interpretations
YES(?,O(n^3))
We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^3)).
Strict Trs:
{ a__terms(N) -> cons(recip(a__sqr(mark(N))), terms(s(N)))
, a__terms(X) -> terms(X)
, a__sqr(X) -> sqr(X)
, a__sqr(s(X)) -> s(add(sqr(X), dbl(X)))
, a__sqr(0()) -> 0()
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(recip(X)) -> recip(mark(X))
, mark(terms(X)) -> a__terms(mark(X))
, mark(s(X)) -> s(X)
, mark(0()) -> 0()
, mark(add(X1, X2)) -> a__add(mark(X1), mark(X2))
, mark(sqr(X)) -> a__sqr(mark(X))
, mark(dbl(X)) -> a__dbl(mark(X))
, mark(nil()) -> nil()
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, a__dbl(X) -> dbl(X)
, a__dbl(s(X)) -> s(s(dbl(X)))
, a__dbl(0()) -> 0()
, a__add(X1, X2) -> add(X1, X2)
, a__add(s(X), Y) -> s(add(X, Y))
, a__add(0(), X) -> mark(X)
, a__first(X1, X2) -> first(X1, X2)
, a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
, a__first(0(), X) -> nil() }
Obligation:
innermost runtime complexity
Answer:
YES(?,O(n^3))
The following argument positions are usable:
Uargs(a__terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1},
Uargs(a__sqr) = {1}, Uargs(mark) = {}, Uargs(terms) = {},
Uargs(s) = {}, Uargs(add) = {}, Uargs(sqr) = {}, Uargs(dbl) = {},
Uargs(a__dbl) = {1}, Uargs(a__add) = {1, 2},
Uargs(a__first) = {1, 2}, Uargs(first) = {}
TcT has computed following constructor-based matrix interpretation
satisfying not(EDA).
[1 2 2] [3]
[a__terms](x1) = [0 1 0] x1 + [2]
[0 0 1] [3]
[1 1 0] [0]
[cons](x1, x2) = [0 1 0] x1 + [1]
[0 0 1] [0]
[1 0 0] [0]
[recip](x1) = [0 1 0] x1 + [0]
[0 0 1] [1]
[1 0 1] [1]
[a__sqr](x1) = [0 1 0] x1 + [0]
[0 0 1] [2]
[1 1 1] [0]
[mark](x1) = [0 1 0] x1 + [0]
[0 0 1] [0]
[1 2 2] [0]
[terms](x1) = [0 1 0] x1 + [2]
[0 0 1] [3]
[0]
[s](x1) = [2]
[2]
[3]
[0] = [2]
[0]
[1 1 2] [1 3 1] [1]
[add](x1, x2) = [0 0 1] x1 + [0 1 0] x2 + [3]
[0 1 0] [0 0 1] [1]
[1 0 1] [0]
[sqr](x1) = [0 1 0] x1 + [0]
[0 0 1] [2]
[1 0 0] [1]
[dbl](x1) = [0 0 1] x1 + [2]
[0 1 0] [0]
[1 0 0] [2]
[a__dbl](x1) = [0 0 1] x1 + [2]
[0 1 0] [0]
[1 1 2] [1 3 1] [2]
[a__add](x1, x2) = [0 0 1] x1 + [0 1 0] x2 + [3]
[0 1 0] [0 0 1] [1]
[1 1 0] [1 2 2] [1]
[a__first](x1, x2) = [0 1 0] x1 + [0 1 0] x2 + [0]
[0 0 1] [0 0 1] [2]
[2]
[nil] = [1]
[0]
[1 1 0] [1 2 2] [0]
[first](x1, x2) = [0 1 0] x1 + [0 1 0] x2 + [0]
[0 0 1] [0 0 1] [2]
This order satisfies following ordering constraints
[a__terms(N)] = [1 2 2] [3]
[0 1 0] N + [2]
[0 0 1] [3]
> [1 2 2] [1]
[0 1 0] N + [1]
[0 0 1] [3]
= [cons(recip(a__sqr(mark(N))), terms(s(N)))]
[a__terms(X)] = [1 2 2] [3]
[0 1 0] X + [2]
[0 0 1] [3]
> [1 2 2] [0]
[0 1 0] X + [2]
[0 0 1] [3]
= [terms(X)]
[a__sqr(X)] = [1 0 1] [1]
[0 1 0] X + [0]
[0 0 1] [2]
> [1 0 1] [0]
[0 1 0] X + [0]
[0 0 1] [2]
= [sqr(X)]
[a__sqr(s(X))] = [3]
[2]
[4]
> [0]
[2]
[2]
= [s(add(sqr(X), dbl(X)))]
[a__sqr(0())] = [4]
[2]
[2]
> [3]
[2]
[0]
= [0()]
[mark(cons(X1, X2))] = [1 2 1] [1]
[0 1 0] X1 + [1]
[0 0 1] [0]
> [1 2 1] [0]
[0 1 0] X1 + [1]
