We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ from(X) -> cons(X, n__from(s(X)))
, from(X) -> n__from(X)
, first(X1, X2) -> n__first(X1, X2)
, first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
, first(0(), Z) -> nil()
, activate(X) -> X
, activate(n__from(X)) -> from(X)
, activate(n__first(X1, X2)) -> first(X1, X2)
, sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
, sel(0(), cons(X, Z)) -> X }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We add the following weak dependency pairs:
Strict DPs:
{ from^#(X) -> c_1()
, from^#(X) -> c_2()
, first^#(X1, X2) -> c_3()
, first^#(s(X), cons(Y, Z)) -> c_4(activate^#(Z))
, first^#(0(), Z) -> c_5()
, activate^#(X) -> c_6()
, activate^#(n__from(X)) -> c_7(from^#(X))
, activate^#(n__first(X1, X2)) -> c_8(first^#(X1, X2))
, sel^#(s(X), cons(Y, Z)) -> c_9(sel^#(X, activate(Z)))
, sel^#(0(), cons(X, Z)) -> c_10() }
and mark the set of starting terms.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ from^#(X) -> c_1()
, from^#(X) -> c_2()
, first^#(X1, X2) -> c_3()
, first^#(s(X), cons(Y, Z)) -> c_4(activate^#(Z))
, first^#(0(), Z) -> c_5()
, activate^#(X) -> c_6()
, activate^#(n__from(X)) -> c_7(from^#(X))
, activate^#(n__first(X1, X2)) -> c_8(first^#(X1, X2))
, sel^#(s(X), cons(Y, Z)) -> c_9(sel^#(X, activate(Z)))
, sel^#(0(), cons(X, Z)) -> c_10() }
Strict Trs:
{ from(X) -> cons(X, n__from(s(X)))
, from(X) -> n__from(X)
, first(X1, X2) -> n__first(X1, X2)
, first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
, first(0(), Z) -> nil()
, activate(X) -> X
, activate(n__from(X)) -> from(X)
, activate(n__first(X1, X2)) -> first(X1, X2)
, sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
, sel(0(), cons(X, Z)) -> X }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We replace rewrite rules by usable rules:
Strict Usable Rules:
{ from(X) -> cons(X, n__from(s(X)))
, from(X) -> n__from(X)
, first(X1, X2) -> n__first(X1, X2)
, first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
, first(0(), Z) -> nil()
, activate(X) -> X
, activate(n__from(X)) -> from(X)
, activate(n__first(X1, X2)) -> first(X1, X2) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ from^#(X) -> c_1()
, from^#(X) -> c_2()
, first^#(X1, X2) -> c_3()
, first^#(s(X), cons(Y, Z)) -> c_4(activate^#(Z))
, first^#(0(), Z) -> c_5()
, activate^#(X) -> c_6()
, activate^#(n__from(X)) -> c_7(from^#(X))
, activate^#(n__first(X1, X2)) -> c_8(first^#(X1, X2))
, sel^#(s(X), cons(Y, Z)) -> c_9(sel^#(X, activate(Z)))
, sel^#(0(), cons(X, Z)) -> c_10() }
Strict Trs:
{ from(X) -> cons(X, n__from(s(X)))
, from(X) -> n__from(X)
, first(X1, X2) -> n__first(X1, X2)
, first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
, first(0(), Z) -> nil()
, activate(X) -> X
, activate(n__from(X)) -> from(X)
, activate(n__first(X1, X2)) -> first(X1, X2) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The weightgap principle applies (using the following constant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(cons) = {2}, Uargs(n__first) = {2}, Uargs(c_4) = {1},
Uargs(c_7) = {1}, Uargs(c_8) = {1}, Uargs(sel^#) = {2},
Uargs(c_9) = {1}
TcT has computed the following constructor-restricted matrix
interpretation.
