We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
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(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(a__terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1},
Uargs(a__sqr) = {1}, Uargs(a__dbl) = {1}, Uargs(a__add) = {1, 2},
Uargs(a__first) = {1, 2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__terms](x1) = [1] x1 + [4]
[cons](x1, x2) = [1] x1 + [0]
[recip](x1) = [1] x1 + [0]
[a__sqr](x1) = [1] x1 + [0]
[mark](x1) = [0]
[terms](x1) = [1] x1 + [0]
[s](x1) = [0]
[0] = [0]
[add](x1, x2) = [0]
[sqr](x1) = [1] x1 + [0]
[dbl](x1) = [0]
[a__dbl](x1) = [1] x1 + [0]
[a__add](x1, x2) = [1] x1 + [1] x2 + [0]
[a__first](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
[first](x1, x2) = [0]
The order satisfies the following ordering constraints:
[a__terms(N)] = [1] N + [4]
> [0]
= [cons(recip(a__sqr(mark(N))), terms(s(N)))]
[a__terms(X)] = [1] X + [4]
> [1] X + [0]
= [terms(X)]
[a__sqr(X)] = [1] X + [0]
>= [1] X + [0]
= [sqr(X)]
[a__sqr(s(X))] = [0]
>= [0]
= [s(add(sqr(X), dbl(X)))]
[a__sqr(0())] = [0]
>= [0]
= [0()]
[mark(cons(X1, X2))] = [0]
>= [0]
= [cons(mark(X1), X2)]
[mark(recip(X))] = [0]
>= [0]
= [recip(mark(X))]
[mark(terms(X))] = [0]
? [4]
= [a__terms(mark(X))]
[mark(s(X))] = [0]
>= [0]
= [s(X)]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(add(X1, X2))] = [0]
>= [0]
= [a__add(mark(X1), mark(X2))]
[mark(sqr(X))] = [0]
>= [0]
= [a__sqr(mark(X))]
[mark(dbl(X))] = [0]
>= [0]
= [a__dbl(mark(X))]
[mark(nil())] = [0]
>= [0]
= [nil()]
[mark(first(X1, X2))] = [0]
>= [0]
= [a__first(mark(X1), mark(X2))]
[a__dbl(X)] = [1] X + [0]
>= [0]
= [dbl(X)]
[a__dbl(s(X))] = [0]
>= [0]
= [s(s(dbl(X)))]
[a__dbl(0())] = [0]
>= [0]
= [0()]
[a__add(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [0]
= [add(X1, X2)]
[a__add(s(X), Y)] = [1] Y + [0]
>= [0]
= [s(add(X, Y))]
[a__add(0(), X)] = [1] X + [0]
>= [0]
= [mark(X)]
[a__first(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [0]
= [first(X1, X2)]
[a__first(s(X), cons(Y, Z))] = [1] Y + [0]
>= [0]
= [cons(mark(Y), first(X, Z))]
[a__first(0(), X)] = [1] X + [0]
>= [0]
= [nil()]
Further, it can be verified that all rules not oriented are covered by the weightgap condition.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ 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() }
Weak Trs:
{ a__terms(N) -> cons(recip(a__sqr(mark(N))), terms(s(N)))
, a__terms(X) -> terms(X) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(a__terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1},
Uargs(a__sqr) = {1}, Uargs(a__dbl) = {1}, Uargs(a__add) = {1, 2},
Uargs(a__first) = {1, 2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__terms](x1) = [1] x1 + [4]
[cons](x1, x2) = [1] x1 + [0]
[recip](x1) = [1] x1 + [0]
[a__sqr](x1) = [1] x1 + [1]
[mark](x1) = [0]
[terms](x1) = [1] x1 + [0]
[s](x1) = [0]
[0] = [0]
[add](x1, x2) = [0]
[sqr](x1) = [0]
[dbl](x1) = [1] x1 + [0]
[a__dbl](x1) = [1] x1 + [0]
[a__add](x1, x2) = [1] x1 + [1] x2 + [0]
[a__first](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
[first](x1, x2) = [0]
The order satisfies the following ordering constraints:
[a__terms(N)] = [1] N + [4]
> [1]
= [cons(recip(a__sqr(mark(N))), terms(s(N)))]
[a__terms(X)] = [1] X + [4]
> [1] X + [0]
= [terms(X)]
[a__sqr(X)] = [1] X + [1]
> [0]
= [sqr(X)]
[a__sqr(s(X))] = [1]
> [0]
= [s(add(sqr(X), dbl(X)))]
[a__sqr(0())] = [1]
> [0]
= [0()]
[mark(cons(X1, X2))] = [0]
>= [0]
= [cons(mark(X1), X2)]
[mark(recip(X))] = [0]
>= [0]
= [recip(mark(X))]
[mark(terms(X))] = [0]
? [4]
= [a__terms(mark(X))]
[mark(s(X))] = [0]
>= [0]
= [s(X)]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(add(X1, X2))] = [0]
>= [0]
= [a__add(mark(X1), mark(X2))]
[mark(sqr(X))] = [0]
