LMPO
YES(?,ELEMENTARY)
We consider the following Problem:
Strict Trs:
{ terms(N) -> cons(recip(sqr(N)))
, sqr(0()) -> 0()
, sqr(s()) -> s()
, dbl(0()) -> 0()
, dbl(s()) -> s()
, add(0(), X) -> X
, add(s(), Y) -> s()
, first(0(), X) -> nil()
, first(s(), cons(Y)) -> cons(Y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,ELEMENTARY)
Proof:
The input was oriented with the instance of
Lightweight Multiset Path Order () as induced by the safe mapping
safe(terms) = {1}, safe(cons) = {1}, safe(recip) = {1},
safe(sqr) = {1}, safe(0) = {}, safe(s) = {}, safe(dbl) = {},
safe(add) = {}, safe(first) = {2}, safe(nil) = {}
and precedence
terms > sqr .
Following symbols are considered recursive:
{terms, sqr}
The recursion depth is 2 .
For your convenience, here are the oriented rules in predicative
notation (possibly applying argument filtering):
Strict DPs: {}
Weak DPs : {}
Strict Trs:
{ terms(; N) -> cons(; recip(; sqr(; N)))
, sqr(; 0()) -> 0()
, sqr(; s()) -> s()
, dbl(0();) -> 0()
, dbl(s();) -> s()
, add(0(), X;) -> X
, add(s(), Y;) -> s()
, first(0(); X) -> nil()
, first(s(); cons(; Y)) -> cons(; Y)}
Weak Trs : {}
Hurray, we answered YES(?,ELEMENTARY)
MPO
YES(?,PRIMREC)
We consider the following Problem:
Strict Trs:
{ terms(N) -> cons(recip(sqr(N)))
, sqr(0()) -> 0()
, sqr(s()) -> s()
, dbl(0()) -> 0()
, dbl(s()) -> s()
, add(0(), X) -> X
, add(s(), Y) -> s()
, first(0(), X) -> nil()
, first(s(), cons(Y)) -> cons(Y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,PRIMREC)
Proof:
The input was oriented with the instance of
'multiset path orders' as induced by the precedence
terms > cons, terms > recip, terms > sqr, 0 > nil .
Hurray, we answered YES(?,PRIMREC)
POP*
YES(?,POLY)
We consider the following Problem:
Strict Trs:
{ terms(N) -> cons(recip(sqr(N)))
, sqr(0()) -> 0()
, sqr(s()) -> s()
, dbl(0()) -> 0()
, dbl(s()) -> s()
, add(0(), X) -> X
, add(s(), Y) -> s()
, first(0(), X) -> nil()
, first(s(), cons(Y)) -> cons(Y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,POLY)
Proof:
The input was oriented with the instance of
Polynomial Path Order () as induced by the safe mapping
safe(terms) = {1}, safe(cons) = {1}, safe(recip) = {1},
safe(sqr) = {1}, safe(0) = {}, safe(s) = {}, safe(dbl) = {},
safe(add) = {}, safe(first) = {2}, safe(nil) = {}
and precedence
terms > sqr .
Following symbols are considered recursive:
{terms, sqr}
The recursion depth is 2 .
For your convenience, here are the oriented rules in predicative
notation (possibly applying argument filtering):
Strict DPs: {}
Weak DPs : {}
Strict Trs:
{ terms(; N) -> cons(; recip(; sqr(; N)))
, sqr(; 0()) -> 0()
, sqr(; s()) -> s()
, dbl(0();) -> 0()
, dbl(s();) -> s()
, add(0(), X;) -> X
, add(s(), Y;) -> s()
, first(0(); X) -> nil()
, first(s(); cons(; Y)) -> cons(; Y)}
Weak Trs : {}
Hurray, we answered YES(?,POLY)
POP* (PS)
YES(?,POLY)
We consider the following Problem:
Strict Trs:
{ terms(N) -> cons(recip(sqr(N)))
, sqr(0()) -> 0()
, sqr(s()) -> s()
, dbl(0()) -> 0()
, dbl(s()) -> s()
, add(0(), X) -> X
, add(s(), Y) -> s()
, first(0(), X) -> nil()
, first(s(), cons(Y)) -> cons(Y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,POLY)
Proof:
The input was oriented with the instance of
Polynomial Path Order (PS) as induced by the safe mapping
safe(terms) = {1}, safe(cons) = {1}, safe(recip) = {1},
safe(sqr) = {1}, safe(0) = {}, safe(s) = {}, safe(dbl) = {},
safe(add) = {}, safe(first) = {2}, safe(nil) = {}
and precedence
terms > sqr .
