Nuprl Lemma : cbva-sqequal0
∀[a:Base]. let x ⟵ a in 0 ~ 0 supposing has-valueall(a)
Proof
Definitions occuring in Statement :
has-valueall: has-valueall(a),
callbyvalueall: callbyvalueall,
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,
callbyvalueall: callbyvalueall,
has-value: (a)↓,
has-valueall: has-valueall(a),
prop: ℙ
Lemmas referenced :
has-valueall_wf_base,
base_wf
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation,
introduction,
cut,
sqequalRule,
callbyvalueReduce,
hypothesis,
sqequalAxiom,
lemma_by_obid,
sqequalHypSubstitution,
isectElimination,
thin,
hypothesisEquality,
isect_memberEquality,
because_Cache,
equalityTransitivity,
equalitySymmetry
Latex:
\mforall{}[a:Base]. let x \mleftarrow{}{} a in 0 \msim{} 0 supposing has-valueall(a)
Date html generated:
2016_05_13-PM-03_28_42
Last ObjectModification:
2015_12_26-AM-09_48_10
Theory : arithmetic
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