We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, bits(0()) -> 0()
, bits(s(x)) -> s(bits(half(s(x)))) }
Obligation:
runtime complexity
Answer:
YES(O(1),O(n^1))
The input is overlay and right-linear. Switching to innermost
rewriting.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, bits(0()) -> 0()
, bits(s(x)) -> s(bits(half(s(x)))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We add the following weak dependency pairs:
Strict DPs:
{ half^#(0()) -> c_1()
, half^#(s(0())) -> c_2()
, half^#(s(s(x))) -> c_3(half^#(x))
, bits^#(0()) -> c_4()
, bits^#(s(x)) -> c_5(bits^#(half(s(x)))) }
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:
{ half^#(0()) -> c_1()
, half^#(s(0())) -> c_2()
, half^#(s(s(x))) -> c_3(half^#(x))
, bits^#(0()) -> c_4()
, bits^#(s(x)) -> c_5(bits^#(half(s(x)))) }
Strict Trs:
{ half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, bits(0()) -> 0()
, bits(s(x)) -> s(bits(half(s(x)))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We replace rewrite rules by usable rules:
Strict Usable Rules:
{ half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ half^#(0()) -> c_1()
, half^#(s(0())) -> c_2()
, half^#(s(s(x))) -> c_3(half^#(x))
, bits^#(0()) -> c_4()
, bits^#(s(x)) -> c_5(bits^#(half(s(x)))) }
Strict Trs:
{ half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x)) }
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(s) = {1}, Uargs(c_3) = {1}, Uargs(bits^#) = {1},
Uargs(c_5) = {1}
TcT has computed the following constructor-restricted matrix
interpretation.
[half](x1) = [1 0] x1 + [1]
[0 0] [1]
[0] = [2]
[0]
[s](x1) = [1 1] x1 + [2]
[0 0] [0]
[half^#](x1) = [0]
[0]
[c_1] = [0]
[0]
[c_2] = [0]
[0]
[c_3](x1) = [1 0] x1 + [0]
[0 1] [0]
[bits^#](x1) = [2 0] x1 + [0]
[0 0] [0]
[c_4] = [0]
[0]
[c_5](x1) = [1 0] x1 + [0]
[0 1] [0]
The order satisfies the following ordering constraints:
[half(0())] = [3]
[1]
> [2]
[0]
= [0()]
[half(s(0()))] = [5]
[1]
> [2]
[0]
= [0()]
[half(s(s(x)))] = [1 1] x + [5]
[0 0] [1]
> [1 0] x + [4]
[0 0] [0]
= [s(half(x))]
[half^#(0())] = [0]
[0]
>= [0]
[0]
= [c_1()]
[half^#(s(0()))] = [0]
[0]
>= [0]
[0]
= [c_2()]
[half^#(s(s(x)))] = [0]
[0]
>= [0]
[0]
= [c_3(half^#(x))]
[bits^#(0())] = [4]
[0]
> [0]
[0]
= [c_4()]
[bits^#(s(x))] = [2 2] x + [4]
[0 0] [0]
? [2 2] x + [6]
[0 0] [0]
= [c_5(bits^#(half(s(x))))]
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:
{ half^#(0()) -> c_1()
, half^#(s(0())) -> c_2()
, half^#(s(s(x))) -> c_3(half^#(x))
, bits^#(s(x)) -> c_5(bits^#(half(s(x)))) }
Weak DPs: { bits^#(0()) -> c_4() }
Weak Trs:
{ half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We estimate the number of application of {1,2} by applications of
Pre({1,2}) = {3}. Here rules are labeled as follows:
DPs:
{ 1: half^#(0()) -> c_1()
, 2: half^#(s(0())) -> c_2()
, 3: half^#(s(s(x))) -> c_3(half^#(x))
, 4: bits^#(s(x)) -> c_5(bits^#(half(s(x))))
, 5: bits^#(0()) -> c_4() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ half^#(s(s(x))) -> c_3(half^#(x))
, bits^#(s(x)) -> c_5(bits^#(half(s(x)))) }
Weak DPs:
{ half^#(0()) -> c_1()
, half^#(s(0())) -> c_2()
, bits^#(0()) -> c_4() }
Weak Trs:
{ half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x)) }
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.
{ half^#(0()) -> c_1()
, half^#(s(0())) -> c_2()
, bits^#(0()) -> c_4() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ half^#(s(s(x))) -> c_3(half^#(x))
, bits^#(s(x)) -> c_5(bits^#(half(s(x)))) }
Weak Trs:
{ half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
DPs:
{ 1: half^#(s(s(x))) -> c_3(half^#(x)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_3) = {1}, Uargs(c_5) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[half](x1) = [0]
[0] = [0]
[s](x1) = [1] x1 + [1]
[half^#](x1) = [1] x1 + [7]
[c_3](x1) = [1] x1 + [0]
[bits^#](x1) = [2]
[c_5](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[half(0())] = [0]
>= [0]
= [0()]
[half(s(0()))] = [0]
>= [0]
= [0()]
[half(s(s(x)))] = [0]
? [1]
= [s(half(x))]
[half^#(s(s(x)))] = [1] x + [9]
> [1] x + [7]
= [c_3(half^#(x))]
[bits^#(s(x))] = [2]
>= [2]
= [c_5(bits^#(half(s(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(n^1)).
Strict DPs: { bits^#(s(x)) -> c_5(bits^#(half(s(x)))) }
Weak DPs: { half^#(s(s(x))) -> c_3(half^#(x)) }
Weak Trs:
{ half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x)) }
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.
{ half^#(s(s(x))) -> c_3(half^#(x)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs: { bits^#(s(x)) -> c_5(bits^#(half(s(x)))) }
Weak Trs:
{ half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.
DPs:
{ 1: bits^#(s(x)) -> c_5(bits^#(half(s(x)))) }
Trs:
{ half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_5) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(1)).
[half](x1) = [1 0] x1 + [0]
[1 0] [0]
[0] = [0]
[0]
[s](x1) = [1 0] x1 + [3]
[1 0] [4]
[half^#](x1) = [0]
[0]
[c_3](x1) = [0]
[0]
[bits^#](x1) = [1 2] x1 + [2]
[0 0] [1]
[c_5](x1) = [1 1] x1 + [0]
[0 0] [0]
The order satisfies the following ordering constraints:
[half(0())] = [0]
[0]
>= [0]
[0]
= [0()]
[half(s(0()))] = [3]
[3]
> [0]
[0]
= [0()]
[half(s(s(x)))] = [1 0] x + [6]
[1 0] [6]
> [1 0] x + [3]
[1 0] [4]
= [s(half(x))]
[bits^#(s(x))] = [3 0] x + [13]
[0 0] [1]
> [3 0] x + [12]
[0 0] [0]
= [c_5(bits^#(half(s(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: { bits^#(s(x)) -> c_5(bits^#(half(s(x)))) }
Weak Trs:
{ half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(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.
{ bits^#(s(x)) -> c_5(bits^#(half(s(x)))) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(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^1))