We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).
Strict Trs:
{ times(x, 0()) -> 0()
, times(x, s(y)) -> plus(times(x, y), x)
, plus(x, 0()) -> x
, plus(x, s(y)) -> s(plus(x, y))
, plus(0(), x) -> x
, plus(s(x), y) -> s(plus(x, y)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^3))
We add the following dependency tuples:
Strict DPs:
{ times^#(x, 0()) -> c_1()
, times^#(x, s(y)) -> c_2(plus^#(times(x, y), x), times^#(x, y))
, plus^#(x, 0()) -> c_3()
, plus^#(x, s(y)) -> c_4(plus^#(x, y))
, plus^#(0(), x) -> c_5()
, plus^#(s(x), y) -> c_6(plus^#(x, y)) }
and mark the set of starting terms.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).
Strict DPs:
{ times^#(x, 0()) -> c_1()
, times^#(x, s(y)) -> c_2(plus^#(times(x, y), x), times^#(x, y))
, plus^#(x, 0()) -> c_3()
, plus^#(x, s(y)) -> c_4(plus^#(x, y))
, plus^#(0(), x) -> c_5()
, plus^#(s(x), y) -> c_6(plus^#(x, y)) }
Weak Trs:
{ times(x, 0()) -> 0()
, times(x, s(y)) -> plus(times(x, y), x)
, plus(x, 0()) -> x
, plus(x, s(y)) -> s(plus(x, y))
, plus(0(), x) -> x
, plus(s(x), y) -> s(plus(x, y)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^3))
We estimate the number of application of {1,3,5} by applications of
Pre({1,3,5}) = {2,4,6}. Here rules are labeled as follows:
DPs:
{ 1: times^#(x, 0()) -> c_1()
, 2: times^#(x, s(y)) -> c_2(plus^#(times(x, y), x), times^#(x, y))
, 3: plus^#(x, 0()) -> c_3()
, 4: plus^#(x, s(y)) -> c_4(plus^#(x, y))
, 5: plus^#(0(), x) -> c_5()
, 6: plus^#(s(x), y) -> c_6(plus^#(x, y)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).
Strict DPs:
{ times^#(x, s(y)) -> c_2(plus^#(times(x, y), x), times^#(x, y))
, plus^#(x, s(y)) -> c_4(plus^#(x, y))
, plus^#(s(x), y) -> c_6(plus^#(x, y)) }
Weak DPs:
{ times^#(x, 0()) -> c_1()
, plus^#(x, 0()) -> c_3()
, plus^#(0(), x) -> c_5() }
Weak Trs:
{ times(x, 0()) -> 0()
, times(x, s(y)) -> plus(times(x, y), x)
, plus(x, 0()) -> x
, plus(x, s(y)) -> s(plus(x, y))
, plus(0(), x) -> x
, plus(s(x), y) -> s(plus(x, y)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^3))
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ times^#(x, 0()) -> c_1()
, plus^#(x, 0()) -> c_3()
, plus^#(0(), x) -> c_5() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).
Strict DPs:
{ times^#(x, s(y)) -> c_2(plus^#(times(x, y), x), times^#(x, y))
, plus^#(x, s(y)) -> c_4(plus^#(x, y))
, plus^#(s(x), y) -> c_6(plus^#(x, y)) }
Weak Trs:
{ times(x, 0()) -> 0()
, times(x, s(y)) -> plus(times(x, y), x)
, plus(x, 0()) -> x
, plus(x, s(y)) -> s(plus(x, y))
, plus(0(), x) -> x
, plus(s(x), y) -> s(plus(x, y)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^3))
We decompose the input problem according to the dependency graph
into the upper component
{ times^#(x, s(y)) -> c_2(plus^#(times(x, y), x), times^#(x, y)) }
and lower component
{ plus^#(x, s(y)) -> c_4(plus^#(x, y))
, plus^#(s(x), y) -> c_6(plus^#(x, y)) }
Further, following extension rules are added to the lower
component.
