Nuprl Lemma : filter_trivial2
∀[T:Type]. ∀[P:T ⟶ 𝔹]. ∀[L:T List]. filter(P;L) = L ∈ (T List) supposing (∀x∈L.↑P[x])
Proof
Definitions occuring in Statement :
l_all: (∀x∈L.P[x]),
filter: filter(P;l),
list: T List,
assert: ↑b,
bool: 𝔹,
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
so_apply: x[s],
function: x:A ⟶ B[x],
universe: Type,
equal: s = t ∈ T
Definitions unfolded in proof :
prop: ℙ,
so_apply: x[s],
so_lambda: λ2x.t[x],
uimplies: b supposing a,
member: t ∈ T,
uall: ∀[x:A]. B[x]
Lemmas referenced :
istype-universe,
bool_wf,
list_wf,
l_member_wf,
assert_wf,
l_all_wf,
filter_trivial
Rules used in proof :
universeEquality,
instantiate,
Error :functionIsType,
Error :inhabitedIsType,
Error :isectIsTypeImplies,
axiomEquality,
Error :isect_memberEquality_alt,
Error :setIsType,
rename,
setElimination,
applyEquality,
Error :lambdaEquality_alt,
Error :universeIsType,
hypothesis,
independent_isectElimination,
hypothesisEquality,
thin,
isectElimination,
sqequalHypSubstitution,
extract_by_obid,
sqequalRule,
cut,
introduction,
Error :isect_memberFormation_alt,
sqequalReflexivity,
computationStep,
sqequalTransitivity,
sqequalSubstitution
Latex:
\mforall{}[T:Type]. \mforall{}[P:T {}\mrightarrow{} \mBbbB{}]. \mforall{}[L:T List]. filter(P;L) = L supposing (\mforall{}x\mmember{}L.\muparrow{}P[x])
Date html generated:
2019_06_20-PM-01_05_36
Last ObjectModification:
2019_06_20-PM-00_45_15
Theory : list_0
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