Metamath Proof Explorer


Theorem dfttc4lem1

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

Ref Expression
Hypotheses dfttc4lem1.1 ⊢ B = x | ∃ y A ∩ y ≠ ∅ ∧ ∀ z ∈ y z ∩ y = ∅ → z = x
dfttc4lem1.2 ⊢ C ∈ V
dfttc4lem1.3 ⊢ D ∈ V
Assertion dfttc4lem1 ⊢ A ∩ C ≠ ∅ ∧ ∀ z ∈ C z ∩ C = ∅ → z = D → D ∈ B

Proof

Step Hyp Ref Expression
1 dfttc4lem1.1 ⊢ B = x | ∃ y A ∩ y ≠ ∅ ∧ ∀ z ∈ y z ∩ y = ∅ → z = x
2 dfttc4lem1.2 ⊢ C ∈ V
3 dfttc4lem1.3 ⊢ D ∈ V
4 ineq2 ⊢ y = C → A ∩ y = A ∩ C
5 4 neeq1d ⊢ y = C → A ∩ y ≠ ∅ ↔ A ∩ C ≠ ∅
6 ineq2 ⊢ y = C → z ∩ y = z ∩ C
7 6 eqeq1d ⊢ y = C → z ∩ y = ∅ ↔ z ∩ C = ∅
8 7 imbi1d ⊢ y = C → z ∩ y = ∅ → z = D ↔ z ∩ C = ∅ → z = D
9 8 raleqbi1dv ⊢ y = C → ∀ z ∈ y z ∩ y = ∅ → z = D ↔ ∀ z ∈ C z ∩ C = ∅ → z = D
10 5 9 anbi12d ⊢ y = C → A ∩ y ≠ ∅ ∧ ∀ z ∈ y z ∩ y = ∅ → z = D ↔ A ∩ C ≠ ∅ ∧ ∀ z ∈ C z ∩ C = ∅ → z = D
11 2 10 spcev ⊢ A ∩ C ≠ ∅ ∧ ∀ z ∈ C z ∩ C = ∅ → z = D → ∃ y A ∩ y ≠ ∅ ∧ ∀ z ∈ y z ∩ y = ∅ → z = D
12 eqeq2 ⊢ x = D → z = x ↔ z = D
13 12 imbi2d ⊢ x = D → z ∩ y = ∅ → z = x ↔ z ∩ y = ∅ → z = D
14 13 ralbidv ⊢ x = D → ∀ z ∈ y z ∩ y = ∅ → z = x ↔ ∀ z ∈ y z ∩ y = ∅ → z = D
15 14 anbi2d ⊢ x = D → A ∩ y ≠ ∅ ∧ ∀ z ∈ y z ∩ y = ∅ → z = x ↔ A ∩ y ≠ ∅ ∧ ∀ z ∈ y z ∩ y = ∅ → z = D
16 15 exbidv ⊢ x = D → ∃ y A ∩ y ≠ ∅ ∧ ∀ z ∈ y z ∩ y = ∅ → z = x ↔ ∃ y A ∩ y ≠ ∅ ∧ ∀ z ∈ y z ∩ y = ∅ → z = D
17 3 16 1 elab2 ⊢ D ∈ B ↔ ∃ y A ∩ y ≠ ∅ ∧ ∀ z ∈ y z ∩ y = ∅ → z = D
18 11 17 sylibr ⊢ A ∩ C ≠ ∅ ∧ ∀ z ∈ C z ∩ C = ∅ → z = D → D ∈ B