We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ p(0()) -> 0()
, p(s(x)) -> x
, le(0(), y) -> true()
, le(s(x), 0()) -> false()
, le(s(x), s(y)) -> le(x, y)
, minus(x, 0()) -> x
, minus(x, s(y)) -> if(le(x, s(y)), 0(), p(minus(x, p(s(y)))))
, if(true(), x, y) -> x
, if(false(), x, y) -> y }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We add the following dependency tuples:
Strict DPs:
{ p^#(0()) -> c_1()
, p^#(s(x)) -> c_2()
, le^#(0(), y) -> c_3()
, le^#(s(x), 0()) -> c_4()
, le^#(s(x), s(y)) -> c_5(le^#(x, y))
, minus^#(x, 0()) -> c_6()
, minus^#(x, s(y)) ->
c_7(if^#(le(x, s(y)), 0(), p(minus(x, p(s(y))))),
le^#(x, s(y)),
p^#(minus(x, p(s(y)))),
minus^#(x, p(s(y))),
p^#(s(y)))
, if^#(true(), x, y) -> c_8()
, if^#(false(), x, y) -> c_9() }
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:
{ p^#(0()) -> c_1()
, p^#(s(x)) -> c_2()
, le^#(0(), y) -> c_3()
, le^#(s(x), 0()) -> c_4()
, le^#(s(x), s(y)) -> c_5(le^#(x, y))
, minus^#(x, 0()) -> c_6()
, minus^#(x, s(y)) ->
c_7(if^#(le(x, s(y)), 0(), p(minus(x, p(s(y))))),
le^#(x, s(y)),
p^#(minus(x, p(s(y)))),
minus^#(x, p(s(y))),
p^#(s(y)))
, if^#(true(), x, y) -> c_8()
, if^#(false(), x, y) -> c_9() }
Weak Trs:
{ p(0()) -> 0()
, p(s(x)) -> x
, le(0(), y) -> true()
, le(s(x), 0()) -> false()
, le(s(x), s(y)) -> le(x, y)
, minus(x, 0()) -> x
, minus(x, s(y)) -> if(le(x, s(y)), 0(), p(minus(x, p(s(y)))))
, if(true(), x, y) -> x
, if(false(), x, y) -> y }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We estimate the number of application of {1,2,3,4,6,8,9} by
applications of Pre({1,2,3,4,6,8,9}) = {5,7}. Here rules are
labeled as follows:
DPs:
{ 1: p^#(0()) -> c_1()
, 2: p^#(s(x)) -> c_2()
, 3: le^#(0(), y) -> c_3()
, 4: le^#(s(x), 0()) -> c_4()
, 5: le^#(s(x), s(y)) -> c_5(le^#(x, y))
, 6: minus^#(x, 0()) -> c_6()
, 7: minus^#(x, s(y)) ->
c_7(if^#(le(x, s(y)), 0(), p(minus(x, p(s(y))))),
le^#(x, s(y)),
p^#(minus(x, p(s(y)))),
minus^#(x, p(s(y))),
p^#(s(y)))
, 8: if^#(true(), x, y) -> c_8()
, 9: if^#(false(), x, y) -> c_9() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ le^#(s(x), s(y)) -> c_5(le^#(x, y))
, minus^#(x, s(y)) ->
c_7(if^#(le(x, s(y)), 0(), p(minus(x, p(s(y))))),
le^#(x, s(y)),
p^#(minus(x, p(s(y)))),
minus^#(x, p(s(y))),
p^#(s(y))) }
Weak DPs:
{ p^#(0()) -> c_1()
, p^#(s(x)) -> c_2()
, le^#(0(), y) -> c_3()
, le^#(s(x), 0()) -> c_4()
, minus^#(x, 0()) -> c_6()
, if^#(true(), x, y) -> c_8()
, if^#(false(), x, y) -> c_9() }
Weak Trs:
{ p(0()) -> 0()
, p(s(x)) -> x
, le(0(), y) -> true()
, le(s(x), 0()) -> false()
, le(s(x), s(y)) -> le(x, y)
, minus(x, 0()) -> x
, minus(x, s(y)) -> if(le(x, s(y)), 0(), p(minus(x, p(s(y)))))
, if(true(), x, y) -> x
