Nuprl Lemma : genrec-ap-unroll
∀[g,x:Top]. (letrec rec(n)=g[n;rec] in rec(x) ~ g[x;λx.letrec rec(n)=g[n;rec] in rec(x)])
Proof
Definitions occuring in Statement :
genrec-ap: genrec-ap,
uall: ∀[x:A]. B[x],
top: Top,
so_apply: x[s1;s2],
lambda: λx.A[x],
sqequal: s ~ t
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
genrec-ap: genrec-ap
Lemmas referenced :
istype-top
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
Error :isect_memberFormation_alt,
introduction,
cut,
sqequalRule,
hypothesis,
axiomSqEquality,
Error :inhabitedIsType,
hypothesisEquality,
sqequalHypSubstitution,
Error :isect_memberEquality_alt,
isectElimination,
thin,
Error :isectIsTypeImplies,
extract_by_obid
Latex:
\mforall{}[g,x:Top]. (letrec rec(n)=g[n;rec] in rec(x) \msim{} g[x;\mlambda{}x.letrec rec(n)=g[n;rec] in rec(x)])
Date html generated:
2019_06_20-AM-11_33_36
Last ObjectModification:
2019_04_01-PM-02_37_42
Theory : int_1
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