Theorem dfttc4lem1 37238

Description: Lemma for dfttc4 37240. (Contributed by Matthew House, 6-Apr-2026.)

Hypotheses

Ref	Expression
dfttc4lem1.1	⊢ 𝐵 = {𝑥 ∣ ∃𝑦((𝐴 ∩ 𝑦) ≠ ∅ ∧ ∀𝑧 ∈ 𝑦 ((𝑧 ∩ 𝑦) = ∅ → 𝑧 = 𝑥))}
dfttc4lem1.2	⊢ 𝐶 ∈ V
dfttc4lem1.3	⊢ 𝐷 ∈ V

Assertion

Ref	Expression
dfttc4lem1	⊢ (((𝐴 ∩ 𝐶) ≠ ∅ ∧ ∀𝑧 ∈ 𝐶 ((𝑧 ∩ 𝐶) = ∅ → 𝑧 = 𝐷)) → 𝐷 ∈ 𝐵)

Distinct variable groups:   𝑥,𝐴,𝑦   𝑦,𝐶,𝑧   𝑥,𝐷,𝑦,𝑧

Allowed substitution hints:   𝐴(𝑧)   𝐵(𝑥, 𝑦, 𝑧)   𝐶(𝑥)

Proof of Theorem dfttc4lem1

Step	Hyp	Ref	Expression
1		dfttc4lem1.2	. . 3 ⊢ 𝐶 ∈ V
2		ineq2 4159	. . . . 5 ⊢ (𝑦 = 𝐶 → (𝐴 ∩ 𝑦) = (𝐴 ∩ 𝐶))
3	2	neeq1d 3014	. . . 4 ⊢ (𝑦 = 𝐶 → ((𝐴 ∩ 𝑦) ≠ ∅ ↔ (𝐴 ∩ 𝐶) ≠ ∅))
4		ineq2 4159	. . . . . . 7 ⊢ (𝑦 = 𝐶 → (𝑧 ∩ 𝑦) = (𝑧 ∩ 𝐶))
5	4	eqeq1d 2762	. . . . . 6 ⊢ (𝑦 = 𝐶 → ((𝑧 ∩ 𝑦) = ∅ ↔ (𝑧 ∩ 𝐶) = ∅))
6	5	imbi1d 344	. . . . 5 ⊢ (𝑦 = 𝐶 → (((𝑧 ∩ 𝑦) = ∅ → 𝑧 = 𝐷) ↔ ((𝑧 ∩ 𝐶) = ∅ → 𝑧 = 𝐷)))
7	6	raleqbi1dv 3329	. . . 4 ⊢ (𝑦 = 𝐶 → (∀𝑧 ∈ 𝑦 ((𝑧 ∩ 𝑦) = ∅ → 𝑧 = 𝐷) ↔ ∀𝑧 ∈ 𝐶 ((𝑧 ∩ 𝐶) = ∅ → 𝑧 = 𝐷)))
8	3, 7	anbi12d 644	. . 3 ⊢ (𝑦 = 𝐶 → (((𝐴 ∩ 𝑦) ≠ ∅ ∧ ∀𝑧 ∈ 𝑦 ((𝑧 ∩ 𝑦) = ∅ → 𝑧 = 𝐷)) ↔ ((𝐴 ∩ 𝐶) ≠ ∅ ∧ ∀𝑧 ∈ 𝐶 ((𝑧 ∩ 𝐶) = ∅ → 𝑧 = 𝐷))))
9	1, 8	spcev 3560	. 2 ⊢ (((𝐴 ∩ 𝐶) ≠ ∅ ∧ ∀𝑧 ∈ 𝐶 ((𝑧 ∩ 𝐶) = ∅ → 𝑧 = 𝐷)) → ∃𝑦((𝐴 ∩ 𝑦) ≠ ∅ ∧ ∀𝑧 ∈ 𝑦 ((𝑧 ∩ 𝑦) = ∅ → 𝑧 = 𝐷)))
10		dfttc4lem1.3	. . 3 ⊢ 𝐷 ∈ V
11		eqeq2 2772	. . . . . . 7 ⊢ (𝑥 = 𝐷 → (𝑧 = 𝑥 ↔ 𝑧 = 𝐷))
12	11	imbi2d 343	. . . . . 6 ⊢ (𝑥 = 𝐷 → (((𝑧 ∩ 𝑦) = ∅ → 𝑧 = 𝑥) ↔ ((𝑧 ∩ 𝑦) = ∅ → 𝑧 = 𝐷)))
13	12	ralbidv 3185	. . . . 5 ⊢ (𝑥 = 𝐷 → (∀𝑧 ∈ 𝑦 ((𝑧 ∩ 𝑦) = ∅ → 𝑧 = 𝑥) ↔ ∀𝑧 ∈ 𝑦 ((𝑧 ∩ 𝑦) = ∅ → 𝑧 = 𝐷)))
14	13	anbi2d 642	. . . 4 ⊢ (𝑥 = 𝐷 → (((𝐴 ∩ 𝑦) ≠ ∅ ∧ ∀𝑧 ∈ 𝑦 ((𝑧 ∩ 𝑦) = ∅ → 𝑧 = 𝑥)) ↔ ((𝐴 ∩ 𝑦) ≠ ∅ ∧ ∀𝑧 ∈ 𝑦 ((𝑧 ∩ 𝑦) = ∅ → 𝑧 = 𝐷))))
15	14	exbidv 1954	. . 3 ⊢ (𝑥 = 𝐷 → (∃𝑦((𝐴 ∩ 𝑦) ≠ ∅ ∧ ∀𝑧 ∈ 𝑦 ((𝑧 ∩ 𝑦) = ∅ → 𝑧 = 𝑥)) ↔ ∃𝑦((𝐴 ∩ 𝑦) ≠ ∅ ∧ ∀𝑧 ∈ 𝑦 ((𝑧 ∩ 𝑦) = ∅ → 𝑧 = 𝐷))))
16		dfttc4lem1.1	. . 3 ⊢ 𝐵 = {𝑥 ∣ ∃𝑦((𝐴 ∩ 𝑦) ≠ ∅ ∧ ∀𝑧 ∈ 𝑦 ((𝑧 ∩ 𝑦) = ∅ → 𝑧 = 𝑥))}
17	10, 15, 16	elab2 3635	. 2 ⊢ (𝐷 ∈ 𝐵 ↔ ∃𝑦((𝐴 ∩ 𝑦) ≠ ∅ ∧ ∀𝑧 ∈ 𝑦 ((𝑧 ∩ 𝑦) = ∅ → 𝑧 = 𝐷)))
18	9, 17	sylibr 237	1 ⊢ (((𝐴 ∩ 𝐶) ≠ ∅ ∧ ∀𝑧 ∈ 𝐶 ((𝑧 ∩ 𝐶) = ∅ → 𝑧 = 𝐷)) → 𝐷 ∈ 𝐵)

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570  ∃wex 1812   ∈ wcel 2145  {cab 2738   ≠ wne 2955  ∀wral 3076  Vcvv 3450   ∩ cin 3897  ∅c0 4278

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-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-ne 2956  df-ral 3077  df-rex 3087  df-rab 3413  df-in 3905

This theorem is used by:  dfttc4lem2  37239