Nuprl Lemma : copath-tl-cons
∀[b:Top]. ∀[p:ℕ × Top]. (copath-tl(copath-cons(b;p)) ~ p)
Proof
Definitions occuring in Statement :
copath-cons: copath-cons(b;x),
copath-tl: copath-tl(x),
nat: ℕ,
uall: ∀[x:A]. B[x],
top: Top,
product: x:A × B[x],
sqequal: s ~ t
Definitions unfolded in proof :
uall: ∀[x:A]. B[x],
member: t ∈ T,
copath-cons: copath-cons(b;x),
copath-tl: copath-tl(x),
pi2: snd(t),
nat: ℕ
Lemmas referenced :
add-subtract-cancel,
nat_wf,
top_wf
Rules used in proof :
sqequalSubstitution,
sqequalTransitivity,
computationStep,
sqequalReflexivity,
isect_memberFormation,
introduction,
cut,
productElimination,
thin,
sqequalRule,
extract_by_obid,
sqequalHypSubstitution,
isectElimination,
setElimination,
rename,
hypothesisEquality,
hypothesis,
natural_numberEquality,
sqequalAxiom,
productEquality,
isect_memberEquality,
because_Cache
Latex:
\mforall{}[b:Top]. \mforall{}[p:\mBbbN{} \mtimes{} Top]. (copath-tl(copath-cons(b;p)) \msim{} p)
Date html generated:
2018_07_25-PM-01_40_04
Last ObjectModification:
2018_06_04-PM-06_56_49
Theory : co-recursion
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