Following symbols are considered recursive:
{terms, sqr}
The recursion depth is 2 .
For your convenience, here are the oriented rules in predicative
notation (possibly applying argument filtering):
Strict DPs: {}
Weak DPs : {}
Strict Trs:
{ terms(; N) -> cons(; recip(; sqr(; N)))
, sqr(; 0()) -> 0()
, sqr(; s()) -> s()
, dbl(0();) -> 0()
, dbl(s();) -> s()
, add(0(), X;) -> X
, add(s(), Y;) -> s()
, first(0(); X) -> nil()
, first(s(); cons(; Y)) -> cons(; Y)}
Weak Trs : {}
Hurray, we answered YES(?,POLY)
Small POP*
YES(?,O(1))
We consider the following Problem:
Strict Trs:
{ terms(N) -> cons(recip(sqr(N)))
, sqr(0()) -> 0()
, sqr(s()) -> s()
, dbl(0()) -> 0()
, dbl(s()) -> s()
, add(0(), X) -> X
, add(s(), Y) -> s()
, first(0(), X) -> nil()
, first(s(), cons(Y)) -> cons(Y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(1))
Proof:
The input was oriented with the instance of
Small Polynomial Path Order (WSC,
Nat 0-bounded) as induced by the safe mapping
safe(terms) = {1}, safe(cons) = {1}, safe(recip) = {1},
safe(sqr) = {1}, safe(0) = {}, safe(s) = {}, safe(dbl) = {},
safe(add) = {}, safe(first) = {2}, safe(nil) = {}
and precedence
terms > sqr .
Following symbols are considered recursive:
{}
The recursion depth is 0 .
For your convenience, here are the oriented rules in predicative
notation (possibly applying argument filtering):
Strict DPs: {}
Weak DPs : {}
Strict Trs:
{ terms(; N) -> cons(; recip(; sqr(; N)))
, sqr(; 0()) -> 0()
, sqr(; s()) -> s()
, dbl(0();) -> 0()
, dbl(s();) -> s()
, add(0(), X;) -> X
, add(s(), Y;) -> s()
, first(0(); X) -> nil()
, first(s(); cons(; Y)) -> cons(; Y)}
Weak Trs : {}
Hurray, we answered YES(?,O(1))
Small POP* (PS)
YES(?,O(1))
We consider the following Problem:
Strict Trs:
{ terms(N) -> cons(recip(sqr(N)))
, sqr(0()) -> 0()
, sqr(s()) -> s()
, dbl(0()) -> 0()
, dbl(s()) -> s()
, add(0(), X) -> X
, add(s(), Y) -> s()
, first(0(), X) -> nil()
, first(s(), cons(Y)) -> cons(Y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(1))
Proof:
The input was oriented with the instance of
Small Polynomial Path Order (WSC,
PS,
Nat 0-bounded) as induced by the safe mapping
safe(terms) = {}, safe(cons) = {1}, safe(recip) = {1},
safe(sqr) = {1}, safe(0) = {}, safe(s) = {}, safe(dbl) = {},
safe(add) = {}, safe(first) = {2}, safe(nil) = {}
and precedence
terms > sqr .
Following symbols are considered recursive:
{}
The recursion depth is 0 .
For your convenience, here are the oriented rules in predicative
notation (possibly applying argument filtering):
Strict DPs: {}
Weak DPs : {}
Strict Trs:
{ terms(N;) -> cons(; recip(; sqr(; N)))
, sqr(; 0()) -> 0()
, sqr(; s()) -> s()
, dbl(0();) -> 0()
, dbl(s();) -> s()
, add(0(), X;) -> X
, add(s(), Y;) -> s()
, first(0(); X) -> nil()
, first(s(); cons(; Y)) -> cons(; Y)}
Weak Trs : {}
Hurray, we answered YES(?,O(1))