Metamath Proof Explorer


Theorem fness

Description: A cover is finer than its subcovers. (Contributed by Jeff Hankins, 11-Oct-2009)

Ref Expression
Hypotheses fness.1 ⊢ X = ⋃ A
fness.2 ⊢ Y = ⋃ B
Assertion fness ⊢ B ∈ C ∧ A ⊆ B ∧ X = Y → A Fne B

Proof

Step Hyp Ref Expression
1 fness.1 ⊢ X = ⋃ A
2 fness.2 ⊢ Y = ⋃ B
3 simp3 ⊢ B ∈ C ∧ A ⊆ B ∧ X = Y → X = Y
4 ssel2 ⊢ A ⊆ B ∧ x ∈ A → x ∈ B
5 4 3adant3 ⊢ A ⊆ B ∧ x ∈ A ∧ y ∈ x → x ∈ B
6 simp3 ⊢ A ⊆ B ∧ x ∈ A ∧ y ∈ x → y ∈ x
7 ssid ⊢ x ⊆ x
8 6 7 jctir ⊢ A ⊆ B ∧ x ∈ A ∧ y ∈ x → y ∈ x ∧ x ⊆ x
9 elequ2 ⊢ z = x → y ∈ z ↔ y ∈ x
10 sseq1 ⊢ z = x → z ⊆ x ↔ x ⊆ x
11 9 10 anbi12d ⊢ z = x → y ∈ z ∧ z ⊆ x ↔ y ∈ x ∧ x ⊆ x
12 11 rspcev ⊢ x ∈ B ∧ y ∈ x ∧ x ⊆ x → ∃ z ∈ B y ∈ z ∧ z ⊆ x
13 5 8 12 syl2anc ⊢ A ⊆ B ∧ x ∈ A ∧ y ∈ x → ∃ z ∈ B y ∈ z ∧ z ⊆ x
14 13 3expib ⊢ A ⊆ B → x ∈ A ∧ y ∈ x → ∃ z ∈ B y ∈ z ∧ z ⊆ x
15 14 ralrimivv ⊢ A ⊆ B → ∀ x ∈ A ∀ y ∈ x ∃ z ∈ B y ∈ z ∧ z ⊆ x
16 15 3ad2ant2 ⊢ B ∈ C ∧ A ⊆ B ∧ X = Y → ∀ x ∈ A ∀ y ∈ x ∃ z ∈ B y ∈ z ∧ z ⊆ x
17 1 2 isfne2 ⊢ B ∈ C → A Fne B ↔ X = Y ∧ ∀ x ∈ A ∀ y ∈ x ∃ z ∈ B y ∈ z ∧ z ⊆ x
18 17 3ad2ant1 ⊢ B ∈ C ∧ A ⊆ B ∧ X = Y → A Fne B ↔ X = Y ∧ ∀ x ∈ A ∀ y ∈ x ∃ z ∈ B y ∈ z ∧ z ⊆ x
19 3 16 18 mpbir2and ⊢ B ∈ C ∧ A ⊆ B ∧ X = Y → A Fne B