[from](x1) = [2]
[2]
[cons](x1, x2) = [1 1] x2 + [0]
[0 1] [0]
[n__from](x1) = [0]
[1]
[s](x1) = [1 1] x1 + [2]
[0 0] [2]
[first](x1, x2) = [1 1] x1 + [1 0] x2 + [1]
[0 0] [0 0] [2]
[0] = [0]
[0]
[nil] = [0]
[0]
[n__first](x1, x2) = [1 1] x1 + [1 0] x2 + [0]
[0 0] [0 0] [2]
[activate](x1) = [1 1] x1 + [2]
[0 1] [1]
[from^#](x1) = [1]
[0]
[c_1] = [0]
[0]
[c_2] = [0]
[0]
[first^#](x1, x2) = [0]
[0]
[c_3] = [0]
[0]
[c_4](x1) = [1 0] x1 + [0]
[0 1] [0]
[activate^#](x1) = [0]
[0]
[c_5] = [0]
[0]
[c_6] = [0]
[0]
[c_7](x1) = [1 0] x1 + [1]
[0 1] [0]
[c_8](x1) = [1 0] x1 + [0]
[0 1] [0]
[sel^#](x1, x2) = [2 2] x2 + [0]
[0 0] [0]
[c_9](x1) = [1 0] x1 + [0]
[0 1] [0]
[c_10] = [0]
[0]
The order satisfies the following ordering constraints:
[from(X)] = [2]
[2]
> [1]
[1]
= [cons(X, n__from(s(X)))]
[from(X)] = [2]
[2]
> [0]
[1]
= [n__from(X)]
[first(X1, X2)] = [1 1] X1 + [1 0] X2 + [1]
[0 0] [0 0] [2]
> [1 1] X1 + [1 0] X2 + [0]
[0 0] [0 0] [2]
= [n__first(X1, X2)]
[first(s(X), cons(Y, Z))] = [1 1] X + [1 1] Z + [5]
[0 0] [0 0] [2]
> [1 1] X + [1 1] Z + [4]
[0 0] [0 0] [2]
= [cons(Y, n__first(X, activate(Z)))]
[first(0(), Z)] = [1 0] Z + [1]
[0 0] [2]
> [0]
[0]
= [nil()]
[activate(X)] = [1 1] X + [2]
[0 1] [1]
> [1 0] X + [0]
[0 1] [0]
= [X]
[activate(n__from(X))] = [3]
[2]
> [2]
[2]
= [from(X)]
[activate(n__first(X1, X2))] = [1 1] X1 + [1 0] X2 + [4]
[0 0] [0 0] [3]
> [1 1] X1 + [1 0] X2 + [1]
[0 0] [0 0] [2]
= [first(X1, X2)]
[from^#(X)] = [1]
[0]
> [0]
[0]
= [c_1()]
[from^#(X)] = [1]
[0]
> [0]
[0]
= [c_2()]
[first^#(X1, X2)] = [0]
[0]
>= [0]
[0]
= [c_3()]
[first^#(s(X), cons(Y, Z))] = [0]
[0]
>= [0]
[0]
= [c_4(activate^#(Z))]
[first^#(0(), Z)] = [0]
[0]
>= [0]
[0]
= [c_5()]
[activate^#(X)] = [0]
[0]
>= [0]
[0]
= [c_6()]
[activate^#(n__from(X))] = [0]
[0]
? [2]
[0]
= [c_7(from^#(X))]
[activate^#(n__first(X1, X2))] = [0]
[0]
>= [0]
[0]
= [c_8(first^#(X1, X2))]
[sel^#(s(X), cons(Y, Z))] = [2 4] Z + [0]
[0 0] [0]
? [2 4] Z + [6]
[0 0] [0]
= [c_9(sel^#(X, activate(Z)))]
[sel^#(0(), cons(X, Z))] = [2 4] Z + [0]
[0 0] [0]
>= [0]
[0]
= [c_10()]
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^1)).
Strict DPs:
{ first^#(X1, X2) -> c_3()
, first^#(s(X), cons(Y, Z)) -> c_4(activate^#(Z))
, first^#(0(), Z) -> c_5()
, activate^#(X) -> c_6()
, activate^#(n__from(X)) -> c_7(from^#(X))
, activate^#(n__first(X1, X2)) -> c_8(first^#(X1, X2))
, sel^#(s(X), cons(Y, Z)) -> c_9(sel^#(X, activate(Z)))
, sel^#(0(), cons(X, Z)) -> c_10() }
Weak DPs:
{ from^#(X) -> c_1()
, from^#(X) -> c_2() }
Weak Trs:
{ from(X) -> cons(X, n__from(s(X)))
, from(X) -> n__from(X)
, first(X1, X2) -> n__first(X1, X2)
, first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
, first(0(), Z) -> nil()
, activate(X) -> X
, activate(n__from(X)) -> from(X)
, activate(n__first(X1, X2)) -> first(X1, X2) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We estimate the number of application of {1,3,4,5,8} by
applications of Pre({1,3,4,5,8}) = {2,6,7}. Here rules are labeled
as follows:
DPs:
{ 1: first^#(X1, X2) -> c_3()
, 2: first^#(s(X), cons(Y, Z)) -> c_4(activate^#(Z))
, 3: first^#(0(), Z) -> c_5()
, 4: activate^#(X) -> c_6()
, 5: activate^#(n__from(X)) -> c_7(from^#(X))
, 6: activate^#(n__first(X1, X2)) -> c_8(first^#(X1, X2))
, 7: sel^#(s(X), cons(Y, Z)) -> c_9(sel^#(X, activate(Z)))
, 8: sel^#(0(), cons(X, Z)) -> c_10()
, 9: from^#(X) -> c_1()
, 10: from^#(X) -> c_2() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ first^#(s(X), cons(Y, Z)) -> c_4(activate^#(Z))
, activate^#(n__first(X1, X2)) -> c_8(first^#(X1, X2))
, sel^#(s(X), cons(Y, Z)) -> c_9(sel^#(X, activate(Z))) }
Weak DPs:
{ from^#(X) -> c_1()
, from^#(X) -> c_2()
, first^#(X1, X2) -> c_3()
, first^#(0(), Z) -> c_5()
, activate^#(X) -> c_6()
, activate^#(n__from(X)) -> c_7(from^#(X))
, sel^#(0(), cons(X, Z)) -> c_10() }
Weak Trs:
{ from(X) -> cons(X, n__from(s(X)))
, from(X) -> n__from(X)
, first(X1, X2) -> n__first(X1, X2)
, first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
, first(0(), Z) -> nil()
, activate(X) -> X
, activate(n__from(X)) -> from(X)
, activate(n__first(X1, X2)) -> first(X1, X2) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ from^#(X) -> c_1()
, from^#(X) -> c_2()
, first^#(X1, X2) -> c_3()
, first^#(0(), Z) -> c_5()
, activate^#(X) -> c_6()
, activate^#(n__from(X)) -> c_7(from^#(X))
, sel^#(0(), cons(X, Z)) -> c_10() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ first^#(s(X), cons(Y, Z)) -> c_4(activate^#(Z))
, activate^#(n__first(X1, X2)) -> c_8(first^#(X1, X2))
, sel^#(s(X), cons(Y, Z)) -> c_9(sel^#(X, activate(Z))) }
Weak Trs:
{ from(X) -> cons(X, n__from(s(X)))
, from(X) -> n__from(X)
, first(X1, X2) -> n__first(X1, X2)
, first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
, first(0(), Z) -> nil()
, activate(X) -> X
, activate(n__from(X)) -> from(X)
, activate(n__first(X1, X2)) -> first(X1, X2) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'Small Polynomial Path Order (PS,1-bounded)'
to orient following rules strictly.
