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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