Metamath Proof Explorer


Theorem ipolub0

Description: The LUB of the empty set is the intersection of the base. (Contributed by Zhi Wang, 30-Sep-2024)

Ref Expression
Hypotheses ipoglb0.i
|- I = ( toInc ` F )
ipolub0.u
|- ( ph -> U = ( lub ` I ) )
ipolub0.f
|- ( ph -> |^| F e. F )
ipolub0.v
|- ( ph -> F e. V )
Assertion ipolub0
|- ( ph -> ( U ` (/) ) = |^| F )

Proof

Step Hyp Ref Expression
1 ipoglb0.i
 |-  I = ( toInc ` F )
2 ipolub0.u
 |-  ( ph -> U = ( lub ` I ) )
3 ipolub0.f
 |-  ( ph -> |^| F e. F )
4 ipolub0.v
 |-  ( ph -> F e. V )
5 0ss
 |-  (/) C_ F
6 5 a1i
 |-  ( ph -> (/) C_ F )
7 uni0
 |-  U. (/) = (/)
8 0ss
 |-  (/) C_ x
9 7 8 eqsstri
 |-  U. (/) C_ x
10 9 a1i
 |-  ( x e. F -> U. (/) C_ x )
11 10 rabeqc
 |-  { x e. F | U. (/) C_ x } = F
12 11 eqcomi
 |-  F = { x e. F | U. (/) C_ x }
13 12 inteqi
 |-  |^| F = |^| { x e. F | U. (/) C_ x }
14 13 a1i
 |-  ( ph -> |^| F = |^| { x e. F | U. (/) C_ x } )
15 1 4 6 2 14 3 ipolub
 |-  ( ph -> ( U ` (/) ) = |^| F )