We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ even(Cons(x', Cons(x, xs))) -> even(xs)
, even(Cons(x, Nil())) -> False()
, even(Nil()) -> True()
, lte(Cons(x', xs'), Cons(x, xs)) -> lte(xs', xs)
, lte(Cons(x, xs), Nil()) -> False()
, lte(Nil(), y) -> True()
, notEmpty(Cons(x, xs)) -> True()
, notEmpty(Nil()) -> False()
, goal(x, y) -> and(lte(x, y), even(x)) }
Weak Trs:
{ and(True(), True()) -> True()
, and(True(), False()) -> False()
, and(False(), True()) -> False()
, and(False(), False()) -> False() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We add the following weak dependency pairs:
Strict DPs:
{ even^#(Cons(x', Cons(x, xs))) -> c_1(even^#(xs))
, even^#(Cons(x, Nil())) -> c_2()
, even^#(Nil()) -> c_3()
, lte^#(Cons(x', xs'), Cons(x, xs)) -> c_4(lte^#(xs', xs))
, lte^#(Cons(x, xs), Nil()) -> c_5()
, lte^#(Nil(), y) -> c_6()
, notEmpty^#(Cons(x, xs)) -> c_7()
, notEmpty^#(Nil()) -> c_8()
, goal^#(x, y) -> c_9(and^#(lte(x, y), even(x))) }
Weak DPs:
{ and^#(True(), True()) -> c_10()
, and^#(True(), False()) -> c_11()
, and^#(False(), True()) -> c_12()
, and^#(False(), False()) -> c_13() }
and mark the set of starting terms.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ even^#(Cons(x', Cons(x, xs))) -> c_1(even^#(xs))
, even^#(Cons(x, Nil())) -> c_2()
, even^#(Nil()) -> c_3()
, lte^#(Cons(x', xs'), Cons(x, xs)) -> c_4(lte^#(xs', xs))
, lte^#(Cons(x, xs), Nil()) -> c_5()
, lte^#(Nil(), y) -> c_6()
, notEmpty^#(Cons(x, xs)) -> c_7()
, notEmpty^#(Nil()) -> c_8()
, goal^#(x, y) -> c_9(and^#(lte(x, y), even(x))) }
Strict Trs:
{ even(Cons(x', Cons(x, xs))) -> even(xs)
, even(Cons(x, Nil())) -> False()
, even(Nil()) -> True()
, lte(Cons(x', xs'), Cons(x, xs)) -> lte(xs', xs)
, lte(Cons(x, xs), Nil()) -> False()
, lte(Nil(), y) -> True()
, notEmpty(Cons(x, xs)) -> True()
, notEmpty(Nil()) -> False()
, goal(x, y) -> and(lte(x, y), even(x)) }
Weak DPs:
{ and^#(True(), True()) -> c_10()
, and^#(True(), False()) -> c_11()
, and^#(False(), True()) -> c_12()
, and^#(False(), False()) -> c_13() }
Weak Trs:
{ and(True(), True()) -> True()
, and(True(), False()) -> False()
, and(False(), True()) -> False()
, and(False(), False()) -> False() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We replace rewrite rules by usable rules:
Strict Usable Rules:
{ even(Cons(x', Cons(x, xs))) -> even(xs)
, even(Cons(x, Nil())) -> False()
, even(Nil()) -> True()
, lte(Cons(x', xs'), Cons(x, xs)) -> lte(xs', xs)
, lte(Cons(x, xs), Nil()) -> False()
, lte(Nil(), y) -> True() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ even^#(Cons(x', Cons(x, xs))) -> c_1(even^#(xs))
, even^#(Cons(x, Nil())) -> c_2()
, even^#(Nil()) -> c_3()
, lte^#(Cons(x', xs'), Cons(x, xs)) -> c_4(lte^#(xs', xs))
, lte^#(Cons(x, xs), Nil()) -> c_5()
, lte^#(Nil(), y) -> c_6()
, notEmpty^#(Cons(x, xs)) -> c_7()
, notEmpty^#(Nil()) -> c_8()
, goal^#(x, y) -> c_9(and^#(lte(x, y), even(x))) }
Strict Trs:
{ even(Cons(x', Cons(x, xs))) -> even(xs)
, even(Cons(x, Nil())) -> False()
, even(Nil()) -> True()
, lte(Cons(x', xs'), Cons(x, xs)) -> lte(xs', xs)
, lte(Cons(x, xs), Nil()) -> False()
, lte(Nil(), y) -> True() }
Weak DPs:
{ and^#(True(), True()) -> c_10()
, and^#(True(), False()) -> c_11()
, and^#(False(), True()) -> c_12()
, and^#(False(), False()) -> c_13() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
The weightgap principle applies (using the following constant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(c_1) = {1}, Uargs(c_4) = {1}, Uargs(c_9) = {1},
Uargs(and^#) = {1, 2}
TcT has computed the following constructor-restricted matrix
interpretation.