? [1]
= [a__sqr(mark(X))]
[mark(dbl(X))] = [0]
>= [0]
= [a__dbl(mark(X))]
[mark(nil())] = [0]
>= [0]
= [nil()]
[mark(first(X1, X2))] = [0]
>= [0]
= [a__first(mark(X1), mark(X2))]
[a__dbl(X)] = [1] X + [0]
>= [1] X + [0]
= [dbl(X)]
[a__dbl(s(X))] = [0]
>= [0]
= [s(s(dbl(X)))]
[a__dbl(0())] = [0]
>= [0]
= [0()]
[a__add(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [0]
= [add(X1, X2)]
[a__add(s(X), Y)] = [1] Y + [0]
>= [0]
= [s(add(X, Y))]
[a__add(0(), X)] = [1] X + [0]
>= [0]
= [mark(X)]
[a__first(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [0]
= [first(X1, X2)]
[a__first(s(X), cons(Y, Z))] = [1] Y + [0]
>= [0]
= [cons(mark(Y), first(X, Z))]
[a__first(0(), X)] = [1] X + [0]
>= [0]
= [nil()]
Further, it can be verified that all rules not oriented are covered by the weightgap condition.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ 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() }
Weak 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() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(a__terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1},
Uargs(a__sqr) = {1}, Uargs(a__dbl) = {1}, Uargs(a__add) = {1, 2},
Uargs(a__first) = {1, 2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__terms](x1) = [1] x1 + [4]
[cons](x1, x2) = [1] x1 + [0]
[recip](x1) = [1] x1 + [0]
[a__sqr](x1) = [1] x1 + [1]
[mark](x1) = [0]
[terms](x1) = [1] x1 + [0]
[s](x1) = [1]
[0] = [0]
[add](x1, x2) = [1] x1 + [1] x2 + [2]
[sqr](x1) = [0]
[dbl](x1) = [1] x1 + [0]
[a__dbl](x1) = [1] x1 + [0]
[a__add](x1, x2) = [1] x1 + [1] x2 + [4]
[a__first](x1, x2) = [1] x1 + [1] x2 + [0]
[nil] = [0]
[first](x1, x2) = [1] x2 + [0]
The order satisfies the following ordering constraints:
[a__terms(N)] = [1] N + [4]
> [1]
= [cons(recip(a__sqr(mark(N))), terms(s(N)))]
[a__terms(X)] = [1] X + [4]
> [1] X + [0]
= [terms(X)]
[a__sqr(X)] = [1] X + [1]
> [0]
= [sqr(X)]
[a__sqr(s(X))] = [2]
> [1]
= [s(add(sqr(X), dbl(X)))]
[a__sqr(0())] = [1]
> [0]
= [0()]
[mark(cons(X1, X2))] = [0]
>= [0]
= [cons(mark(X1), X2)]
[mark(recip(X))] = [0]
>= [0]
= [recip(mark(X))]
[mark(terms(X))] = [0]
? [4]
= [a__terms(mark(X))]
[mark(s(X))] = [0]
? [1]
= [s(X)]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(add(X1, X2))] = [0]
? [4]
= [a__add(mark(X1), mark(X2))]
[mark(sqr(X))] = [0]
? [1]
= [a__sqr(mark(X))]
[mark(dbl(X))] = [0]
>= [0]
= [a__dbl(mark(X))]
[mark(nil())] = [0]
>= [0]
= [nil()]
[mark(first(X1, X2))] = [0]
>= [0]
= [a__first(mark(X1), mark(X2))]
[a__dbl(X)] = [1] X + [0]
>= [1] X + [0]
= [dbl(X)]
[a__dbl(s(X))] = [1]
>= [1]
= [s(s(dbl(X)))]
[a__dbl(0())] = [0]
>= [0]
= [0()]
[a__add(X1, X2)] = [1] X1 + [1] X2 + [4]
> [1] X1 + [1] X2 + [2]
= [add(X1, X2)]
[a__add(s(X), Y)] = [1] Y + [5]
> [1]
= [s(add(X, Y))]
[a__add(0(), X)] = [1] X + [4]
> [0]
= [mark(X)]
[a__first(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X2 + [0]
= [first(X1, X2)]
[a__first(s(X), cons(Y, Z))] = [1] Y + [1]
> [0]
= [cons(mark(Y), first(X, Z))]
[a__first(0(), X)] = [1] X + [0]
>= [0]
= [nil()]
Further, it can be verified that all rules not oriented are covered by the weightgap condition.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ 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__first(X1, X2) -> first(X1, X2)
, a__first(0(), X) -> nil() }
Weak 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()
, a__add(X1, X2) -> add(X1, X2)
, a__add(s(X), Y) -> s(add(X, Y))
, a__add(0(), X) -> mark(X)
, a__first(s(X), cons(Y, Z)) -> cons(mark(Y), first(X, Z)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(a__terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1},
Uargs(a__sqr) = {1}, Uargs(a__dbl) = {1}, Uargs(a__add) = {1, 2},