DPs:
{ 1: first^#(s(X), cons(Y, Z)) -> c_4(activate^#(Z))
, 2: activate^#(n__first(X1, X2)) -> c_8(first^#(X1, X2))
, 3: sel^#(s(X), cons(Y, Z)) -> c_9(sel^#(X, activate(Z))) }
Sub-proof:
----------
The input was oriented with the instance of 'Small Polynomial Path
Order (PS,1-bounded)' as induced by the safe mapping
safe(from) = {1}, safe(cons) = {1, 2}, safe(n__from) = {1},
safe(s) = {1}, safe(first) = {}, safe(0) = {}, safe(nil) = {},
safe(n__first) = {1, 2}, safe(activate) = {}, safe(first^#) = {1},
safe(c_4) = {}, safe(activate^#) = {}, safe(c_8) = {},
safe(sel^#) = {2}, safe(c_9) = {}
and precedence
first > activate, sel^# > activate, first^# ~ activate^# .
Following symbols are considered recursive:
{first^#, activate^#, sel^#}
The recursion depth is 1.
Further, following argument filtering is employed:
pi(from) = [1], pi(cons) = [2], pi(n__from) = 1, pi(s) = [1],
pi(first) = [], pi(0) = [], pi(nil) = [], pi(n__first) = [2],
pi(activate) = [1], pi(first^#) = [2], pi(c_4) = [1],
pi(activate^#) = [1], pi(c_8) = [1], pi(sel^#) = [1], pi(c_9) = [1]
Usable defined function symbols are a subset of:
{first^#, activate^#, sel^#}
For your convenience, here are the satisfied ordering constraints:
pi(first^#(s(X), cons(Y, Z))) = first^#(cons(; Z);)
> c_4(activate^#(Z;);)
= pi(c_4(activate^#(Z)))
pi(activate^#(n__first(X1, X2))) = activate^#(n__first(; X2);)
> c_8(first^#(X2;);)
= pi(c_8(first^#(X1, X2)))
pi(sel^#(s(X), cons(Y, Z))) = sel^#(s(; X);)
> c_9(sel^#(X;);)
= pi(c_9(sel^#(X, activate(Z))))
The strictly oriented rules are moved into the weak component.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak DPs:
{ first^#(s(X), cons(Y, Z)) -> c_4(activate^#(Z))
, activate^#(n__first(X1, X2)) -> c_8(first^#(X1, X2))
, sel^#(s(X), cons(Y, Z)) -> c_9(sel^#(X, activate(Z))) }
Weak Trs:
{ from(X) -> cons(X, n__from(s(X)))
, from(X) -> n__from(X)
, first(X1, X2) -> n__first(X1, X2)
, first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
, first(0(), Z) -> nil()
, activate(X) -> X
, activate(n__from(X)) -> from(X)
, activate(n__first(X1, X2)) -> first(X1, X2) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ first^#(s(X), cons(Y, Z)) -> c_4(activate^#(Z))
, activate^#(n__first(X1, X2)) -> c_8(first^#(X1, X2))
, sel^#(s(X), cons(Y, Z)) -> c_9(sel^#(X, activate(Z))) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ from(X) -> cons(X, n__from(s(X)))
, from(X) -> n__from(X)
, first(X1, X2) -> n__first(X1, X2)
, first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z)))
, first(0(), Z) -> nil()
, activate(X) -> X
, activate(n__from(X)) -> from(X)
, activate(n__first(X1, X2)) -> first(X1, X2) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
No rule is usable, rules are removed from the input problem.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Rules: Empty
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
Empty rules are trivially bounded
Hurray, we answered YES(O(1),O(n^2))