YES(?,O(n^1))

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^1)).

Strict Trs:
  { minus(0(), Y) -> 0()
  , minus(s(X), s(Y)) -> minus(X, Y)
  , geq(X, 0()) -> true()
  , geq(0(), s(Y)) -> false()
  , geq(s(X), s(Y)) -> geq(X, Y)
  , div(0(), s(Y)) -> 0()
  , div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
  , if(true(), X, Y) -> X
  , if(false(), X, Y) -> Y }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

We add following dependency tuples:

Strict DPs:
  { minus^#(0(), Y) -> c_1()
  , minus^#(s(X), s(Y)) -> c_2(minus^#(X, Y))
  , geq^#(X, 0()) -> c_3()
  , geq^#(0(), s(Y)) -> c_4()
  , geq^#(s(X), s(Y)) -> c_5(geq^#(X, Y))
  , div^#(0(), s(Y)) -> c_6()
  , div^#(s(X), s(Y)) ->
    c_7(if^#(geq(X, Y), s(div(minus(X, Y), s(Y))), 0()),
        geq^#(X, Y),
        div^#(minus(X, Y), s(Y)),
        minus^#(X, 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(n^1)).

Strict DPs:
  { minus^#(0(), Y) -> c_1()
  , minus^#(s(X), s(Y)) -> c_2(minus^#(X, Y))
  , geq^#(X, 0()) -> c_3()
  , geq^#(0(), s(Y)) -> c_4()
  , geq^#(s(X), s(Y)) -> c_5(geq^#(X, Y))
  , div^#(0(), s(Y)) -> c_6()
  , div^#(s(X), s(Y)) ->
    c_7(if^#(geq(X, Y), s(div(minus(X, Y), s(Y))), 0()),
        geq^#(X, Y),
        div^#(minus(X, Y), s(Y)),
        minus^#(X, Y))
  , if^#(true(), X, Y) -> c_8()
  , if^#(false(), X, Y) -> c_9() }
Weak Trs:
  { minus(0(), Y) -> 0()
  , minus(s(X), s(Y)) -> minus(X, Y)
  , geq(X, 0()) -> true()
  , geq(0(), s(Y)) -> false()
  , geq(s(X), s(Y)) -> geq(X, Y)
  , div(0(), s(Y)) -> 0()
  , div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
  , if(true(), X, Y) -> X
  , if(false(), X, Y) -> Y }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

We estimate the number of application of {1,3,4,6,8,9} by
applications of Pre({1,3,4,6,8,9}) = {2,5,7}. Here rules are
labeled as follows:

  DPs:
    { 1: minus^#(0(), Y) -> c_1()
    , 2: minus^#(s(X), s(Y)) -> c_2(minus^#(X, Y))
    , 3: geq^#(X, 0()) -> c_3()
    , 4: geq^#(0(), s(Y)) -> c_4()
    , 5: geq^#(s(X), s(Y)) -> c_5(geq^#(X, Y))
    , 6: div^#(0(), s(Y)) -> c_6()
    , 7: div^#(s(X), s(Y)) ->
         c_7(if^#(geq(X, Y), s(div(minus(X, Y), s(Y))), 0()),
             geq^#(X, Y),
             div^#(minus(X, Y), s(Y)),
             minus^#(X, 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(n^1)).

Strict DPs:
  { minus^#(s(X), s(Y)) -> c_2(minus^#(X, Y))
  , geq^#(s(X), s(Y)) -> c_5(geq^#(X, Y))
  , div^#(s(X), s(Y)) ->
    c_7(if^#(geq(X, Y), s(div(minus(X, Y), s(Y))), 0()),
        geq^#(X, Y),
        div^#(minus(X, Y), s(Y)),
        minus^#(X, Y)) }
Weak DPs:
  { minus^#(0(), Y) -> c_1()
  , geq^#(X, 0()) -> c_3()
  , geq^#(0(), s(Y)) -> c_4()
  , div^#(0(), s(Y)) -> c_6()
  , if^#(true(), X, Y) -> c_8()
  , if^#(false(), X, Y) -> c_9() }
Weak Trs:
  { minus(0(), Y) -> 0()
  , minus(s(X), s(Y)) -> minus(X, Y)
  , geq(X, 0()) -> true()
  , geq(0(), s(Y)) -> false()
  , geq(s(X), s(Y)) -> geq(X, Y)
  , div(0(), s(Y)) -> 0()
  , div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
  , if(true(), X, Y) -> X
  , if(false(), X, Y) -> Y }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ minus^#(0(), Y) -> c_1()
, geq^#(X, 0()) -> c_3()
, geq^#(0(), s(Y)) -> c_4()
, div^#(0(), s(Y)) -> 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(n^1)).