Uargs(a__first) = {1, 2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__terms](x1) = [1] x1 + [1]
[cons](x1, x2) = [1] x1 + [0]
[recip](x1) = [1] x1 + [0]
[a__sqr](x1) = [1] x1 + [0]
[mark](x1) = [1] x1 + [0]
[terms](x1) = [1] x1 + [1]
[s](x1) = [0]
[0] = [0]
[add](x1, x2) = [1] x1 + [1] x2 + [0]
[sqr](x1) = [1] x1 + [0]
[dbl](x1) = [1] x1 + [0]
[a__dbl](x1) = [1] x1 + [0]
[a__add](x1, x2) = [1] x1 + [1] x2 + [0]
[a__first](x1, x2) = [1] x1 + [1] x2 + [1]
[nil] = [0]
[first](x1, x2) = [1] x1 + [1] x2 + [0]
The order satisfies the following ordering constraints:
[a__terms(N)] = [1] N + [1]
> [1] N + [0]
= [cons(recip(a__sqr(mark(N))), terms(s(N)))]
[a__terms(X)] = [1] X + [1]
>= [1] X + [1]
= [terms(X)]
[a__sqr(X)] = [1] X + [0]
>= [1] X + [0]
= [sqr(X)]
[a__sqr(s(X))] = [0]
>= [0]
= [s(add(sqr(X), dbl(X)))]
[a__sqr(0())] = [0]
>= [0]
= [0()]
[mark(cons(X1, X2))] = [1] X1 + [0]
>= [1] X1 + [0]
= [cons(mark(X1), X2)]
[mark(recip(X))] = [1] X + [0]
>= [1] X + [0]
= [recip(mark(X))]
[mark(terms(X))] = [1] X + [1]
>= [1] X + [1]
= [a__terms(mark(X))]
[mark(s(X))] = [0]
>= [0]
= [s(X)]
[mark(0())] = [0]
>= [0]
= [0()]
[mark(add(X1, X2))] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [a__add(mark(X1), mark(X2))]
[mark(sqr(X))] = [1] X + [0]
>= [1] X + [0]
= [a__sqr(mark(X))]
[mark(dbl(X))] = [1] X + [0]
>= [1] X + [0]
= [a__dbl(mark(X))]
[mark(nil())] = [0]
>= [0]
= [nil()]
[mark(first(X1, X2))] = [1] X1 + [1] X2 + [0]
? [1] X1 + [1] X2 + [1]
= [a__first(mark(X1), mark(X2))]
[a__dbl(X)] = [1] X + [0]
>= [1] X + [0]
= [dbl(X)]
[a__dbl(s(X))] = [0]
>= [0]
= [s(s(dbl(X)))]
[a__dbl(0())] = [0]
>= [0]
= [0()]
[a__add(X1, X2)] = [1] X1 + [1] X2 + [0]
>= [1] X1 + [1] X2 + [0]
= [add(X1, X2)]
[a__add(s(X), Y)] = [1] Y + [0]
>= [0]
= [s(add(X, Y))]
[a__add(0(), X)] = [1] X + [0]
>= [1] X + [0]
= [mark(X)]
[a__first(X1, X2)] = [1] X1 + [1] X2 + [1]
> [1] X1 + [1] X2 + [0]
= [first(X1, X2)]
[a__first(s(X), cons(Y, Z))] = [1] Y + [1]
> [1] Y + [0]
= [cons(mark(Y), first(X, Z))]
[a__first(0(), X)] = [1] X + [1]
> [0]
= [nil()]
Further, it can be verified that all rules not oriented are covered by the weightgap condition.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ 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() }
Weak 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()
, 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(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(a__terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1},
Uargs(a__sqr) = {1}, Uargs(a__dbl) = {1}, Uargs(a__add) = {1, 2},
Uargs(a__first) = {1, 2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__terms](x1) = [1] x1 + [3]
[cons](x1, x2) = [1] x1 + [0]
[recip](x1) = [1] x1 + [0]
[a__sqr](x1) = [1] x1 + [0]
[mark](x1) = [1]
[terms](x1) = [1] x1 + [0]
[s](x1) = [1]
[0] = [1]
[add](x1, x2) = [0]
[sqr](x1) = [0]
[dbl](x1) = [1] x1 + [0]
[a__dbl](x1) = [1] x1 + [4]
[a__add](x1, x2) = [1] x1 + [1] x2 + [7]
[a__first](x1, x2) = [1] x1 + [1] x2 + [7]
[nil] = [0]
[first](x1, x2) = [1] x2 + [0]
The order satisfies the following ordering constraints:
[a__terms(N)] = [1] N + [3]
> [1]
= [cons(recip(a__sqr(mark(N))), terms(s(N)))]
[a__terms(X)] = [1] X + [3]
> [1] X + [0]
= [terms(X)]
[a__sqr(X)] = [1] X + [0]
>= [0]
= [sqr(X)]
[a__sqr(s(X))] = [1]
>= [1]
= [s(add(sqr(X), dbl(X)))]
[a__sqr(0())] = [1]
>= [1]
= [0()]
[mark(cons(X1, X2))] = [1]
>= [1]
= [cons(mark(X1), X2)]
[mark(recip(X))] = [1]
>= [1]
= [recip(mark(X))]
[mark(terms(X))] = [1]
? [4]
= [a__terms(mark(X))]
[mark(s(X))] = [1]
>= [1]
= [s(X)]
[mark(0())] = [1]
>= [1]
= [0()]
[mark(add(X1, X2))] = [1]
? [9]
= [a__add(mark(X1), mark(X2))]
[mark(sqr(X))] = [1]
>= [1]
= [a__sqr(mark(X))]
[mark(dbl(X))] = [1]
? [5]
= [a__dbl(mark(X))]
[mark(nil())] = [1]
> [0]
= [nil()]
