Nuprl Lemma : dM-lift-0
∀[I,J:fset(ℕ)]. ∀[f:I ⟶ J]. ((dM-lift(I;J;f) 0) = 0 ∈ Point(dM(I)))
Proof
Definitions occuring in Statement :
dM-lift: dM-lift(I;J;f),
names-hom: I ⟶ J,
dM0: 0,
dM: dM(I),
lattice-point: Point(l),
fset: fset(T),
nat: ℕ,
uall: ∀[x:A]. B[x],
apply: f a,
equal: s = t ∈ T
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
top: Top
Lemmas referenced :
dM-lift-0-sq,
dM0_wf,
names-hom_wf,
fset_wf,
nat_wf
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation,
introduction,
cut,
sqequalRule,
extract_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
isect_memberEquality,
voidElimination,
voidEquality,
hypothesis,
hypothesisEquality,
axiomEquality,
because_Cache
Latex:
\mforall{}[I,J:fset(\mBbbN{})]. \mforall{}[f:I {}\mrightarrow{} J]. ((dM-lift(I;J;f) 0) = 0)
Date html generated:
2018_05_23-AM-08_27_51
Last ObjectModification:
2018_05_20-PM-05_35_58
Theory : cubical!type!theory
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