Nuprl Lemma : callbyvalueall-single-sqle

[F,G:Base]. ∀[a:Top].  let x ⟵ in F[x] ≤ let x ⟵ [a] in G[x] supposing ∀b:Base. (F[b] ≤ G[[b]])


Proof




Definitions occuring in Statement :  cons: [a b] nil: [] callbyvalueall: callbyvalueall uimplies: supposing a uall: [x:A]. B[x] top: Top so_apply: x[s] all: x:A. B[x] base: Base sqle: s ≤ t
Definitions unfolded in proof :  uall: [x:A]. B[x] so_lambda: λ2x.t[x] member: t ∈ T top: Top so_apply: x[s] callbyvalueall: callbyvalueall uimplies: supposing a has-valueall: has-valueall(a) implies:  Q all: x:A. B[x] prop:
Lemmas referenced :  top_wf sqle_wf_base base_wf all_wf has-valueall_wf_base evalall-sqequal cbv_sqle callbyvalueall-single
Rules used in proof :  sqequalSubstitution sqequalRule sqequalReflexivity cut lemma_by_obid sqequalHypSubstitution sqequalTransitivity computationStep isectElimination thin isect_memberEquality voidElimination voidEquality hypothesisEquality hypothesis sqleRule baseApply closedConclusion baseClosed independent_isectElimination lambdaFormation dependent_functionElimination because_Cache sqleReflexivity lambdaEquality isect_memberFormation introduction axiomSqleEquality equalityTransitivity equalitySymmetry

Latex:
\mforall{}[F,G:Base].  \mforall{}[a:Top].    let  x  \mleftarrow{}{}  a  in  F[x]  \mleq{}  let  x  \mleftarrow{}{}  [a]  in  G[x]  supposing  \mforall{}b:Base.  (F[b]  \mleq{}  G[[b]])



Date html generated: 2016_05_15-PM-02_08_54
Last ObjectModification: 2016_01_15-PM-10_21_40

Theory : untyped!computation


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