Nuprl Lemma : natrec-unroll
∀[g,n:Top]. (letrec f(n)=g[n;f] in f n ~ g[n;letrec f(n)=g[n;f] in f])
Proof
Definitions occuring in Statement :
natrec: natrec,
uall: ∀[x:A]. B[x],
top: Top,
so_apply: x[s1;s2],
apply: f a,
sqequal: s ~ t
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
natrec: natrec,
genrec: genrec
Lemmas referenced :
top_wf
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation,
introduction,
cut,
sqequalRule,
hypothesis,
sqequalAxiom,
lemma_by_obid,
sqequalHypSubstitution,
isect_memberEquality,
isectElimination,
thin,
hypothesisEquality,
because_Cache
Latex:
\mforall{}[g,n:Top]. (letrec f(n)=g[n;f] in f n \msim{} g[n;letrec f(n)=g[n;f] in f])
Date html generated:
2016_05_13-PM-04_03_16
Last ObjectModification:
2015_12_26-AM-10_56_04
Theory : int_1
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