{ times^#(x, s(y)) -> times^#(x, y)
, times^#(x, s(y)) -> plus^#(times(x, y), x) }
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:
{ times^#(x, s(y)) -> c_2(plus^#(times(x, y), x), times^#(x, y)) }
Weak Trs:
{ times(x, 0()) -> 0()
, times(x, s(y)) -> plus(times(x, y), x)
, plus(x, 0()) -> x
, plus(x, s(y)) -> s(plus(x, y))
, plus(0(), x) -> x
, plus(s(x), y) -> s(plus(x, y)) }
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: times^#(x, s(y)) ->
c_2(plus^#(times(x, y), x), times^#(x, y)) }
Sub-proof:
----------
The input was oriented with the instance of 'Small Polynomial Path
Order (PS,1-bounded)' as induced by the safe mapping
safe(times) = {}, safe(0) = {}, safe(s) = {1}, safe(plus) = {2},
safe(times^#) = {}, safe(c_2) = {}, safe(plus^#) = {}
and precedence
empty .
Following symbols are considered recursive:
{times^#}
The recursion depth is 1.
Further, following argument filtering is employed:
pi(times) = 2, pi(0) = [], pi(s) = [1], pi(plus) = [1, 2],
pi(times^#) = [1, 2], pi(c_2) = [1, 2], pi(plus^#) = []
Usable defined function symbols are a subset of:
{times^#, plus^#}
For your convenience, here are the satisfied ordering constraints:
pi(times^#(x, s(y))) = times^#(x, s(; y);)
> c_2(plus^#(), times^#(x, y;);)
= pi(c_2(plus^#(times(x, y), x), times^#(x, 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:
{ times^#(x, s(y)) -> c_2(plus^#(times(x, y), x), times^#(x, y)) }
Weak Trs:
{ times(x, 0()) -> 0()
, times(x, s(y)) -> plus(times(x, y), x)
, plus(x, 0()) -> x
, plus(x, s(y)) -> s(plus(x, y))
, plus(0(), x) -> x
, plus(s(x), y) -> s(plus(x, y)) }
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.
{ times^#(x, s(y)) -> c_2(plus^#(times(x, y), x), times^#(x, y)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ times(x, 0()) -> 0()
, times(x, s(y)) -> plus(times(x, y), x)
, plus(x, 0()) -> x
, plus(x, s(y)) -> s(plus(x, y))
, plus(0(), x) -> x
, plus(s(x), y) -> s(plus(x, y)) }
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^2)).
Strict DPs:
{ plus^#(x, s(y)) -> c_4(plus^#(x, y))
, plus^#(s(x), y) -> c_6(plus^#(x, y)) }
Weak DPs:
{ times^#(x, s(y)) -> times^#(x, y)
, times^#(x, s(y)) -> plus^#(times(x, y), x) }
Weak Trs:
{ times(x, 0()) -> 0()
, times(x, s(y)) -> plus(times(x, y), x)
, plus(x, 0()) -> x
, plus(x, s(y)) -> s(plus(x, y))
, plus(0(), x) -> x
, plus(s(x), y) -> s(plus(x, y)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
DPs:
{ 1: plus^#(x, s(y)) -> c_4(plus^#(x, y))
, 4: times^#(x, s(y)) -> plus^#(times(x, y), x) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_4) = {1}, Uargs(c_6) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[times](x1, x2) = [0]
[0] = [0]
[s](x1) = [1] x1 + [1]
[plus](x1, x2) = [0]
[times^#](x1, x2) = [7] x1 + [1]
[plus^#](x1, x2) = [1] x2 + [0]
[c_4](x1) = [1] x1 + [0]
[c_6](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[times(x, 0())] = [0]
>= [0]
= [0()]
[times(x, s(y))] = [0]
>= [0]
= [plus(times(x, y), x)]
[plus(x, 0())] = [0]
? [1] x + [0]
= [x]
[plus(x, s(y))] = [0]
? [1]
= [s(plus(x, y))]
[plus(0(), x)] = [0]
? [1] x + [0]
= [x]
[plus(s(x), y)] = [0]
? [1]
= [s(plus(x, y))]
[times^#(x, s(y))] = [7] x + [1]
>= [7] x + [1]
= [times^#(x, y)]
[times^#(x, s(y))] = [7] x + [1]
> [1] x + [0]
= [plus^#(times(x, y), x)]
[plus^#(x, s(y))] = [1] y + [1]
> [1] y + [0]
= [c_4(plus^#(x, y))]
[plus^#(s(x), y)] = [1] y + [0]
>= [1] y + [0]
= [c_6(plus^#(x, 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(n^2)).