, if(false(), x, y) -> y }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ p^#(0()) -> c_1()
, p^#(s(x)) -> c_2()
, le^#(0(), y) -> c_3()
, le^#(s(x), 0()) -> c_4()
, minus^#(x, 0()) -> c_6()
, if^#(true(), x, y) -> c_8()
, if^#(false(), x, y) -> c_9() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ le^#(s(x), s(y)) -> c_5(le^#(x, y))
, minus^#(x, s(y)) ->
c_7(if^#(le(x, s(y)), 0(), p(minus(x, p(s(y))))),
le^#(x, s(y)),
p^#(minus(x, p(s(y)))),
minus^#(x, p(s(y))),
p^#(s(y))) }
Weak Trs:
{ p(0()) -> 0()
, p(s(x)) -> x
, le(0(), y) -> true()
, le(s(x), 0()) -> false()
, le(s(x), s(y)) -> le(x, y)
, minus(x, 0()) -> x
, minus(x, s(y)) -> if(le(x, s(y)), 0(), p(minus(x, p(s(y)))))
, if(true(), x, y) -> x
, if(false(), x, y) -> y }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:
{ minus^#(x, s(y)) ->
c_7(if^#(le(x, s(y)), 0(), p(minus(x, p(s(y))))),
le^#(x, s(y)),
p^#(minus(x, p(s(y)))),
minus^#(x, p(s(y))),
p^#(s(y))) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ le^#(s(x), s(y)) -> c_1(le^#(x, y))
, minus^#(x, s(y)) -> c_2(le^#(x, s(y)), minus^#(x, p(s(y)))) }
Weak Trs:
{ p(0()) -> 0()
, p(s(x)) -> x
, le(0(), y) -> true()
, le(s(x), 0()) -> false()
, le(s(x), s(y)) -> le(x, y)
, minus(x, 0()) -> x
, minus(x, s(y)) -> if(le(x, s(y)), 0(), p(minus(x, p(s(y)))))
, if(true(), x, y) -> x
, if(false(), x, y) -> y }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We replace rewrite rules by usable rules:
Weak Usable Rules:
{ p(0()) -> 0()
, p(s(x)) -> x }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ le^#(s(x), s(y)) -> c_1(le^#(x, y))
, minus^#(x, s(y)) -> c_2(le^#(x, s(y)), minus^#(x, p(s(y)))) }
Weak Trs:
{ p(0()) -> 0()
, p(s(x)) -> x }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We decompose the input problem according to the dependency graph
into the upper component
{ minus^#(x, s(y)) -> c_2(le^#(x, s(y)), minus^#(x, p(s(y)))) }
and lower component
{ le^#(s(x), s(y)) -> c_1(le^#(x, y)) }
Further, following extension rules are added to the lower
component.
{ minus^#(x, s(y)) -> le^#(x, s(y))
, minus^#(x, s(y)) -> minus^#(x, p(s(y))) }
TcT solves the upper component with certificate YES(O(1),O(n^1)).
Sub-proof:
----------
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ minus^#(x, s(y)) -> c_2(le^#(x, s(y)), minus^#(x, p(s(y)))) }
Weak Trs:
{ p(0()) -> 0()
, p(s(x)) -> x }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 3' to
orient following rules strictly.
DPs:
{ 1: minus^#(x, s(y)) -> c_2(le^#(x, s(y)), minus^#(x, p(s(y)))) }
Trs:
{ p(0()) -> 0()
, p(s(x)) -> x }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_2) = {2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(1)).
