Nuprl Lemma : test-decide-normalize

  (case of inl(_) => inr(y) => B[y] case of inl(x) => (inl x) inr(y) => B[y] (inr ))


Definitions occuring in Statement :  uall: [x:A]. B[x] top: Top so_apply: x[s] decide: case of inl(x) => s[x] inr(y) => t[y] inr: inr  inl: inl x add: m natural_number: $n sqequal: t
Definitions unfolded in proof :  top: Top member: t ∈ T all: x:A. B[x] implies:  Q has-value: (a)↓ uall: [x:A]. B[x] prop:
Lemmas referenced :  top_wf equal_wf has-value_wf_base is-exception_wf
Rules used in proof :  sqequalSubstitution sqequalRule sqequalReflexivity cut sqequalTransitivity computationStep isect_memberEquality voidElimination voidEquality thin introduction extract_by_obid hypothesis lambdaFormation sqequalSqle divergentSqle callbyvalueDecide sqequalHypSubstitution hypothesisEquality unionEquality unionElimination sqleReflexivity isectElimination equalityTransitivity equalitySymmetry dependent_functionElimination independent_functionElimination decideExceptionCases axiomSqleEquality exceptionSqequal baseApply closedConclusion baseClosed isect_memberFormation sqequalAxiom because_Cache

    (case  a  of  inl($_{}$)  =>  a  +  1  |  inr(y)  =>  B[y]  +  a  \msim{}  case  a
      of  inl(x)  =>
      (inl  x)  +  1
      |  inr(y)  =>
      B[y]  +  (inr  y  ))

Date html generated: 2017_04_14-AM-07_35_19
Last ObjectModification: 2017_02_27-PM-03_08_13

Theory : fun_1

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