[even](x1) = [0 1] x1 + [2]
[0 0] [0]
[lte](x1, x2) = [0 1] x1 + [0 1] x2 + [2]
[0 0] [0 0] [0]
[True] = [0]
[0]
[Cons](x1, x2) = [0 0] x2 + [0]
[0 1] [1]
[Nil] = [0]
[2]
[False] = [0]
[0]
[even^#](x1) = [0]
[0]
[c_1](x1) = [1 0] x1 + [0]
[0 1] [0]
[c_2] = [0]
[0]
[c_3] = [0]
[0]
[lte^#](x1, x2) = [0]
[0]
[c_4](x1) = [1 0] x1 + [0]
[0 1] [0]
[c_5] = [0]
[0]
[c_6] = [0]
[0]
[notEmpty^#](x1) = [0]
[0]
[c_7] = [0]
[0]
[c_8] = [0]
[0]
[goal^#](x1, x2) = [2 2] x1 + [1 2] x2 + [0]
[0 0] [0 0] [0]
[c_9](x1) = [1 0] x1 + [0]
[0 1] [0]
[and^#](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
[c_10] = [0]
[0]
[c_11] = [0]
[0]
[c_12] = [0]
[0]
[c_13] = [0]
[0]
The order satisfies the following ordering constraints:
[even(Cons(x', Cons(x, xs)))] = [0 1] xs + [4]
[0 0] [0]
> [0 1] xs + [2]
[0 0] [0]
= [even(xs)]
[even(Cons(x, Nil()))] = [5]
[0]
> [0]
[0]
= [False()]
[even(Nil())] = [4]
[0]
> [0]
[0]
= [True()]
[lte(Cons(x', xs'), Cons(x, xs))] = [0 1] xs' + [0 1] xs + [4]
[0 0] [0 0] [0]
> [0 1] xs' + [0 1] xs + [2]
[0 0] [0 0] [0]
= [lte(xs', xs)]
[lte(Cons(x, xs), Nil())] = [0 1] xs + [5]
[0 0] [0]
> [0]
[0]
= [False()]
[lte(Nil(), y)] = [0 1] y + [4]
[0 0] [0]
> [0]
[0]
= [True()]
[even^#(Cons(x', Cons(x, xs)))] = [0]
[0]
>= [0]
[0]
= [c_1(even^#(xs))]
[even^#(Cons(x, Nil()))] = [0]
[0]
>= [0]
[0]
= [c_2()]
[even^#(Nil())] = [0]
[0]
>= [0]
[0]
= [c_3()]
[lte^#(Cons(x', xs'), Cons(x, xs))] = [0]
[0]
>= [0]
[0]
= [c_4(lte^#(xs', xs))]
[lte^#(Cons(x, xs), Nil())] = [0]
[0]
>= [0]
[0]
= [c_5()]
[lte^#(Nil(), y)] = [0]
[0]
>= [0]
[0]
= [c_6()]
[notEmpty^#(Cons(x, xs))] = [0]
[0]
>= [0]
[0]
= [c_7()]
[notEmpty^#(Nil())] = [0]
[0]
>= [0]
[0]
= [c_8()]
[goal^#(x, y)] = [2 2] x + [1 2] y + [0]
[0 0] [0 0] [0]
? [0 2] x + [0 1] y + [4]
[0 0] [0 0] [0]
= [c_9(and^#(lte(x, y), even(x)))]
[and^#(True(), True())] = [0]
[0]
>= [0]
[0]
= [c_10()]
[and^#(True(), False())] = [0]
[0]
>= [0]
[0]
= [c_11()]
[and^#(False(), True())] = [0]
[0]
>= [0]
[0]
= [c_12()]
[and^#(False(), False())] = [0]
[0]
>= [0]
[0]
= [c_13()]
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:
{ even^#(Cons(x', Cons(x, xs))) -> c_1(even^#(xs))
, even^#(Cons(x, Nil())) -> c_2()
, even^#(Nil()) -> c_3()
, lte^#(Cons(x', xs'), Cons(x, xs)) -> c_4(lte^#(xs', xs))
, lte^#(Cons(x, xs), Nil()) -> c_5()
, lte^#(Nil(), y) -> c_6()
, notEmpty^#(Cons(x, xs)) -> c_7()
, notEmpty^#(Nil()) -> c_8()
, goal^#(x, y) -> c_9(and^#(lte(x, y), even(x))) }
Weak DPs:
{ and^#(True(), True()) -> c_10()
, and^#(True(), False()) -> c_11()
, and^#(False(), True()) -> c_12()
, and^#(False(), False()) -> c_13() }
Weak Trs:
{ even(Cons(x', Cons(x, xs))) -> even(xs)
, even(Cons(x, Nil())) -> False()
, even(Nil()) -> True()
, lte(Cons(x', xs'), Cons(x, xs)) -> lte(xs', xs)
, lte(Cons(x, xs), Nil()) -> False()
, lte(Nil(), y) -> True() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We estimate the number of application of {2,3,5,6,7,8,9} by
applications of Pre({2,3,5,6,7,8,9}) = {1,4}. Here rules are
labeled as follows:
DPs:
{ 1: even^#(Cons(x', Cons(x, xs))) -> c_1(even^#(xs))
, 2: even^#(Cons(x, Nil())) -> c_2()
, 3: even^#(Nil()) -> c_3()
, 4: lte^#(Cons(x', xs'), Cons(x, xs)) -> c_4(lte^#(xs', xs))
, 5: lte^#(Cons(x, xs), Nil()) -> c_5()
, 6: lte^#(Nil(), y) -> c_6()
, 7: notEmpty^#(Cons(x, xs)) -> c_7()
, 8: notEmpty^#(Nil()) -> c_8()
, 9: goal^#(x, y) -> c_9(and^#(lte(x, y), even(x)))
