Nuprl Lemma : split-at-first-gap

[T:Type]
  ∀f:T ⟶ ℤ. ∀L:T List.
    (∃XY:{T List × (T List)| let X,Y XY 
                             in (L (X Y) ∈ (T List))
                                ∧ (∀i:ℕ||X|| 1. ((f X[i 1]) ((f X[i]) 1) ∈ ℤ))
                                ∧ ((¬↑null(L))
                                   ((¬↑null(X)) ∧ ¬((f hd(Y)) ((f last(X)) 1) ∈ ℤsupposing ||Y|| ≥ ))})


Proof




Definitions occuring in Statement :  last: last(L) select: L[n] hd: hd(l) length: ||as|| null: null(as) append: as bs list: List int_seg: {i..j-} assert: b uimplies: supposing a uall: [x:A]. B[x] ge: i ≥  all: x:A. B[x] sq_exists: x:{A| B[x]} not: ¬A implies:  Q and: P ∧ Q apply: a function: x:A ⟶ B[x] spread: spread def product: x:A × B[x] subtract: m add: m natural_number: $n int: universe: Type equal: t ∈ T
Definitions unfolded in proof :  uall: [x:A]. B[x] all: x:A. B[x] member: t ∈ T so_lambda: λ2y.t[x; y] so_apply: x[s1;s2] implies:  Q
Lemmas referenced :  split-at-first-rel equal_wf decidable__int_equal list_wf
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity isect_memberFormation lambdaFormation cut lemma_by_obid sqequalHypSubstitution isectElimination thin hypothesisEquality sqequalRule lambdaEquality intEquality applyEquality addEquality natural_numberEquality hypothesis independent_functionElimination dependent_functionElimination because_Cache functionEquality universeEquality

Latex:
\mforall{}[T:Type]
    \mforall{}f:T  {}\mrightarrow{}  \mBbbZ{}.  \mforall{}L:T  List.
        (\mexists{}XY:\{T  List  \mtimes{}  (T  List)|  let  X,Y  =  XY 
                                                          in  (L  =  (X  @  Y))
                                                                \mwedge{}  (\mforall{}i:\mBbbN{}||X||  -  1.  ((f  X[i  +  1])  =  ((f  X[i])  +  1)))
                                                                \mwedge{}  ((\mneg{}\muparrow{}null(L))
                                                                    {}\mRightarrow{}  ((\mneg{}\muparrow{}null(X))
                                                                          \mwedge{}  \mneg{}((f  hd(Y))  =  ((f  last(X))  +  1))  supposing  ||Y||  \mgeq{}  1  ))\})



Date html generated: 2016_05_15-PM-04_41_20
Last ObjectModification: 2015_12_27-PM-02_40_24

Theory : general


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