Strict DPs:
  { minus^#(s(X), s(Y)) -> c_2(minus^#(X, Y))
  , geq^#(s(X), s(Y)) -> c_5(geq^#(X, Y))
  , div^#(s(X), s(Y)) ->
    c_7(if^#(geq(X, Y), s(div(minus(X, Y), s(Y))), 0()),
        geq^#(X, Y),
        div^#(minus(X, Y), s(Y)),
        minus^#(X, Y)) }
Weak Trs:
  { minus(0(), Y) -> 0()
  , minus(s(X), s(Y)) -> minus(X, Y)
  , geq(X, 0()) -> true()
  , geq(0(), s(Y)) -> false()
  , geq(s(X), s(Y)) -> geq(X, Y)
  , div(0(), s(Y)) -> 0()
  , div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
  , if(true(), X, Y) -> X
  , if(false(), X, Y) -> Y }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:

  { div^#(s(X), s(Y)) ->
    c_7(if^#(geq(X, Y), s(div(minus(X, Y), s(Y))), 0()),
        geq^#(X, Y),
        div^#(minus(X, Y), s(Y)),
        minus^#(X, Y)) }

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^1)).

Strict DPs:
  { minus^#(s(X), s(Y)) -> c_1(minus^#(X, Y))
  , geq^#(s(X), s(Y)) -> c_2(geq^#(X, Y))
  , div^#(s(X), s(Y)) ->
    c_3(geq^#(X, Y), div^#(minus(X, Y), s(Y)), minus^#(X, Y)) }
Weak Trs:
  { minus(0(), Y) -> 0()
  , minus(s(X), s(Y)) -> minus(X, Y)
  , geq(X, 0()) -> true()
  , geq(0(), s(Y)) -> false()
  , geq(s(X), s(Y)) -> geq(X, Y)
  , div(0(), s(Y)) -> 0()
  , div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0())
  , if(true(), X, Y) -> X
  , if(false(), X, Y) -> Y }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { minus(0(), Y) -> 0()
    , minus(s(X), s(Y)) -> minus(X, Y) }

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^1)).

Strict DPs:
  { minus^#(s(X), s(Y)) -> c_1(minus^#(X, Y))
  , geq^#(s(X), s(Y)) -> c_2(geq^#(X, Y))
  , div^#(s(X), s(Y)) ->
    c_3(geq^#(X, Y), div^#(minus(X, Y), s(Y)), minus^#(X, Y)) }
Weak Trs:
  { minus(0(), Y) -> 0()
  , minus(s(X), s(Y)) -> minus(X, Y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:

  { div^#(s(X), s(Y)) ->
    c_3(geq^#(X, Y), div^#(minus(X, Y), s(Y)), minus^#(X, Y)) }

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^1)).

Strict DPs:
  { minus^#(s(X), s(Y)) -> c_1(minus^#(X, Y))
  , geq^#(s(X), s(Y)) -> c_2(geq^#(X, Y))
  , div^#(s(X), s(Y)) -> c_3(geq^#(X, Y), minus^#(X, Y)) }
Weak Trs:
  { minus(0(), Y) -> 0()
  , minus(s(X), s(Y)) -> minus(X, Y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,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(n^1)).

Strict DPs:
  { minus^#(s(X), s(Y)) -> c_1(minus^#(X, Y))
  , geq^#(s(X), s(Y)) -> c_2(geq^#(X, Y))
  , div^#(s(X), s(Y)) -> c_3(geq^#(X, Y), minus^#(X, Y)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The following argument positions are usable:
  Uargs(c_1) = {1}, Uargs(c_2) = {1}, Uargs(c_3) = {1, 2}

TcT has computed following constructor-based matrix interpretation
satisfying not(EDA).

            [s](x1) = [1] x1 + [2]         
                                           
  [minus^#](x1, x2) = [1] x1 + [2]         
                                           
    [geq^#](x1, x2) = [2] x2 + [0]         
                                           
    [div^#](x1, x2) = [1] x1 + [2] x2 + [1]
                                           
          [c_1](x1) = [1] x1 + [0]         
                                           
          [c_2](x1) = [1] x1 + [1]         
                                           
      [c_3](x1, x2) = [1] x1 + [1] x2 + [3]

This order satisfies following ordering constraints:

  [minus^#(s(X), s(Y))] = [1] X + [4]                      
                        > [1] X + [2]                      
                        = [c_1(minus^#(X, Y))]             
                                                           
    [geq^#(s(X), s(Y))] = [2] Y + [4]                      
                        > [2] Y + [1]                      
                        = [c_2(geq^#(X, Y))]               
                                                           
    [div^#(s(X), s(Y))] = [2] Y + [1] X + [7]              
                        > [2] Y + [1] X + [5]              
                        = [c_3(geq^#(X, Y), minus^#(X, Y))]
                                                           

Hurray, we answered YES(?,O(n^1))