[0 0 1] [0]
= [cons(mark(X1), X2)]
[mark(recip(X))] = [1 1 1] [1]
[0 1 0] X + [0]
[0 0 1] [1]
> [1 1 1] [0]
[0 1 0] X + [0]
[0 0 1] [1]
= [recip(mark(X))]
[mark(terms(X))] = [1 3 3] [5]
[0 1 0] X + [2]
[0 0 1] [3]
> [1 3 3] [3]
[0 1 0] X + [2]
[0 0 1] [3]
= [a__terms(mark(X))]
[mark(s(X))] = [4]
[2]
[2]
> [0]
[2]
[2]
= [s(X)]
[mark(0())] = [5]
[2]
[0]
> [3]
[2]
[0]
= [0()]
[mark(add(X1, X2))] = [1 2 3] [1 4 2] [5]
[0 0 1] X1 + [0 1 0] X2 + [3]
[0 1 0] [0 0 1] [1]
> [1 2 3] [1 4 2] [2]
[0 0 1] X1 + [0 1 0] X2 + [3]
[0 1 0] [0 0 1] [1]
= [a__add(mark(X1), mark(X2))]
[mark(sqr(X))] = [1 1 2] [2]
[0 1 0] X + [0]
[0 0 1] [2]
> [1 1 2] [1]
[0 1 0] X + [0]
[0 0 1] [2]
= [a__sqr(mark(X))]
[mark(dbl(X))] = [1 1 1] [3]
[0 0 1] X + [2]
[0 1 0] [0]
> [1 1 1] [2]
[0 0 1] X + [2]
[0 1 0] [0]
= [a__dbl(mark(X))]
[mark(nil())] = [3]
[1]
[0]
> [2]
[1]
[0]
= [nil()]
[mark(first(X1, X2))] = [1 2 1] [1 3 3] [2]
[0 1 0] X1 + [0 1 0] X2 + [0]
[0 0 1] [0 0 1] [2]
> [1 2 1] [1 3 3] [1]
[0 1 0] X1 + [0 1 0] X2 + [0]
[0 0 1] [0 0 1] [2]
= [a__first(mark(X1), mark(X2))]
[a__dbl(X)] = [1 0 0] [2]
[0 0 1] X + [2]
[0 1 0] [0]
> [1 0 0] [1]
[0 0 1] X + [2]
[0 1 0] [0]
= [dbl(X)]
[a__dbl(s(X))] = [2]
[4]
[2]
> [0]
[2]
[2]
= [s(s(dbl(X)))]
[a__dbl(0())] = [5]
[2]
[2]
> [3]
[2]
[0]
= [0()]
[a__add(X1, X2)] = [1 1 2] [1 3 1] [2]
[0 0 1] X1 + [0 1 0] X2 + [3]
[0 1 0] [0 0 1] [1]
> [1 1 2] [1 3 1] [1]
[0 0 1] X1 + [0 1 0] X2 + [3]
[0 1 0] [0 0 1] [1]
= [add(X1, X2)]
[a__add(s(X), Y)] = [1 3 1] [8]
[0 1 0] Y + [5]
[0 0 1] [3]
> [0]
[2]
[2]
= [s(add(X, Y))]
[a__add(0(), X)] = [1 3 1] [7]
[0 1 0] X + [3]
[0 0 1] [3]
> [1 1 1] [0]
[0 1 0] X + [0]
[0 0 1] [0]
= [mark(X)]
[a__first(X1, X2)] = [1 1 0] [1 2 2] [1]
[0 1 0] X1 + [0 1 0] X2 + [0]
[0 0 1] [0 0 1] [2]
> [1 1 0] [1 2 2] [0]
[0 1 0] X1 + [0 1 0] X2 + [0]
[0 0 1] [0 0 1] [2]
= [first(X1, X2)]
[a__first(s(X), cons(Y, Z))] = [1 3 2] [5]
[0 1 0] Y + [3]
[0 0 1] [4]
> [1 2 1] [0]
[0 1 0] Y + [1]
[0 0 1] [0]
= [cons(mark(Y), first(X, Z))]
[a__first(0(), X)] = [1 2 2] [6]
[0 1 0] X + [2]
[0 0 1] [2]
> [2]
[1]
[0]
= [nil()]
Hurray, we answered YES(?,O(n^3))
lmpo
MAYBE
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict Trs:
{ a__terms(N) -> cons(recip(a__sqr(mark(N))), terms(s(N)))
, a__sqr(0()) -> 0()
, a__sqr(s(X)) -> s(add(sqr(X), dbl(X)))
, a__dbl(0()) -> 0()
, a__dbl(s(X)) -> s(s(dbl(X)))
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
, mark(terms(X)) -> a__terms(mark(X))
, mark(sqr(X)) -> a__sqr(mark(X))
, mark(add(X1, X2)) -> a__add(mark(X1), mark(X2))
, mark(dbl(X)) -> a__dbl(mark(X))
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(recip(X)) -> recip(mark(X))
, mark(s(X)) -> s(X)
, mark(0()) -> 0()
, mark(nil()) -> nil()
, a__terms(X) -> terms(X)
, a__sqr(X) -> sqr(X)
, a__add(X1, X2) -> add(X1, X2)
, a__dbl(X) -> dbl(X)
, a__first(X1, X2) -> first(X1, X2) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The input cannot be shown compatible
Arrrr..
