Nuprl Lemma : finite-Ramsey

∀k,n:ℕ.
  ∃N:ℕ+
   ∀g:ℕN ⟶ ℕN ⟶ ℕk
     ∃f:ℕn ⟶ ℕN. (Inj(ℕn;ℕN;f) ∧ (∀a,b,c,d:ℕn.  (f a < f b ⇒ f c < f d ⇒ ((g (f a) (f b)) = (g (f c) (f d)) ∈ ℤ))))


Proof




Definitions occuring in Statement :  inject: Inj(A;B;f),  int_seg: {i..j-},  nat_plus: ℕ+,  nat: ℕ,  less_than: a < b,  all: ∀x:A. B[x],  exists: ∃x:A. B[x],  implies: P ⇒ Q,  and: P ∧ Q,  apply: f a,  function: x:A ⟶ B[x],  natural_number: $n,  int: ℤ,  equal: s = t ∈ T
Definitions unfolded in proof :  all: ∀x:A. B[x],  member: t ∈ T,  uall: ∀[x:A]. B[x],  nat: ℕ,  exists: ∃x:A. B[x],  and: P ∧ Q,  cand: A c∧ B,  implies: P ⇒ Q,  prop: ℙ,  subtype_rel: A ⊆r B,  int_seg: {i..j-},  nat_plus: ℕ+,  so_lambda: λ2x.t[x],  so_apply: x[s],  squash: ↓T,  true: True,  uimplies: b supposing a,  guard: {T},  iff: P ⇐⇒ Q,  rev_implies: P ⇐ Q
Lemmas referenced :  finite-Ramsey1,  int_seg_wf,  less_than_wf,  inject_wf,  all_wf,  equal_wf,  exists_wf,  nat_wf,  squash_wf,  true_wf,  iff_weakening_equal
Rules used in proof :  cut,  introduction,  extract_by_obid,  sqequalSubstitution,  sqequalTransitivity,  computationStep,  sqequalReflexivity,  lambdaFormation,  hypothesis,  sqequalHypSubstitution,  dependent_functionElimination,  thin,  hypothesisEquality,  lambdaEquality,  isectElimination,  natural_numberEquality,  setElimination,  rename,  productElimination,  dependent_pairFormation,  sqequalRule,  independent_pairFormation,  applyEquality,  functionExtensionality,  because_Cache,  productEquality,  functionEquality,  intEquality,  imageElimination,  equalityTransitivity,  equalitySymmetry,  universeEquality,  independent_functionElimination,  imageMemberEquality,  baseClosed,  independent_isectElimination

Latex:
\mforall{}k,n:\mBbbN{}.
    \mexists{}N:\mBbbN{}\msupplus{}
      \mforall{}g:\mBbbN{}N  {}\mrightarrow{}  \mBbbN{}N  {}\mrightarrow{}  \mBbbN{}k
          \mexists{}f:\mBbbN{}n  {}\mrightarrow{}  \mBbbN{}N
            (Inj(\mBbbN{}n;\mBbbN{}N;f)
            \mwedge{}  (\mforall{}a,b,c,d:\mBbbN{}n.    (f  a  <  f  b  {}\mRightarrow{}  f  c  <  f  d  {}\mRightarrow{}  ((g  (f  a)  (f  b))  =  (g  (f  c)  (f  d))))))



Date html generated: 2017_04_20-AM-07_26_05
Last ObjectModification: 2017_02_27-PM-05_59_29

Theory : continuity


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