Nuprl Lemma : has-valueall-if-has-value-callbyvalueall
∀[a,b:Base]. has-valueall(a) supposing (let x ⟵ a in b[x])↓
Proof
Definitions occuring in Statement :
has-valueall: has-valueall(a),
has-value: (a)↓,
callbyvalueall: callbyvalueall,
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
so_apply: x[s],
base: Base
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
uimplies: b supposing a,
has-valueall: has-valueall(a),
callbyvalueall: callbyvalueall,
has-value: (a)↓,
prop: ℙ
Lemmas referenced :
base_wf,
has-value_wf_base
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation,
introduction,
cut,
sqequalHypSubstitution,
callbyvalueCallbyvalue,
hypothesis,
sqequalRule,
callbyvalueReduce,
axiomSqleEquality,
lemma_by_obid,
isectElimination,
thin,
baseApply,
closedConclusion,
baseClosed,
hypothesisEquality,
isect_memberEquality,
because_Cache,
equalityTransitivity,
equalitySymmetry
Latex:
\mforall{}[a,b:Base]. has-valueall(a) supposing (let x \mleftarrow{}{} a in b[x])\mdownarrow{}
Date html generated:
2016_05_13-PM-03_25_23
Last ObjectModification:
2016_01_14-PM-06_44_46
Theory : call!by!value_1
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