[mark(first(X1, X2))] = [1]
? [9]
= [a__first(mark(X1), mark(X2))]
[a__dbl(X)] = [1] X + [4]
> [1] X + [0]
= [dbl(X)]
[a__dbl(s(X))] = [5]
> [1]
= [s(s(dbl(X)))]
[a__dbl(0())] = [5]
> [1]
= [0()]
[a__add(X1, X2)] = [1] X1 + [1] X2 + [7]
> [0]
= [add(X1, X2)]
[a__add(s(X), Y)] = [1] Y + [8]
> [1]
= [s(add(X, Y))]
[a__add(0(), X)] = [1] X + [8]
> [1]
= [mark(X)]
[a__first(X1, X2)] = [1] X1 + [1] X2 + [7]
> [1] X2 + [0]
= [first(X1, X2)]
[a__first(s(X), cons(Y, Z))] = [1] Y + [8]
> [1]
= [cons(mark(Y), first(X, Z))]
[a__first(0(), X)] = [1] X + [8]
> [0]
= [nil()]
Further, it can be verified that all rules not oriented are covered by the weightgap condition.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ 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(first(X1, X2)) -> a__first(mark(X1), mark(X2)) }
Weak 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(nil()) -> nil()
, 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(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(a__terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1},
Uargs(a__sqr) = {1}, Uargs(a__dbl) = {1}, Uargs(a__add) = {1, 2},
Uargs(a__first) = {1, 2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__terms](x1) = [1] x1 + [7]
[cons](x1, x2) = [1] x1 + [0]
[recip](x1) = [1] x1 + [0]
[a__sqr](x1) = [1] x1 + [0]
[mark](x1) = [1]
[terms](x1) = [1] x1 + [0]
[s](x1) = [1]
[0] = [0]
[add](x1, x2) = [0]
[sqr](x1) = [1] x1 + [0]
[dbl](x1) = [0]
[a__dbl](x1) = [1] x1 + [0]
[a__add](x1, x2) = [1] x1 + [1] x2 + [6]
[a__first](x1, x2) = [1] x1 + [1] x2 + [7]
[nil] = [0]
[first](x1, x2) = [1] x2 + [0]
The order satisfies the following ordering constraints:
[a__terms(N)] = [1] N + [7]
> [1]
= [cons(recip(a__sqr(mark(N))), terms(s(N)))]
[a__terms(X)] = [1] X + [7]
> [1] X + [0]
= [terms(X)]
[a__sqr(X)] = [1] X + [0]
>= [1] X + [0]
= [sqr(X)]
[a__sqr(s(X))] = [1]
>= [1]
= [s(add(sqr(X), dbl(X)))]
[a__sqr(0())] = [0]
>= [0]
= [0()]
[mark(cons(X1, X2))] = [1]
>= [1]
= [cons(mark(X1), X2)]
[mark(recip(X))] = [1]
>= [1]
= [recip(mark(X))]
[mark(terms(X))] = [1]
? [8]
= [a__terms(mark(X))]
[mark(s(X))] = [1]
>= [1]
= [s(X)]
[mark(0())] = [1]
> [0]
= [0()]
[mark(add(X1, X2))] = [1]
? [8]
= [a__add(mark(X1), mark(X2))]
[mark(sqr(X))] = [1]
>= [1]
= [a__sqr(mark(X))]
[mark(dbl(X))] = [1]
>= [1]
= [a__dbl(mark(X))]
[mark(nil())] = [1]
> [0]
= [nil()]
[mark(first(X1, X2))] = [1]
? [9]
= [a__first(mark(X1), mark(X2))]
[a__dbl(X)] = [1] X + [0]
>= [0]
= [dbl(X)]
[a__dbl(s(X))] = [1]
>= [1]
= [s(s(dbl(X)))]
[a__dbl(0())] = [0]
>= [0]
= [0()]
[a__add(X1, X2)] = [1] X1 + [1] X2 + [6]
> [0]
= [add(X1, X2)]
[a__add(s(X), Y)] = [1] Y + [7]
> [1]
= [s(add(X, Y))]
[a__add(0(), X)] = [1] X + [6]
> [1]
= [mark(X)]
[a__first(X1, X2)] = [1] X1 + [1] X2 + [7]
> [1] X2 + [0]
= [first(X1, X2)]
[a__first(s(X), cons(Y, Z))] = [1] Y + [8]
> [1]
= [cons(mark(Y), first(X, Z))]
[a__first(0(), X)] = [1] X + [7]
> [0]
= [nil()]
Further, it can be verified that all rules not oriented are covered by the weightgap condition.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ 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(add(X1, X2)) -> a__add(mark(X1), mark(X2))
, mark(sqr(X)) -> a__sqr(mark(X))
, mark(dbl(X)) -> a__dbl(mark(X))
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2)) }
Weak 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(0()) -> 0()
, mark(nil()) -> nil()
, 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(1),O(n^2))
The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(a__terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1},
Uargs(a__sqr) = {1}, Uargs(a__dbl) = {1}, Uargs(a__add) = {1, 2},
Uargs(a__first) = {1, 2}
TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).
