Nuprl Lemma : evalall-sqle
∀[x:Base]. evalall(x) ≤ x supposing (evalall(x))↓
Proof
Definitions occuring in Statement :
has-value: (a)↓,
evalall: evalall(t),
uimplies: b supposing a,
uall: ∀[x:A]. B[x],
base: Base,
sqle: s ≤ t
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
uimplies: b supposing a,
prop: ℙ
Lemmas referenced :
base_wf,
is-exception_wf,
has-value_wf_base,
evalall-sqequal
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation,
introduction,
cut,
sqequalRule,
lemma_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
hypothesisEquality,
independent_isectElimination,
hypothesis,
divergentSqle,
sqleReflexivity,
axiomSqleEquality,
baseApply,
closedConclusion,
baseClosed,
isect_memberEquality,
because_Cache,
equalityTransitivity,
equalitySymmetry
Latex:
\mforall{}[x:Base]. evalall(x) \mleq{} x supposing (evalall(x))\mdownarrow{}
Date html generated:
2016_05_13-PM-04_07_36
Last ObjectModification:
2016_01_14-PM-07_46_01
Theory : fun_1
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