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