Metamath Proof Explorer


Theorem isfne

Description: The predicate " B is finer than A ". This property is, in a sense, the opposite of refinement, as refinement requires every element to be a subset of an element of the original and fineness requires that every element of the original have a subset in the finer cover containing every point. I do not know of a literature reference for this. (Contributed by Jeff Hankins, 28-Sep-2009)

Ref Expression
Hypotheses isfne.1 ⊢ X = ⋃ A
isfne.2 ⊢ Y = ⋃ B
Assertion isfne ⊢ B ∈ C → A Fne B ↔ X = Y ∧ ∀ x ∈ A x ⊆ ⋃ B ∩ 𝒫 x

Proof

Step Hyp Ref Expression
1 isfne.1 ⊢ X = ⋃ A
2 isfne.2 ⊢ Y = ⋃ B
3 fnerel ⊢ Rel ⁡ Fne
4 3 brrelex1i ⊢ A Fne B → A ∈ V
5 4 anim1i ⊢ A Fne B ∧ B ∈ C → A ∈ V ∧ B ∈ C
6 5 ancoms ⊢ B ∈ C ∧ A Fne B → A ∈ V ∧ B ∈ C
7 simpr ⊢ B ∈ C ∧ X = Y → X = Y
8 7 1 2 3eqtr3g ⊢ B ∈ C ∧ X = Y → ⋃ A = ⋃ B
9 simpr ⊢ B ∈ C ∧ ⋃ A = ⋃ B → ⋃ A = ⋃ B
10 uniexg ⊢ B ∈ C → ⋃ B ∈ V
11 10 adantr ⊢ B ∈ C ∧ ⋃ A = ⋃ B → ⋃ B ∈ V
12 9 11 eqeltrd ⊢ B ∈ C ∧ ⋃ A = ⋃ B → ⋃ A ∈ V
13 uniexb ⊢ A ∈ V ↔ ⋃ A ∈ V
14 12 13 sylibr ⊢ B ∈ C ∧ ⋃ A = ⋃ B → A ∈ V
15 simpl ⊢ B ∈ C ∧ ⋃ A = ⋃ B → B ∈ C
16 14 15 jca ⊢ B ∈ C ∧ ⋃ A = ⋃ B → A ∈ V ∧ B ∈ C
17 8 16 syldan ⊢ B ∈ C ∧ X = Y → A ∈ V ∧ B ∈ C
18 17 adantrr ⊢ B ∈ C ∧ X = Y ∧ ∀ x ∈ A x ⊆ ⋃ B ∩ 𝒫 x → A ∈ V ∧ B ∈ C
19 unieq ⊢ r = A → ⋃ r = ⋃ A
20 19 1 eqtr4di ⊢ r = A → ⋃ r = X
21 20 eqeq1d ⊢ r = A → ⋃ r = ⋃ s ↔ X = ⋃ s
22 raleq ⊢ r = A → ∀ x ∈ r x ⊆ ⋃ s ∩ 𝒫 x ↔ ∀ x ∈ A x ⊆ ⋃ s ∩ 𝒫 x
23 21 22 anbi12d ⊢ r = A → ⋃ r = ⋃ s ∧ ∀ x ∈ r x ⊆ ⋃ s ∩ 𝒫 x ↔ X = ⋃ s ∧ ∀ x ∈ A x ⊆ ⋃ s ∩ 𝒫 x
24 unieq ⊢ s = B → ⋃ s = ⋃ B
25 24 2 eqtr4di ⊢ s = B → ⋃ s = Y
26 25 eqeq2d ⊢ s = B → X = ⋃ s ↔ X = Y
27 ineq1 ⊢ s = B → s ∩ 𝒫 x = B ∩ 𝒫 x
28 27 unieqd ⊢ s = B → ⋃ s ∩ 𝒫 x = ⋃ B ∩ 𝒫 x
29 28 sseq2d ⊢ s = B → x ⊆ ⋃ s ∩ 𝒫 x ↔ x ⊆ ⋃ B ∩ 𝒫 x
30 29 ralbidv ⊢ s = B → ∀ x ∈ A x ⊆ ⋃ s ∩ 𝒫 x ↔ ∀ x ∈ A x ⊆ ⋃ B ∩ 𝒫 x
31 26 30 anbi12d ⊢ s = B → X = ⋃ s ∧ ∀ x ∈ A x ⊆ ⋃ s ∩ 𝒫 x ↔ X = Y ∧ ∀ x ∈ A x ⊆ ⋃ B ∩ 𝒫 x
32 df-fne ⊢ Fne = r s | ⋃ r = ⋃ s ∧ ∀ x ∈ r x ⊆ ⋃ s ∩ 𝒫 x
33 23 31 32 brabg ⊢ A ∈ V ∧ B ∈ C → A Fne B ↔ X = Y ∧ ∀ x ∈ A x ⊆ ⋃ B ∩ 𝒫 x
34 6 18 33 pm5.21nd ⊢ B ∈ C → A Fne B ↔ X = Y ∧ ∀ x ∈ A x ⊆ ⋃ B ∩ 𝒫 x