Theorem disji2 5127

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 2941	. . 3 ⊢ (𝑋 ≠ 𝑌 ↔ ¬ 𝑋 = 𝑌)
2		disjors 5126	. . . . . 6 ⊢ (Disj 𝑥 ∈ 𝐴 𝐵 ↔ ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
3		eqeq1 2741	. . . . . . . 8 ⊢ (𝑦 = 𝑋 → (𝑦 = 𝑧 ↔ 𝑋 = 𝑧))
4		nfcv 2905	. . . . . . . . . . 11 ⊢ Ⅎ𝑥𝑋
5		nfcv 2905	. . . . . . . . . . 11 ⊢ Ⅎ𝑥𝐶
6		disji.1	. . . . . . . . . . 11 ⊢ (𝑥 = 𝑋 → 𝐵 = 𝐶)
7	4, 5, 6	csbhypf 3927	. . . . . . . . . 10 ⊢ (𝑦 = 𝑋 → ⦋𝑦 / 𝑥⦌𝐵 = 𝐶)
8	7	ineq1d 4219	. . . . . . . . 9 ⊢ (𝑦 = 𝑋 → (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = (𝐶 ∩ ⦋𝑧 / 𝑥⦌𝐵))
9	8	eqeq1d 2739	. . . . . . . 8 ⊢ (𝑦 = 𝑋 → ((⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅ ↔ (𝐶 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅))
10	3, 9	orbi12d 919	. . . . . . 7 ⊢ (𝑦 = 𝑋 → ((𝑦 = 𝑧 ∨ (⦋𝑦 / 𝑥⦌𝐵 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅) ↔ (𝑋 = 𝑧 ∨ (𝐶 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅)))
11		eqeq2 2749	. . . . . . . 8 ⊢ (𝑧 = 𝑌 → (𝑋 = 𝑧 ↔ 𝑋 = 𝑌))
12		nfcv 2905	. . . . . . . . . . 11 ⊢ Ⅎ𝑥𝑌
13		nfcv 2905	. . . . . . . . . . 11 ⊢ Ⅎ𝑥𝐷
14		disji.2	. . . . . . . . . . 11 ⊢ (𝑥 = 𝑌 → 𝐵 = 𝐷)
15	12, 13, 14	csbhypf 3927	. . . . . . . . . 10 ⊢ (𝑧 = 𝑌 → ⦋𝑧 / 𝑥⦌𝐵 = 𝐷)
16	15	ineq2d 4220	. . . . . . . . 9 ⊢ (𝑧 = 𝑌 → (𝐶 ∩ ⦋𝑧 / 𝑥⦌𝐵) = (𝐶 ∩ 𝐷))
17	16	eqeq1d 2739	. . . . . . . 8 ⊢ (𝑧 = 𝑌 → ((𝐶 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅ ↔ (𝐶 ∩ 𝐷) = ∅))
18	11, 17	orbi12d 919	. . . . . . 7 ⊢ (𝑧 = 𝑌 → ((𝑋 = 𝑧 ∨ (𝐶 ∩ ⦋𝑧 / 𝑥⦌𝐵) = ∅) ↔ (𝑋 = 𝑌 ∨ (𝐶 ∩ 𝐷) = ∅)))
19	10, 18	rspc2v 3633	. . . . . 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 242	. . . . 5 ⊢ ((𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) → (Disj 𝑥 ∈ 𝐴 𝐵 → (𝑋 = 𝑌 ∨ (𝐶 ∩ 𝐷) = ∅)))
21	20	impcom 407	. . . 4 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (𝑋 = 𝑌 ∨ (𝐶 ∩ 𝐷) = ∅))
22	21	ord 865	. . 3 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (¬ 𝑋 = 𝑌 → (𝐶 ∩ 𝐷) = ∅))
23	1, 22	biimtrid 242	. 2 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (𝑋 ≠ 𝑌 → (𝐶 ∩ 𝐷) = ∅))
24	23	3impia 1118	1 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ 𝑋 ≠ 𝑌) → (𝐶 ∩ 𝐷) = ∅)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 395   ∨ wo 848   ∧ w3a 1087   = wceq 1540   ∈ wcel 2108   ≠ wne 2940  ∀wral 3061  ⦋csb 3899   ∩ cin 3950  ∅c0 4333  Disj wdisj 5110

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2157  ax-12 2177  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2065  df-mo 2540  df-clab 2715  df-cleq 2729  df-clel 2816  df-nfc 2892  df-ne 2941  df-ral 3062  df-rmo 3380  df-rab 3437  df-v 3482  df-sbc 3789  df-csb 3900  df-dif 3954  df-in 3958  df-nul 4334  df-disj 5111

This theorem is referenced by:  disji  5128  disjxiun  5140  voliunlem1  25585  symgcntz  33105  tocyccntz  33164  disjf1  45188