Nuprl Lemma : longest-prefix-is-nil

∀[T:Type]. ∀[L:T List]. ∀[P:(T List) ⟶ 𝔹].
  ∀[L':T List]. (¬↑(P L')) supposing (L' < L and [] < L') supposing longest-prefix(P;L) = [] ∈ (T List)


Proof




Definitions occuring in Statement :  longest-prefix: longest-prefix(P;L),  proper-iseg: L1 < L2,  nil: [],  list: T List,  assert: ↑b,  bool: 𝔹,  uimplies: b supposing a,  uall: ∀[x:A]. B[x],  not: ¬A,  apply: f a,  function: x:A ⟶ B[x],  universe: Type,  equal: s = t ∈ T
Definitions unfolded in proof :  uall: ∀[x:A]. B[x],  member: t ∈ T,  all: ∀x:A. B[x],  subtype_rel: A ⊆r B,  so_lambda: λ2x.t[x],  so_apply: x[s],  uimplies: b supposing a,  listp: A List+,  implies: P ⇒ Q,  not: ¬A,  false: False,  and: P ∧ Q,  or: P ∨ Q,  prop: ℙ,  guard: {T}
Lemmas referenced :  longest-prefix_property,  longest-prefix_wf,  subtype_rel_dep_function,  list_wf,  bool_wf,  listp_wf,  subtype_rel_self,  assert_wf,  proper-iseg_wf,  nil_wf,  equal-wf-T-base,  iseg_wf,  less_than_wf,  length_wf,  or_wf,  length_of_nil_lemma,  all_wf,  not_wf,  equal_wf,  and_wf
Rules used in proof :  cut,  introduction,  extract_by_obid,  sqequalSubstitution,  sqequalTransitivity,  computationStep,  sqequalReflexivity,  isect_memberFormation,  hypothesis,  sqequalHypSubstitution,  isectElimination,  thin,  hypothesisEquality,  dependent_functionElimination,  cumulativity,  applyEquality,  sqequalRule,  lambdaEquality,  independent_isectElimination,  setElimination,  rename,  because_Cache,  lambdaFormation,  productElimination,  unionElimination,  functionExtensionality,  independent_functionElimination,  voidElimination,  isect_memberEquality,  equalityTransitivity,  equalitySymmetry,  baseClosed,  productEquality,  isectEquality,  natural_numberEquality,  applyLambdaEquality,  functionEquality,  universeEquality,  hyp_replacement,  dependent_set_memberEquality,  independent_pairFormation

Latex:
\mforall{}[T:Type].  \mforall{}[L:T  List].  \mforall{}[P:(T  List)  {}\mrightarrow{}  \mBbbB{}].
    \mforall{}[L':T  List].  (\mneg{}\muparrow{}(P  L'))  supposing  (L'  <  L  and  []  <  L')  supposing  longest-prefix(P;L)  =  []



Date html generated: 2018_05_21-PM-06_42_04
Last ObjectModification: 2017_07_26-PM-04_54_04

Theory : general


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