mpo
MAYBE
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict Trs:
{ a__terms(N) -> cons(recip(a__sqr(mark(N))), terms(s(N)))
, a__sqr(0()) -> 0()
, a__sqr(s(X)) -> s(add(sqr(X), dbl(X)))
, a__dbl(0()) -> 0()
, a__dbl(s(X)) -> s(s(dbl(X)))
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
, mark(terms(X)) -> a__terms(mark(X))
, mark(sqr(X)) -> a__sqr(mark(X))
, mark(add(X1, X2)) -> a__add(mark(X1), mark(X2))
, mark(dbl(X)) -> a__dbl(mark(X))
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(recip(X)) -> recip(mark(X))
, mark(s(X)) -> s(X)
, mark(0()) -> 0()
, mark(nil()) -> nil()
, a__terms(X) -> terms(X)
, a__sqr(X) -> sqr(X)
, a__add(X1, X2) -> add(X1, X2)
, a__dbl(X) -> dbl(X)
, a__first(X1, X2) -> first(X1, X2) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The input cannot be shown compatible
Arrrr..
popstar
MAYBE
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict Trs:
{ a__terms(N) -> cons(recip(a__sqr(mark(N))), terms(s(N)))
, a__sqr(0()) -> 0()
, a__sqr(s(X)) -> s(add(sqr(X), dbl(X)))
, a__dbl(0()) -> 0()
, a__dbl(s(X)) -> s(s(dbl(X)))
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
, mark(terms(X)) -> a__terms(mark(X))
, mark(sqr(X)) -> a__sqr(mark(X))
, mark(add(X1, X2)) -> a__add(mark(X1), mark(X2))
, mark(dbl(X)) -> a__dbl(mark(X))
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(recip(X)) -> recip(mark(X))
, mark(s(X)) -> s(X)
, mark(0()) -> 0()
, mark(nil()) -> nil()
, a__terms(X) -> terms(X)
, a__sqr(X) -> sqr(X)
, a__add(X1, X2) -> add(X1, X2)
, a__dbl(X) -> dbl(X)
, a__first(X1, X2) -> first(X1, X2) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The input cannot be shown compatible
Arrrr..
popstar-ps
MAYBE
We are left with following problem, upon which TcT provides the
certificate MAYBE.
Strict Trs:
{ a__terms(N) -> cons(recip(a__sqr(mark(N))), terms(s(N)))
, a__sqr(0()) -> 0()
, a__sqr(s(X)) -> s(add(sqr(X), dbl(X)))
, a__dbl(0()) -> 0()
, a__dbl(s(X)) -> s(s(dbl(X)))
, a__add(0(), X) -> mark(X)
, a__add(s(X), Y) -> s(add(X, Y))
, a__first(0(), X) -> nil()
, a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z))
, mark(terms(X)) -> a__terms(mark(X))
, mark(sqr(X)) -> a__sqr(mark(X))
, mark(add(X1, X2)) -> a__add(mark(X1), mark(X2))
, mark(dbl(X)) -> a__dbl(mark(X))
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
, mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(recip(X)) -> recip(mark(X))
, mark(s(X)) -> s(X)
, mark(0()) -> 0()
, mark(nil()) -> nil()
, a__terms(X) -> terms(X)
, a__sqr(X) -> sqr(X)
, a__add(X1, X2) -> add(X1, X2)
, a__dbl(X) -> dbl(X)
, a__first(X1, X2) -> first(X1, X2) }
Obligation:
innermost runtime complexity
Answer:
MAYBE
The input cannot be shown compatible
Arrrr..