[a__terms](x1) = [1] x1 + [7]
[cons](x1, x2) = [1] x1 + [0]
[recip](x1) = [1] x1 + [0]
[a__sqr](x1) = [1] x1 + [4]
[mark](x1) = [1] x1 + [1]
[terms](x1) = [1] x1 + [0]
[s](x1) = [3]
[0] = [7]
[add](x1, x2) = [1] x1 + [1] x2 + [0]
[sqr](x1) = [1] x1 + [0]
[dbl](x1) = [1] x1 + [0]
[a__dbl](x1) = [1] x1 + [4]
[a__add](x1, x2) = [1] x1 + [1] x2 + [6]
[a__first](x1, x2) = [1] x1 + [1] x2 + [6]
[nil] = [0]
[first](x1, x2) = [1] x1 + [1] x2 + [0]
The order satisfies the following ordering constraints:
[a__terms(N)] = [1] N + [7]
> [1] N + [5]
= [cons(recip(a__sqr(mark(N))), terms(s(N)))]
[a__terms(X)] = [1] X + [7]
> [1] X + [0]
= [terms(X)]
[a__sqr(X)] = [1] X + [4]
> [1] X + [0]
= [sqr(X)]
[a__sqr(s(X))] = [7]
> [3]
= [s(add(sqr(X), dbl(X)))]
[a__sqr(0())] = [11]
> [7]
= [0()]
[mark(cons(X1, X2))] = [1] X1 + [1]
>= [1] X1 + [1]
= [cons(mark(X1), X2)]
[mark(recip(X))] = [1] X + [1]
>= [1] X + [1]
= [recip(mark(X))]
[mark(terms(X))] = [1] X + [1]
? [1] X + [8]
= [a__terms(mark(X))]
[mark(s(X))] = [4]
> [3]
= [s(X)]
[mark(0())] = [8]
> [7]
= [0()]
[mark(add(X1, X2))] = [1] X1 + [1] X2 + [1]
? [1] X1 + [1] X2 + [8]
= [a__add(mark(X1), mark(X2))]
[mark(sqr(X))] = [1] X + [1]
? [1] X + [5]
= [a__sqr(mark(X))]
[mark(dbl(X))] = [1] X + [1]
? [1] X + [5]
= [a__dbl(mark(X))]
[mark(nil())] = [1]
> [0]
= [nil()]
[mark(first(X1, X2))] = [1] X1 + [1] X2 + [1]
? [1] X1 + [1] X2 + [8]
= [a__first(mark(X1), mark(X2))]
[a__dbl(X)] = [1] X + [4]
> [1] X + [0]
= [dbl(X)]
[a__dbl(s(X))] = [7]
> [3]
= [s(s(dbl(X)))]
[a__dbl(0())] = [11]
> [7]
= [0()]
[a__add(X1, X2)] = [1] X1 + [1] X2 + [6]
> [1] X1 + [1] X2 + [0]
= [add(X1, X2)]
[a__add(s(X), Y)] = [1] Y + [9]
> [3]
= [s(add(X, Y))]
[a__add(0(), X)] = [1] X + [13]
> [1] X + [1]
= [mark(X)]
[a__first(X1, X2)] = [1] X1 + [1] X2 + [6]
> [1] X1 + [1] X2 + [0]
= [first(X1, X2)]
[a__first(s(X), cons(Y, Z))] = [1] Y + [9]
> [1] Y + [1]
= [cons(mark(Y), first(X, Z))]
[a__first(0(), X)] = [1] X + [13]
> [0]
= [nil()]
Further, it can be verified that all rules not oriented are covered by the weightgap condition.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(recip(X)) -> recip(mark(X))
, mark(terms(X)) -> a__terms(mark(X))
, 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(first(X1, X2)) -> a__first(mark(X1), mark(X2)) }
Weak 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(s(X)) -> s(X)
, mark(0()) -> 0()
, mark(nil()) -> nil()
, 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(1),O(n^2))
We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.
