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

t,a,b:Base.  ((t)↓  ((if is inr then else a) ∨ (if is inr then else b)))


Proof




Definitions occuring in Statement :  has-value: (a)↓ isinr: isinr def all: x:A. B[x] implies:  Q or: P ∨ Q base: Base sqequal: t
Definitions unfolded in proof :  all: x:A. B[x] implies:  Q has-value: (a)↓ member: t ∈ T uall: [x:A]. B[x] or: P ∨ Q top: Top guard: {T} prop:
Lemmas referenced :  base_wf top_wf is-exception_wf has-value_wf_base
Rules used in proof :  sqequalSubstitution sqequalTransitivity computationStep sqequalReflexivity lambdaFormation isinrCases divergentSqle hypothesis cut lemma_by_obid sqequalHypSubstitution isectElimination thin baseClosed hypothesisEquality sqequalRule inlFormation sqequalIntensionalEquality isect_memberFormation introduction sqequalAxiom isect_memberEquality because_Cache voidElimination voidEquality inrFormation

Latex:
\mforall{}t,a,b:Base.    ((t)\mdownarrow{}  {}\mRightarrow{}  ((if  t  is  inr  then  a  else  b  \msim{}  a)  \mvee{}  (if  t  is  inr  then  a  else  b  \msim{}  b)))



Date html generated: 2016_05_13-PM-03_22_27
Last ObjectModification: 2016_01_14-PM-06_46_53

Theory : call!by!value_1


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