Nuprl Lemma : decide-int-if-has-value

t:Base. ((t)↓  Dec(t ∈ ℤ))


Proof




Definitions occuring in Statement :  has-value: (a)↓ decidable: Dec(P) all: x:A. B[x] implies:  Q member: t ∈ T int: base: Base
Definitions unfolded in proof :  all: x:A. B[x] implies:  Q member: t ∈ T has-value: (a)↓ uall: [x:A]. B[x] decidable: Dec(P) or: P ∨ Q prop: top: Top not: ¬A assert: b ifthenelse: if then else fi  btrue: tt true: True bfalse: ff false: False
Lemmas referenced :  base_wf top_wf member_wf not_wf is-exception_wf has-value_wf_base
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity lambdaFormation introduction isintCases divergentSqle hypothesis cut lemma_by_obid sqequalHypSubstitution isectElimination thin baseClosed hypothesisEquality sqequalRule isintReduceTrue equalityTransitivity equalitySymmetry inlEquality axiomEquality intEquality isect_memberFormation sqequalAxiom isect_memberEquality because_Cache voidElimination voidEquality inrEquality lambdaEquality natural_numberEquality

Latex:
\mforall{}t:Base.  ((t)\mdownarrow{}  {}\mRightarrow{}  Dec(t  \mmember{}  \mBbbZ{}))



Date html generated: 2016_05_13-PM-03_22_18
Last ObjectModification: 2016_01_14-PM-06_47_12

Theory : call!by!value_1


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