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

t:Base. ((t)↓  Dec(t ~ <fst(t), snd(t)>))


Proof




Definitions occuring in Statement :  has-value: (a)↓ decidable: Dec(P) pi1: fst(t) pi2: snd(t) all: x:A. B[x] implies:  Q pair: <a, b> base: Base sqequal: t
Definitions unfolded in proof :  all: x:A. B[x] implies:  Q member: t ∈ T has-value: (a)↓ uall: [x:A]. B[x] pi1: fst(t) pi2: snd(t) decidable: Dec(P) or: P ∨ Q prop: top: Top not: ¬A false: False
Lemmas referenced :  base_wf not_zero_sqequal_one top_wf not_wf is-exception_wf has-value_wf_base
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity lambdaFormation introduction ispairCases divergentSqle hypothesis cut lemma_by_obid sqequalHypSubstitution isectElimination thin baseClosed hypothesisEquality sqequalRule inlEquality sqequalAxiom sqequalIntensionalEquality baseApply closedConclusion isect_memberFormation isect_memberEquality because_Cache voidElimination voidEquality inrEquality lambdaEquality independent_functionElimination

Latex:
\mforall{}t:Base.  ((t)\mdownarrow{}  {}\mRightarrow{}  Dec(t  \msim{}  <fst(t),  snd(t)>))



Date html generated: 2016_05_13-PM-03_22_12
Last ObjectModification: 2016_01_14-PM-06_47_03

Theory : call!by!value_1


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