Nuprl Lemma : valueall-type-has-valueall
∀[T:Type]. ∀[x:T]. has-valueall(x) supposing valueall-type(T)
Proof
Definitions occuring in Statement : 
valueall-type: valueall-type(T)
, 
has-valueall: has-valueall(a)
, 
uimplies: b supposing a
, 
uall: ∀[x:A]. B[x]
, 
universe: Type
Definitions unfolded in proof : 
uall: ∀[x:A]. B[x]
, 
member: t ∈ T
, 
uimplies: b supposing a
, 
has-valueall: has-valueall(a)
, 
has-value: (a)↓
, 
valueall-type: valueall-type(T)
, 
prop: ℙ
Lemmas referenced : 
istype-universe, 
valueall-type_wf, 
sqle_wf_base
Rules used in proof : 
sqequalSubstitution, 
sqequalTransitivity, 
computationStep, 
sqequalReflexivity, 
Error :isect_memberFormation_alt, 
introduction, 
cut, 
sqequalRule, 
sqequalHypSubstitution, 
axiomSqleEquality, 
hypothesis, 
extract_by_obid, 
isectElimination, 
thin, 
hypothesisEquality, 
Error :isect_memberEquality_alt, 
Error :universeIsType, 
because_Cache, 
equalityTransitivity, 
equalitySymmetry, 
universeEquality, 
pointwiseFunctionality, 
independent_isectElimination, 
callbyvalueReduce, 
baseClosed
Latex:
\mforall{}[T:Type].  \mforall{}[x:T].  has-valueall(x)  supposing  valueall-type(T)
Date html generated:
2019_06_20-AM-11_20_47
Last ObjectModification:
2018_10_06-AM-09_24_51
Theory : call!by!value_1
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