Nuprl Lemma : callbyvalueall-single-sqle2

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

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



Date html generated: 2016_05_15-PM-02_08_56
Last ObjectModification: 2016_01_15-PM-10_21_34

Theory : untyped!computation


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