Nuprl Lemma : has-valueall-cons
∀[a,b:Base]. uiff(has-valueall(a) ∧ has-valueall(b);has-valueall([a / b]))
Proof
Definitions occuring in Statement :
cons: [a / b],
has-valueall: has-valueall(a),
uiff: uiff(P;Q),
uall: ∀[x:A]. B[x],
and: P ∧ Q,
base: Base
Definitions unfolded in proof :
has-valueall: has-valueall(a),
uall: ∀[x:A]. B[x],
member: t ∈ T,
uiff: uiff(P;Q),
and: P ∧ Q,
uimplies: b supposing a,
evalall: evalall(t),
cons: [a / b],
has-value: (a)↓,
prop: ℙ
Lemmas referenced :
base_wf,
and_wf,
is-exception_wf,
has-value_wf_base
Rules used in proof :
sqequalSubstitution,
sqequalRule,
sqequalReflexivity,
sqequalTransitivity,
computationStep,
isect_memberFormation,
introduction,
cut,
independent_pairFormation,
sqequalHypSubstitution,
productElimination,
thin,
callbyvalueReduce,
hypothesis,
divergentSqle,
sqleReflexivity,
lemma_by_obid,
isectElimination,
baseClosed,
axiomSqleEquality,
baseApply,
closedConclusion,
hypothesisEquality,
callbyvalueCallbyvalue,
independent_pairEquality,
isect_memberEquality,
because_Cache,
equalityTransitivity,
equalitySymmetry
Latex:
\mforall{}[a,b:Base]. uiff(has-valueall(a) \mwedge{} has-valueall(b);has-valueall([a / b]))
Date html generated:
2016_05_14-AM-06_26_05
Last ObjectModification:
2016_01_14-PM-08_26_50
Theory : list_0
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