[0 0 1] [1]
[p](x1) = [1 1 1] x1 + [0]
[1 1 0] [0]
[0]
[0] = [0]
[0]
[1 0 0] [7]
[s](x1) = [1 1 2] x1 + [0]
[1 0 0] [3]
[0 0 0] [1]
[le^#](x1, x2) = [0 0 1] x2 + [1]
[0 1 0] [0]
[0 0 0] [2 0 0] [0]
[minus^#](x1, x2) = [0 4 4] x1 + [0 0 0] x2 + [1]
[0 0 0] [0 0 0] [1]
[2 0 0] [1 0 3] [0]
[c_2](x1, x2) = [0 0 0] x1 + [0 0 0] x2 + [0]
[0 0 0] [0 0 0] [0]
The order satisfies the following ordering constraints:
[p(0())] = [1]
[0]
[0]
> [0]
[0]
[0]
= [0()]
[p(s(x))] = [1 0 0] [4]
[3 1 2] x + [10]
[2 1 2] [7]
> [1 0 0] [0]
[0 1 0] x + [0]
[0 0 1] [0]
= [x]
[minus^#(x, s(y))] = [0 0 0] [2 0 0] [14]
[0 4 4] x + [0 0 0] y + [1]
[0 0 0] [0 0 0] [1]
> [2 0 0] [13]
[0 0 0] y + [0]
[0 0 0] [0]
= [c_2(le^#(x, s(y)), minus^#(x, p(s(y))))]
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:
{ minus^#(x, s(y)) -> c_2(le^#(x, s(y)), minus^#(x, p(s(y)))) }
Weak Trs:
{ p(0()) -> 0()
, p(s(x)) -> x }
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.
{ minus^#(x, s(y)) -> c_2(le^#(x, s(y)), minus^#(x, p(s(y)))) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ p(0()) -> 0()
, p(s(x)) -> x }
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
We return to the main proof.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs: { le^#(s(x), s(y)) -> c_1(le^#(x, y)) }
Weak DPs:
{ minus^#(x, s(y)) -> le^#(x, s(y))
, minus^#(x, s(y)) -> minus^#(x, p(s(y))) }
Weak Trs:
{ p(0()) -> 0()
, p(s(x)) -> x }
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: le^#(s(x), s(y)) -> c_1(le^#(x, y)) }
Trs: { p(s(x)) -> x }
Sub-proof:
----------
The input was oriented with the instance of 'Small Polynomial Path
Order (PS,1-bounded)' as induced by the safe mapping
safe(p) = {1}, safe(0) = {}, safe(s) = {1}, safe(le^#) = {},
safe(minus^#) = {}, safe(c_1) = {}
and precedence
minus^# > p, le^# ~ minus^# .
Following symbols are considered recursive:
{le^#, minus^#}
The recursion depth is 1.
Further, following argument filtering is employed:
pi(p) = 1, pi(0) = [], pi(s) = [1], pi(le^#) = [1, 2],
pi(minus^#) = [1, 2], pi(c_1) = [1]
Usable defined function symbols are a subset of:
{p, le^#, minus^#}
For your convenience, here are the satisfied ordering constraints:
pi(le^#(s(x), s(y))) = le^#(s(; x), s(; y);)
> c_1(le^#(x, y;);)
= pi(c_1(le^#(x, y)))
pi(minus^#(x, s(y))) = minus^#(x, s(; y);)
>= le^#(x, s(; y);)
= pi(le^#(x, s(y)))
pi(minus^#(x, s(y))) = minus^#(x, s(; y);)
>= minus^#(x, s(; y);)
= pi(minus^#(x, p(s(y))))
pi(p(0())) = 0()
>= 0()
= pi(0())
pi(p(s(x))) = s(; x)
> x
= pi(x)
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:
{ le^#(s(x), s(y)) -> c_1(le^#(x, y))
, minus^#(x, s(y)) -> le^#(x, s(y))
, minus^#(x, s(y)) -> minus^#(x, p(s(y))) }
Weak Trs:
{ p(0()) -> 0()
, p(s(x)) -> x }
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.
{ le^#(s(x), s(y)) -> c_1(le^#(x, y))
, minus^#(x, s(y)) -> le^#(x, s(y))
, minus^#(x, s(y)) -> minus^#(x, p(s(y))) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ p(0()) -> 0()
, p(s(x)) -> x }
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))