Strict DPs: { plus^#(s(x), y) -> c_6(plus^#(x, y)) }
Weak DPs:
{ times^#(x, s(y)) -> times^#(x, y)
, times^#(x, s(y)) -> plus^#(times(x, y), x)
, plus^#(x, s(y)) -> c_4(plus^#(x, y)) }
Weak Trs:
{ times(x, 0()) -> 0()
, times(x, s(y)) -> plus(times(x, y), x)
, plus(x, 0()) -> x
, plus(x, s(y)) -> s(plus(x, y))
, plus(0(), x) -> x
, plus(s(x), y) -> s(plus(x, y)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'custom shape polynomial interpretation' to
orient following rules strictly.
DPs:
{ 1: plus^#(s(x), y) -> c_6(plus^#(x, y))
, 2: times^#(x, s(y)) -> times^#(x, y)
, 3: times^#(x, s(y)) -> plus^#(times(x, y), x) }
Trs:
{ times(x, 0()) -> 0()
, times(x, s(y)) -> plus(times(x, y), x)
, plus(x, 0()) -> x
, plus(0(), x) -> x }
Sub-proof:
----------
The following argument positions are considered usable:
Uargs(c_4) = {1}, Uargs(c_6) = {1}
TcT has computed the following constructor-restricted polynomial
interpretation.
[times](x1, x2) = x1*x2 + 2*x2
[0]() = 2
[s](x1) = 1 + x1
[plus](x1, x2) = x1 + x2
[times^#](x1, x2) = 3 + x1*x2 + 3*x1^2 + 3*x2 + x2^2
[plus^#](x1, x2) = 2 + x1 + x2 + x2^2
[c_4](x1) = 2 + x1
[c_6](x1) = x1
This order satisfies the following ordering constraints.
[times(x, 0())] = 2*x + 4
> 2
= [0()]
[times(x, s(y))] = x + x*y + 2 + 2*y
> x*y + 2*y + x
= [plus(times(x, y), x)]
[plus(x, 0())] = x + 2
> x
= [x]
[plus(x, s(y))] = x + 1 + y
>= 1 + x + y
= [s(plus(x, y))]
[plus(0(), x)] = 2 + x
> x
= [x]
[plus(s(x), y)] = 1 + x + y
>= 1 + x + y
= [s(plus(x, y))]
[times^#(x, s(y))] = 7 + x + x*y + 3*x^2 + 5*y + y^2
> 3 + x*y + 3*x^2 + 3*y + y^2
= [times^#(x, y)]
[times^#(x, s(y))] = 7 + x + x*y + 3*x^2 + 5*y + y^2
> 2 + x*y + 2*y + x + x^2
= [plus^#(times(x, y), x)]
[plus^#(x, s(y))] = 4 + x + 3*y + y^2
>= 4 + x + y + y^2
= [c_4(plus^#(x, y))]
[plus^#(s(x), y)] = 3 + x + y + y^2
> 2 + x + y + y^2
= [c_6(plus^#(x, 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:
{ times^#(x, s(y)) -> times^#(x, y)
, times^#(x, s(y)) -> plus^#(times(x, y), x)
, plus^#(x, s(y)) -> c_4(plus^#(x, y))
, plus^#(s(x), y) -> c_6(plus^#(x, y)) }
Weak Trs:
{ times(x, 0()) -> 0()
, times(x, s(y)) -> plus(times(x, y), x)
, plus(x, 0()) -> x
, plus(x, s(y)) -> s(plus(x, y))
, plus(0(), x) -> x
, plus(s(x), y) -> s(plus(x, y)) }
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.
{ times^#(x, s(y)) -> times^#(x, y)
, times^#(x, s(y)) -> plus^#(times(x, y), x)
, plus^#(x, s(y)) -> c_4(plus^#(x, y))
, plus^#(s(x), y) -> c_6(plus^#(x, y)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ times(x, 0()) -> 0()
, times(x, s(y)) -> plus(times(x, y), x)
, plus(x, 0()) -> x
, plus(x, s(y)) -> s(plus(x, y))
, plus(0(), x) -> x
, plus(s(x), y) -> s(plus(x, y)) }
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^3))