Nuprl Lemma : cbv-sqequal0
∀[a:Base]. eval x = a in 0 ~ 0 supposing (a)↓
Proof
Definitions occuring in Statement :
has-value: (a)↓,
callbyvalue: callbyvalue,
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
natural_number: $n,
base: Base,
sqequal: s ~ t
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
uimplies: b supposing a,
has-value: (a)↓,
implies: P ⇒ Q,
false: False,
prop: ℙ
Lemmas referenced :
base_wf,
exception-not-value,
is-exception_wf,
has-value_wf_base
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation,
introduction,
cut,
sqequalSqle,
divergentSqle,
callbyvalueCallbyvalue,
sqequalHypSubstitution,
hypothesis,
sqequalRule,
callbyvalueReduce,
sqleReflexivity,
lemma_by_obid,
isectElimination,
thin,
baseClosed,
callbyvalueExceptionCases,
axiomSqleEquality,
hypothesisEquality,
independent_isectElimination,
independent_functionElimination,
voidElimination,
baseApply,
closedConclusion,
sqequalAxiom,
isect_memberEquality,
because_Cache,
equalityTransitivity,
equalitySymmetry
Latex:
\mforall{}[a:Base]. eval x = a in 0 \msim{} 0 supposing (a)\mdownarrow{}
Date html generated:
2016_05_13-PM-03_28_41
Last ObjectModification:
2016_01_14-PM-06_41_58
Theory : arithmetic
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