Nuprl Lemma : has-value-bor
∀[a,b:Base]. (a)↓ supposing (a ∨bb)↓
Proof
Definitions occuring in Statement :
bor: p ∨bq,
has-value: (a)↓,
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
base: Base
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
uimplies: b supposing a,
bor: p ∨bq,
ifthenelse: if b then t else f fi ,
has-value: (a)↓,
prop: ℙ
Lemmas referenced :
base_wf,
has-value_wf_base,
top_wf,
union-value-type,
value-type-has-value
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation,
introduction,
cut,
sqequalHypSubstitution,
callbyvalueDecide,
hypothesis,
lemma_by_obid,
isectElimination,
thin,
because_Cache,
independent_isectElimination,
sqequalRule,
axiomSqleEquality,
baseApply,
closedConclusion,
baseClosed,
hypothesisEquality,
isect_memberEquality,
equalityTransitivity,
equalitySymmetry
Latex:
\mforall{}[a,b:Base]. (a)\mdownarrow{} supposing (a \mvee{}\msubb{}b)\mdownarrow{}
Date html generated:
2016_05_13-PM-03_59_43
Last ObjectModification:
2016_01_14-PM-07_20_55
Theory : bool_1
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