Nuprl Lemma : member-has-valueall
∀[a:Base]. Ax ∈ has-valueall(a) supposing has-valueall(a)
Proof
Definitions occuring in Statement : 
has-valueall: has-valueall(a)
, 
uimplies: b supposing a
, 
uall: ∀[x:A]. B[x]
, 
member: t ∈ T
, 
base: Base
, 
axiom: Ax
Definitions unfolded in proof : 
uall: ∀[x:A]. B[x]
, 
member: t ∈ T
, 
uimplies: b supposing a
, 
has-valueall: has-valueall(a)
, 
has-value: (a)↓
, 
prop: ℙ
Lemmas referenced : 
has-valueall_wf_base, 
base_wf
Rules used in proof : 
sqequalSubstitution, 
sqequalTransitivity, 
computationStep, 
sqequalReflexivity, 
isect_memberFormation, 
introduction, 
cut, 
axiomSqleEquality, 
hypothesis, 
sqequalHypSubstitution, 
sqequalRule, 
axiomEquality, 
equalityTransitivity, 
equalitySymmetry, 
lemma_by_obid, 
isectElimination, 
thin, 
hypothesisEquality, 
isect_memberEquality, 
because_Cache
Latex:
\mforall{}[a:Base].  Ax  \mmember{}  has-valueall(a)  supposing  has-valueall(a)
Date html generated:
2016_05_13-PM-03_25_19
Last ObjectModification:
2015_12_26-AM-09_29_22
Theory : call!by!value_1
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