Trs:
{ mark(terms(X)) -> a__terms(mark(X))
, mark(add(X1, X2)) -> a__add(mark(X1), mark(X2))
, mark(first(X1, X2)) -> a__first(mark(X1), mark(X2)) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(a__terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1},
Uargs(a__sqr) = {1}, Uargs(a__dbl) = {1}, Uargs(a__add) = {1, 2},
Uargs(a__first) = {1, 2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[a__terms](x1) = [1 7] x1 + [1]
[0 1] [2]
[cons](x1, x2) = [1 0] x1 + [0]
[0 1] [0]
[recip](x1) = [1 0] x1 + [0]
[0 1] [0]
[a__sqr](x1) = [1 0] x1 + [0]
[0 1] [0]
[mark](x1) = [1 7] x1 + [1]
[0 1] [0]
[terms](x1) = [1 7] x1 + [0]
[0 1] [2]
[s](x1) = [0]
[0]
[0] = [0]
[0]
[add](x1, x2) = [1 5] x1 + [1 7] x2 + [0]
[0 1] [0 1] [1]
[sqr](x1) = [1 0] x1 + [0]
[0 1] [0]
[dbl](x1) = [1 0] x1 + [0]
[0 1] [0]
[a__dbl](x1) = [1 0] x1 + [0]
[0 1] [0]
[a__add](x1, x2) = [1 5] x1 + [1 7] x2 + [1]
[0 1] [0 1] [1]
[a__first](x1, x2) = [1 2] x1 + [1 7] x2 + [5]
[0 1] [0 1] [1]
[nil] = [0]
[0]
[first](x1, x2) = [1 2] x1 + [1 7] x2 + [0]
[0 1] [0 1] [1]
The order satisfies the following ordering constraints:
[a__terms(N)] = [1 7] N + [1]
[0 1] [2]
>= [1 7] N + [1]
[0 1] [0]
= [cons(recip(a__sqr(mark(N))), terms(s(N)))]
[a__terms(X)] = [1 7] X + [1]
[0 1] [2]
> [1 7] X + [0]
[0 1] [2]
= [terms(X)]
[a__sqr(X)] = [1 0] X + [0]
[0 1] [0]
>= [1 0] X + [0]
[0 1] [0]
= [sqr(X)]
[a__sqr(s(X))] = [0]
[0]
>= [0]
[0]
= [s(add(sqr(X), dbl(X)))]
[a__sqr(0())] = [0]
[0]
>= [0]
[0]
= [0()]
[mark(cons(X1, X2))] = [1 7] X1 + [1]
[0 1] [0]
>= [1 7] X1 + [1]
[0 1] [0]
= [cons(mark(X1), X2)]
[mark(recip(X))] = [1 7] X + [1]
[0 1] [0]
>= [1 7] X + [1]
[0 1] [0]
= [recip(mark(X))]
[mark(terms(X))] = [1 14] X + [15]
[0 1] [2]
> [1 14] X + [2]
[0 1] [2]
= [a__terms(mark(X))]
[mark(s(X))] = [1]
[0]
> [0]
[0]
= [s(X)]
[mark(0())] = [1]
[0]
> [0]
[0]
= [0()]
[mark(add(X1, X2))] = [1 12] X1 + [1 14] X2 + [8]
[0 1] [0 1] [1]
> [1 12] X1 + [1 14] X2 + [3]
[0 1] [0 1] [1]
= [a__add(mark(X1), mark(X2))]
[mark(sqr(X))] = [1 7] X + [1]
[0 1] [0]
>= [1 7] X + [1]
[0 1] [0]
= [a__sqr(mark(X))]
[mark(dbl(X))] = [1 7] X + [1]
[0 1] [0]
>= [1 7] X + [1]
[0 1] [0]
= [a__dbl(mark(X))]
[mark(nil())] = [1]
[0]
> [0]
[0]
= [nil()]
[mark(first(X1, X2))] = [1 9] X1 + [1 14] X2 + [8]
[0 1] [0 1] [1]
> [1 9] X1 + [1 14] X2 + [7]
[0 1] [0 1] [1]
= [a__first(mark(X1), mark(X2))]
[a__dbl(X)] = [1 0] X + [0]
[0 1] [0]
>= [1 0] X + [0]
[0 1] [0]
= [dbl(X)]
[a__dbl(s(X))] = [0]
[0]
>= [0]
[0]
= [s(s(dbl(X)))]
[a__dbl(0())] = [0]
[0]
>= [0]
[0]
= [0()]
[a__add(X1, X2)] = [1 5] X1 + [1 7] X2 + [1]
[0 1] [0 1] [1]
> [1 5] X1 + [1 7] X2 + [0]
[0 1] [0 1] [1]
= [add(X1, X2)]
[a__add(s(X), Y)] = [1 7] Y + [1]
[0 1] [1]
> [0]
[0]
= [s(add(X, Y))]
[a__add(0(), X)] = [1 7] X + [1]
[0 1] [1]
>= [1 7] X + [1]
[0 1] [0]
= [mark(X)]
[a__first(X1, X2)] = [1 2] X1 + [1 7] X2 + [5]
[0 1] [0 1] [1]
> [1 2] X1 + [1 7] X2 + [0]
[0 1] [0 1] [1]
= [first(X1, X2)]
[a__first(s(X), cons(Y, Z))] = [1 7] Y + [5]
[0 1] [1]
> [1 7] Y + [1]
[0 1] [0]
= [cons(mark(Y), first(X, Z))]
[a__first(0(), X)] = [1 7] X + [5]
[0 1] [1]
> [0]
[0]
= [nil()]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(recip(X)) -> recip(mark(X))
, mark(sqr(X)) -> a__sqr(mark(X))
, mark(dbl(X)) -> a__dbl(mark(X)) }
Weak 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(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(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(1),O(n^2))
We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.