, 10: and^#(True(), True()) -> c_10()
, 11: and^#(True(), False()) -> c_11()
, 12: and^#(False(), True()) -> c_12()
, 13: and^#(False(), False()) -> c_13() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ even^#(Cons(x', Cons(x, xs))) -> c_1(even^#(xs))
, lte^#(Cons(x', xs'), Cons(x, xs)) -> c_4(lte^#(xs', xs)) }
Weak DPs:
{ even^#(Cons(x, Nil())) -> c_2()
, even^#(Nil()) -> c_3()
, lte^#(Cons(x, xs), Nil()) -> c_5()
, lte^#(Nil(), y) -> c_6()
, notEmpty^#(Cons(x, xs)) -> c_7()
, notEmpty^#(Nil()) -> c_8()
, goal^#(x, y) -> c_9(and^#(lte(x, y), even(x)))
, and^#(True(), True()) -> c_10()
, and^#(True(), False()) -> c_11()
, and^#(False(), True()) -> c_12()
, and^#(False(), False()) -> c_13() }
Weak Trs:
{ even(Cons(x', Cons(x, xs))) -> even(xs)
, even(Cons(x, Nil())) -> False()
, even(Nil()) -> True()
, lte(Cons(x', xs'), Cons(x, xs)) -> lte(xs', xs)
, lte(Cons(x, xs), Nil()) -> False()
, lte(Nil(), y) -> True() }
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.
{ even^#(Cons(x, Nil())) -> c_2()
, even^#(Nil()) -> c_3()
, lte^#(Cons(x, xs), Nil()) -> c_5()
, lte^#(Nil(), y) -> c_6()
, notEmpty^#(Cons(x, xs)) -> c_7()
, notEmpty^#(Nil()) -> c_8()
, goal^#(x, y) -> c_9(and^#(lte(x, y), even(x)))
, and^#(True(), True()) -> c_10()
, and^#(True(), False()) -> c_11()
, and^#(False(), True()) -> c_12()
, and^#(False(), False()) -> c_13() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ even^#(Cons(x', Cons(x, xs))) -> c_1(even^#(xs))
, lte^#(Cons(x', xs'), Cons(x, xs)) -> c_4(lte^#(xs', xs)) }
Weak Trs:
{ even(Cons(x', Cons(x, xs))) -> even(xs)
, even(Cons(x, Nil())) -> False()
, even(Nil()) -> True()
, lte(Cons(x', xs'), Cons(x, xs)) -> lte(xs', xs)
, lte(Cons(x, xs), Nil()) -> False()
, lte(Nil(), y) -> True() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^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(n^1)).
Strict DPs:
{ even^#(Cons(x', Cons(x, xs))) -> c_1(even^#(xs))
, lte^#(Cons(x', xs'), Cons(x, xs)) -> c_4(lte^#(xs', xs)) }
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: even^#(Cons(x', Cons(x, xs))) -> c_1(even^#(xs))
, 2: lte^#(Cons(x', xs'), Cons(x, xs)) -> c_4(lte^#(xs', xs)) }
Sub-proof:
----------
The input was oriented with the instance of 'Small Polynomial Path
Order (PS,1-bounded)' as induced by the safe mapping
safe(Cons) = {1, 2}, safe(even^#) = {}, safe(c_1) = {},
safe(lte^#) = {2}, safe(c_4) = {}
and precedence
empty .
Following symbols are considered recursive:
{even^#, lte^#}
The recursion depth is 1.
Further, following argument filtering is employed:
pi(Cons) = [2], pi(even^#) = [1], pi(c_1) = [1], pi(lte^#) = [1],
pi(c_4) = [1]
Usable defined function symbols are a subset of:
{even^#, lte^#}
For your convenience, here are the satisfied ordering constraints:
pi(even^#(Cons(x', Cons(x, xs)))) = even^#(Cons(; Cons(; xs));)
> c_1(even^#(xs;);)
= pi(c_1(even^#(xs)))
pi(lte^#(Cons(x', xs'), Cons(x, xs))) = lte^#(Cons(; xs');)
> c_4(lte^#(xs';);)
= pi(c_4(lte^#(xs', xs)))
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:
{ even^#(Cons(x', Cons(x, xs))) -> c_1(even^#(xs))
, lte^#(Cons(x', xs'), Cons(x, xs)) -> c_4(lte^#(xs', xs)) }
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.
{ even^#(Cons(x', Cons(x, xs))) -> c_1(even^#(xs))
, lte^#(Cons(x', xs'), Cons(x, xs)) -> c_4(lte^#(xs', xs)) }
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^1))