↔ ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
3		equequ1 2057	. . . . . . . 8 ⊢ (𝑦 = 𝑥 → (𝑦 = 𝑧 ↔ 𝑥 = 𝑧))
4		csbeq1 3850	. . . . . . . . . . 11 ⊢ (𝑦 = 𝑥 → ⦋𝑦 / 𝑥⦌𝐵 = ⦋𝑥 / 𝑥⦌𝐵)
5		csbid 3860	. . . . . . . . . . 11 ⊢ ⦋𝑥 / 𝑥⦌𝐵 = 𝐵
6	4, 5	eqtrdi 2811	. . . . . . . . . 10 ⊢ (𝑦 = 𝑥 → ⦋𝑦 / 𝑥⦌𝐵 = 𝐵)
7	6	ineq1d 4115	. . . . . . . . 9 ⊢ (𝑦 = 𝑥 → (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = (𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵))
8	7	eqeq1d 2762	. . . . . . . 8 ⊢ (𝑦 = 𝑥 → ((⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅ ↔ (𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
9	3, 8	orbi12d 932	. . . . . . 7 ⊢ (𝑦 = 𝑥 → ((𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅) ↔ (𝑥 = 𝑧 ∨ (𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅)))
10		eqeq2 2772	. . . . . . . 8 ⊢ (𝑧 = 𝑌 → (𝑥 = 𝑧 ↔ 𝑥 = 𝑌))
11		nfcv 2922	. . . . . . . . . . 11 ⊢ Ⅎ𝑥𝑌
12		disjif.1	. . . . . . . . . . 11 ⊢ Ⅎ𝑥𝐶
13		disjif.2	. . . . . . . . . . 11 ⊢ (𝑥 = 𝑌 → 𝐵 = 𝐶)
14	11, 12, 13	csbhypf 3875	. . . . . . . . . 10 ⊢ (𝑧 = 𝑌 → ⦋𝑧 / 𝑥⦌𝐵 = 𝐶)
15	14	ineq2d 4116	. . . . . . . . 9 ⊢ (𝑧 = 𝑌 → (𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = (𝐵 ∩ 𝐶))
16	15	eqeq1d 2762	. . . . . . . 8 ⊢ (𝑧 = 𝑌 → ((𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅ ↔ (𝐵 ∩ 𝐶) = ∅))
17	10, 16	orbi12d 932	. . . . . . 7 ⊢ (𝑧 = 𝑌 → ((𝑥 = 𝑧 ∨ (𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅) ↔ (𝑥 = 𝑌 ∨ (𝐵 ∩ 𝐶) = ∅)))
18	9, 17	rspc2v 3586	. . . . . 6 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) → (∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅) → (𝑥 = 𝑌 ∨ (𝐵 ∩ 𝐶) = ∅)))
19	2, 18	biimtrid 245	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) → (Disj 𝑥 ∈ 𝐴 𝐵 → (𝑥 = 𝑌 ∨ (𝐵 ∩ 𝐶) = ∅)))
20	19	impcom 413	. . . 4 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (𝑥 = 𝑌 ∨ (𝐵 ∩ 𝐶) = ∅))
21	20	ord 878	. . 3 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (¬ 𝑥 = 𝑌 → (𝐵 ∩ 𝐶) = ∅))
22	1, 21	biimtrid 245	. 2 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (𝑥 ≠ 𝑌 → (𝐵 ∩ 𝐶) = ∅))
23	22	3impia 1135	1 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ 𝑥 ≠ 𝑌) → (𝐵 ∩ 𝐶) = ∅)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   ∨ wo 861   ∧ w3a 1103   = wceq 1569   ∈ wcel 2144  Ⅎwnfc 2907   ≠ wne 2955  ∀wral 3076  ⦋csb 3847   ∩ cin 4088  ∅c0 4277  Disj wdisj 5069

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-nf 1816  df-sb 2099  df-mo 2564  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-ne 2956  df-ral 3077  df-rmo 3365  df-rab 3413  df-v 3452  df-sbc 3739  df-csb 3848  df-dif 4046  df-in 4090  df-nul 4278  df-disj 5070

This theorem is used by:  disjif  33221