Trs:
{ mark(cons(X1, X2)) -> cons(mark(X1), X2)
, mark(recip(X)) -> recip(mark(X)) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(a__terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1},
Uargs(a__sqr) = {1}, Uargs(a__dbl) = {1}, Uargs(a__add) = {1, 2},
Uargs(a__first) = {1, 2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[a__terms](x1) = [1 3] x1 + [6]
[0 1] [7]
[cons](x1, x2) = [1 2] x1 + [2]
[0 1] [5]
[recip](x1) = [1 0] x1 + [0]
[0 1] [2]
[a__sqr](x1) = [1 0] x1 + [0]
[0 1] [0]
[mark](x1) = [1 1] x1 + [0]
[0 1] [0]
[terms](x1) = [1 3] x1 + [4]
[0 1] [7]
[s](x1) = [0]
[0]
[0] = [0]
[0]
[add](x1, x2) = [1 0] x1 + [1 1] x2 + [0]
[0 1] [0 1] [0]
[sqr](x1) = [1 0] x1 + [0]
[0 1] [0]
[dbl](x1) = [1 0] x1 + [0]
[0 1] [0]
[a__dbl](x1) = [1 0] x1 + [0]
[0 1] [0]
[a__add](x1, x2) = [1 0] x1 + [1 1] x2 + [0]
[0 1] [0 1] [0]
[a__first](x1, x2) = [1 0] x1 + [1 1] x2 + [0]
[0 1] [0 1] [0]
[nil] = [0]
[0]
[first](x1, x2) = [1 0] x1 + [1 1] x2 + [0]
[0 1] [0 1] [0]
The order satisfies the following ordering constraints:
[a__terms(N)] = [1 3] N + [6]
[0 1] [7]
>= [1 3] N + [6]
[0 1] [7]
= [cons(recip(a__sqr(mark(N))), terms(s(N)))]
[a__terms(X)] = [1 3] X + [6]
[0 1] [7]
> [1 3] X + [4]
[0 1] [7]
= [terms(X)]
[a__sqr(X)] = [1 0] X + [0]
[0 1] [0]
>= [1 0] X + [0]
[0 1] [0]
= [sqr(X)]
[a__sqr(s(X))] = [0]
[0]
>= [0]
[0]
= [s(add(sqr(X), dbl(X)))]
[a__sqr(0())] = [0]
[0]
>= [0]
[0]
= [0()]
[mark(cons(X1, X2))] = [1 3] X1 + [7]
[0 1] [5]
> [1 3] X1 + [2]
[0 1] [5]
= [cons(mark(X1), X2)]
[mark(recip(X))] = [1 1] X + [2]
[0 1] [2]
> [1 1] X + [0]
[0 1] [2]
= [recip(mark(X))]
[mark(terms(X))] = [1 4] X + [11]
[0 1] [7]
> [1 4] X + [6]
[0 1] [7]
= [a__terms(mark(X))]
[mark(s(X))] = [0]
[0]
>= [0]
[0]
= [s(X)]
[mark(0())] = [0]
[0]
>= [0]
[0]
= [0()]
[mark(add(X1, X2))] = [1 1] X1 + [1 2] X2 + [0]
[0 1] [0 1] [0]
>= [1 1] X1 + [1 2] X2 + [0]
[0 1] [0 1] [0]
= [a__add(mark(X1), mark(X2))]
[mark(sqr(X))] = [1 1] X + [0]
[0 1] [0]
>= [1 1] X + [0]
[0 1] [0]
= [a__sqr(mark(X))]
[mark(dbl(X))] = [1 1] X + [0]
[0 1] [0]
>= [1 1] X + [0]
[0 1] [0]
= [a__dbl(mark(X))]
[mark(nil())] = [0]
[0]
>= [0]
[0]
= [nil()]
[mark(first(X1, X2))] = [1 1] X1 + [1 2] X2 + [0]
[0 1] [0 1] [0]
>= [1 1] X1 + [1 2] X2 + [0]
[0 1] [0 1] [0]
= [a__first(mark(X1), mark(X2))]
[a__dbl(X)] = [1 0] X + [0]
[0 1] [0]
>= [1 0] X + [0]
[0 1] [0]
= [dbl(X)]
[a__dbl(s(X))] = [0]
[0]
>= [0]
[0]
= [s(s(dbl(X)))]
[a__dbl(0())] = [0]
[0]
>= [0]
[0]
= [0()]
[a__add(X1, X2)] = [1 0] X1 + [1 1] X2 + [0]
[0 1] [0 1] [0]
>= [1 0] X1 + [1 1] X2 + [0]
[0 1] [0 1] [0]
= [add(X1, X2)]
[a__add(s(X), Y)] = [1 1] Y + [0]
[0 1] [0]
>= [0]
[0]
= [s(add(X, Y))]
[a__add(0(), X)] = [1 1] X + [0]
[0 1] [0]
>= [1 1] X + [0]
[0 1] [0]
= [mark(X)]
[a__first(X1, X2)] = [1 0] X1 + [1 1] X2 + [0]
[0 1] [0 1] [0]
>= [1 0] X1 + [1 1] X2 + [0]
[0 1] [0 1] [0]
= [first(X1, X2)]
[a__first(s(X), cons(Y, Z))] = [1 3] Y + [7]
[0 1] [5]
> [1 3] Y + [2]
[0 1] [5]
= [cons(mark(Y), first(X, Z))]
[a__first(0(), X)] = [1 1] X + [0]
[0 1] [0]
>= [0]
[0]
= [nil()]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ mark(sqr(X)) -> a__sqr(mark(X))
, mark(dbl(X)) -> a__dbl(mark(X)) }
Weak 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(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(1),O(n^2))
We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.
Trs:
{ mark(sqr(X)) -> a__sqr(mark(X))
, mark(dbl(X)) -> a__dbl(mark(X)) }
The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).
