Theorem disji2f 28772

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 2782	. . 3 ⊢ (𝑥 ≠ 𝑌 ↔ ¬ 𝑥 = 𝑌)
2		disjors 4568	. . . . . 6 ⊢ (Disj 𝑥 ∈ 𝐴 𝐵 ↔ ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
3		equequ1 1939	. . . . . . . 8 ⊢ (𝑦 = 𝑥 → (𝑦 = 𝑧 ↔ 𝑥 = 𝑧))
4		csbeq1 3502	. . . . . . . . . . 11 ⊢ (𝑦 = 𝑥 → ⦋𝑦 / 𝑥⦌𝐵 = ⦋𝑥 / 𝑥⦌𝐵)
5		csbid 3507	. . . . . . . . . . 11 ⊢ ⦋𝑥 / 𝑥⦌𝐵 = 𝐵
6	4, 5	syl6eq 2660	. . . . . . . . . 10 ⊢ (𝑦 = 𝑥 → ⦋𝑦 / 𝑥⦌𝐵 = 𝐵)
7	6	ineq1d 3775	. . . . . . . . 9 ⊢ (𝑦 = 𝑥 → (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = (𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵))
8	7	eqeq1d 2612	. . . . . . . 8 ⊢ (𝑦 = 𝑥 → ((⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅ ↔ (𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
9	3, 8	orbi12d 742	. . . . . . 7 ⊢ (𝑦 = 𝑥 → ((𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅) ↔ (𝑥 = 𝑧 ∨ (𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅)))
10		eqeq2 2621	. . . . . . . 8 ⊢ (𝑧 = 𝑌 → (𝑥 = 𝑧 ↔ 𝑥 = 𝑌))
11		nfcv 2751	. . . . . . . . . . 11 ⊢ Ⅎ𝑥𝑌
12		disjif.1	. . . . . . . . . . 11 ⊢ Ⅎ𝑥𝐶
13		disjif.2	. . . . . . . . . . 11 ⊢ (𝑥 = 𝑌 → 𝐵 = 𝐶)
14	11, 12, 13	csbhypf 3518	. . . . . . . . . 10 ⊢ (𝑧 = 𝑌 → ⦋𝑧 / 𝑥⦌𝐵 = 𝐶)
15	14	ineq2d 3776	. . . . . . . . 9 ⊢ (𝑧 = 𝑌 → (𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = (𝐵 ∩ 𝐶))
16	15	eqeq1d 2612	. . . . . . . 8 ⊢ (𝑧 = 𝑌 → ((𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅ ↔ (𝐵 ∩ 𝐶) = ∅))
17	10, 16	orbi12d 742	. . . . . . 7 ⊢ (𝑧 = 𝑌 → ((𝑥 = 𝑧 ∨ (𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅) ↔ (𝑥 = 𝑌 ∨ (𝐵 ∩ 𝐶) = ∅)))
18	9, 17	rspc2v 3293	. . . . . 6 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) → (∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅) → (𝑥 = 𝑌 ∨ (𝐵 ∩ 𝐶) = ∅)))
19	2, 18	syl5bi 231	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) → (Disj 𝑥 ∈ 𝐴 𝐵 → (𝑥 = 𝑌 ∨ (𝐵 ∩ 𝐶) = ∅)))
20	19	impcom 445	. . . 4 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (𝑥 = 𝑌 ∨ (𝐵 ∩ 𝐶) = ∅))
21	20	ord 391	. . 3 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (¬ 𝑥 = 𝑌 → (𝐵 ∩ 𝐶) = ∅))
22	1, 21	syl5bi 231	. 2 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (𝑥 ≠ 𝑌 → (𝐵 ∩ 𝐶) = ∅))
23	22	3impia 1253	1 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ 𝑥 ≠ 𝑌) → (𝐵 ∩ 𝐶) = ∅)

Syntax hints:  ¬ wn 3   → wi 4   ∨ wo 382   ∧ wa 383   ∧ w3a 1031   = wceq 1475   ∈ wcel 1977  Ⅎwnfc 2738   ≠ wne 2780  ∀wral 2896  ⦋csb 3499   ∩ cin 3539  ∅c0 3874  Disj wdisj 4553

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-3an 1033  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-eu 2462  df-mo 2463  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-ne 2782  df-ral 2901  df-rex 2902  df-reu 2903  df-rmo 2904  df-v 3175  df-sbc 3403  df-csb 3500  df-dif 3543  df-in 3547  df-nul 3875  df-disj 4554

This theorem is referenced by:  disjif  28773