YES(O(1),O(1))
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Strict Trs:
{ main(S(x1)) -> x1
, main(0()) -> 0() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
We add the following innermost weak dependency pairs:
Strict DPs:
{ main^#(S(x1)) -> c_1()
, main^#(0()) -> c_2() }
and mark the set of starting terms.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Strict DPs:
{ main^#(S(x1)) -> c_1()
, main^#(0()) -> c_2() }
Strict Trs:
{ main(S(x1)) -> x1
, main(0()) -> 0() }
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)).
Strict DPs:
{ main^#(S(x1)) -> c_1()
, main^#(0()) -> c_2() }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(1))
The weightgap principle applies (using the following constant
growth matrix-interpretation)
The following argument positions are usable:
none
TcT has computed the following constructor-restricted matrix
interpretation.
[S](x1) = [2]
[1]
[0] = [1]
[1]
[main^#](x1) = [1 1] x1 + [2]
[1 1] [1]
[c_1] = [0]
[0]
[c_2] = [1]
[1]
The following symbols are considered usable
{main^#}
The order satisfies the following ordering constraints:
[main^#(S(x1))] = [5]
[4]
> [0]
[0]
= [c_1()]
[main^#(0())] = [4]
[3]
> [1]
[1]
= [c_2()]
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(1)).
Weak DPs:
{ main^#(S(x1)) -> c_1()
, main^#(0()) -> c_2() }
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.
{ main^#(S(x1)) -> c_1()
, main^#(0()) -> c_2() }
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(1))