Sub-proof:
----------
The following argument positions are usable:
Uargs(a__terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1},
Uargs(a__sqr) = {1}, Uargs(a__dbl) = {1}, Uargs(a__add) = {1, 2},
Uargs(a__first) = {1, 2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[a__terms](x1) = [1 7] x1 + [7]
[0 1] [4]
[cons](x1, x2) = [1 0] x1 + [0]
[0 1] [0]
[recip](x1) = [1 4] x1 + [0]
[0 1] [2]
[a__sqr](x1) = [1 0] x1 + [1]
[0 1] [1]
[mark](x1) = [1 2] x1 + [1]
[0 1] [0]
[terms](x1) = [1 7] x1 + [4]
[0 1] [4]
[s](x1) = [0]
[0]
[0] = [0]
[0]
[add](x1, x2) = [1 3] x1 + [1 5] x2 + [4]
[0 1] [0 1] [1]
[sqr](x1) = [1 0] x1 + [1]
[0 1] [1]
[dbl](x1) = [1 0] x1 + [0]
[0 1] [1]
[a__dbl](x1) = [1 0] x1 + [0]
[0 1] [1]
[a__add](x1, x2) = [1 3] x1 + [1 5] x2 + [4]
[0 1] [0 1] [1]
[a__first](x1, x2) = [1 6] x1 + [1 7] x2 + [4]
[0 1] [0 1] [1]
[nil] = [0]
[0]
[first](x1, x2) = [1 6] x1 + [1 7] x2 + [4]
[0 1] [0 1] [1]
The order satisfies the following ordering constraints:
[a__terms(N)] = [1 7] N + [7]
[0 1] [4]
> [1 6] N + [6]
[0 1] [3]
= [cons(recip(a__sqr(mark(N))), terms(s(N)))]
[a__terms(X)] = [1 7] X + [7]
[0 1] [4]
> [1 7] X + [4]
[0 1] [4]
= [terms(X)]
[a__sqr(X)] = [1 0] X + [1]
[0 1] [1]
>= [1 0] X + [1]
[0 1] [1]
= [sqr(X)]
[a__sqr(s(X))] = [1]
[1]
> [0]
[0]
= [s(add(sqr(X), dbl(X)))]
[a__sqr(0())] = [1]
[1]
> [0]
[0]
= [0()]
[mark(cons(X1, X2))] = [1 2] X1 + [1]
[0 1] [0]
>= [1 2] X1 + [1]
[0 1] [0]
= [cons(mark(X1), X2)]
[mark(recip(X))] = [1 6] X + [5]
[0 1] [2]
> [1 6] X + [1]
[0 1] [2]
= [recip(mark(X))]
[mark(terms(X))] = [1 9] X + [13]
[0 1] [4]
> [1 9] X + [8]
[0 1] [4]
= [a__terms(mark(X))]
[mark(s(X))] = [1]
[0]
> [0]
[0]
= [s(X)]
[mark(0())] = [1]
[0]
> [0]
[0]
= [0()]
[mark(add(X1, X2))] = [1 5] X1 + [1 7] X2 + [7]
[0 1] [0 1] [1]
> [1 5] X1 + [1 7] X2 + [6]
[0 1] [0 1] [1]
= [a__add(mark(X1), mark(X2))]
[mark(sqr(X))] = [1 2] X + [4]
[0 1] [1]
> [1 2] X + [2]
[0 1] [1]
= [a__sqr(mark(X))]
[mark(dbl(X))] = [1 2] X + [3]
[0 1] [1]
> [1 2] X + [1]
[0 1] [1]
= [a__dbl(mark(X))]
[mark(nil())] = [1]
[0]
> [0]
[0]
= [nil()]
[mark(first(X1, X2))] = [1 8] X1 + [1 9] X2 + [7]
[0 1] [0 1] [1]
> [1 8] X1 + [1 9] X2 + [6]
[0 1] [0 1] [1]
= [a__first(mark(X1), mark(X2))]
[a__dbl(X)] = [1 0] X + [0]
[0 1] [1]
>= [1 0] X + [0]
[0 1] [1]
= [dbl(X)]
[a__dbl(s(X))] = [0]
[1]
>= [0]
[0]
= [s(s(dbl(X)))]
[a__dbl(0())] = [0]
[1]
>= [0]
[0]
= [0()]
[a__add(X1, X2)] = [1 3] X1 + [1 5] X2 + [4]
[0 1] [0 1] [1]
>= [1 3] X1 + [1 5] X2 + [4]
[0 1] [0 1] [1]
= [add(X1, X2)]
[a__add(s(X), Y)] = [1 5] Y + [4]
[0 1] [1]
> [0]
[0]
= [s(add(X, Y))]
[a__add(0(), X)] = [1 5] X + [4]
[0 1] [1]
> [1 2] X + [1]
[0 1] [0]
= [mark(X)]
[a__first(X1, X2)] = [1 6] X1 + [1 7] X2 + [4]
[0 1] [0 1] [1]
>= [1 6] X1 + [1 7] X2 + [4]
[0 1] [0 1] [1]
= [first(X1, X2)]
[a__first(s(X), cons(Y, Z))] = [1 7] Y + [4]
[0 1] [1]
> [1 2] Y + [1]
[0 1] [0]
= [cons(mark(Y), first(X, Z))]
[a__first(0(), X)] = [1 7] X + [4]
[0 1] [1]
> [0]
[0]
= [nil()]
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak 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(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^2))