YES(?,O(n^3))

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^3)).

Strict Trs:
  { min(X, 0()) -> X
  , min(s(X), s(Y)) -> min(X, Y)
  , quot(0(), s(Y)) -> 0()
  , quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
  , log(s(0())) -> 0()
  , log(s(s(X))) -> s(log(s(quot(X, s(s(0())))))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^3))

We add following dependency tuples:

Strict DPs:
  { min^#(X, 0()) -> c_1()
  , min^#(s(X), s(Y)) -> c_2(min^#(X, Y))
  , quot^#(0(), s(Y)) -> c_3()
  , quot^#(s(X), s(Y)) -> c_4(quot^#(min(X, Y), s(Y)), min^#(X, Y))
  , log^#(s(0())) -> c_5()
  , log^#(s(s(X))) ->
    c_6(log^#(s(quot(X, s(s(0()))))), quot^#(X, s(s(0())))) }

and mark the set of starting terms.

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^3)).

Strict DPs:
  { min^#(X, 0()) -> c_1()
  , min^#(s(X), s(Y)) -> c_2(min^#(X, Y))
  , quot^#(0(), s(Y)) -> c_3()
  , quot^#(s(X), s(Y)) -> c_4(quot^#(min(X, Y), s(Y)), min^#(X, Y))
  , log^#(s(0())) -> c_5()
  , log^#(s(s(X))) ->
    c_6(log^#(s(quot(X, s(s(0()))))), quot^#(X, s(s(0())))) }
Weak Trs:
  { min(X, 0()) -> X
  , min(s(X), s(Y)) -> min(X, Y)
  , quot(0(), s(Y)) -> 0()
  , quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
  , log(s(0())) -> 0()
  , log(s(s(X))) -> s(log(s(quot(X, s(s(0())))))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,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: min^#(X, 0()) -> c_1()
    , 2: min^#(s(X), s(Y)) -> c_2(min^#(X, Y))
    , 3: quot^#(0(), s(Y)) -> c_3()
    , 4: quot^#(s(X), s(Y)) ->
         c_4(quot^#(min(X, Y), s(Y)), min^#(X, Y))
    , 5: log^#(s(0())) -> c_5()
    , 6: log^#(s(s(X))) ->
         c_6(log^#(s(quot(X, s(s(0()))))), quot^#(X, s(s(0())))) }

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^3)).

Strict DPs:
  { min^#(s(X), s(Y)) -> c_2(min^#(X, Y))
  , quot^#(s(X), s(Y)) -> c_4(quot^#(min(X, Y), s(Y)), min^#(X, Y))
  , log^#(s(s(X))) ->
    c_6(log^#(s(quot(X, s(s(0()))))), quot^#(X, s(s(0())))) }
Weak DPs:
  { min^#(X, 0()) -> c_1()
  , quot^#(0(), s(Y)) -> c_3()
  , log^#(s(0())) -> c_5() }
Weak Trs:
  { min(X, 0()) -> X
  , min(s(X), s(Y)) -> min(X, Y)
  , quot(0(), s(Y)) -> 0()
  , quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
  , log(s(0())) -> 0()
  , log(s(s(X))) -> s(log(s(quot(X, s(s(0())))))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^3))

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

{ min^#(X, 0()) -> c_1()
, quot^#(0(), s(Y)) -> c_3()
, log^#(s(0())) -> c_5() }

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^3)).

Strict DPs:
  { min^#(s(X), s(Y)) -> c_2(min^#(X, Y))
  , quot^#(s(X), s(Y)) -> c_4(quot^#(min(X, Y), s(Y)), min^#(X, Y))
  , log^#(s(s(X))) ->
    c_6(log^#(s(quot(X, s(s(0()))))), quot^#(X, s(s(0())))) }
Weak Trs:
  { min(X, 0()) -> X
  , min(s(X), s(Y)) -> min(X, Y)
  , quot(0(), s(Y)) -> 0()
  , quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
  , log(s(0())) -> 0()
  , log(s(s(X))) -> s(log(s(quot(X, s(s(0())))))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^3))

We replace rewrite rules by usable rules:

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

We are left with following problem, upon which TcT provides the
certificate YES(?,O(n^3)).

Strict DPs:
  { min^#(s(X), s(Y)) -> c_2(min^#(X, Y))
  , quot^#(s(X), s(Y)) -> c_4(quot^#(min(X, Y), s(Y)), min^#(X, Y))
  , log^#(s(s(X))) ->
    c_6(log^#(s(quot(X, s(s(0()))))), quot^#(X, s(s(0())))) }
Weak Trs:
  { min(X, 0()) -> X
  , min(s(X), s(Y)) -> min(X, Y)
  , quot(0(), s(Y)) -> 0()
  , quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^3))

The following argument positions are usable:
  Uargs(c_2) = {1}, Uargs(c_4) = {1, 2}, Uargs(c_6) = {1, 2}

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

                     [1 0 0]      [0]             
     [min](x1, x2) = [0 1 0] x1 + [0]             
                     [0 0 1]      [0]             
                                                  
                     [0]                          
               [0] = [0]                          
                     [0]                          
                                                  
                     [1 0 1]      [0]             
           [s](x1) = [0 1 0] x1 + [1]             
                     [0 1 1]      [1]             
                                                  
                     [1 0 0]      [0]             
    [quot](x1, x2) = [0 1 0] x1 + [0]             
                     [0 0 1]      [0]             
                                                  
                     [0 1 0]      [0]             
   [min^#](x1, x2) = [0 0 0] x1 + [0]             
                     [0 0 0]      [0]             
                                                  
                     [1 0 0]      [0]             
         [c_2](x1) = [0 0 0] x1 + [0]             
                     [0 0 0]      [0]             
                                                  
                     [0 0 1]      [1]             
  [quot^#](x1, x2) = [0 0 0] x1 + [0]             
                     [0 0 0]      [0]             
                                                  
                     [1 0 0]      [1 0 0]      [0]
     [c_4](x1, x2) = [0 0 0] x1 + [0 0 0] x2 + [0]
                     [0 0 0]      [0 0 0]      [0]
                                                  
                     [1 1 0]      [0]             
       [log^#](x1) = [0 0 0] x1 + [0]             
                     [0 0 0]      [0]             
                                                  
                     [1 0 0]      [1 0 0]      [0]
     [c_6](x1, x2) = [0 0 0] x1 + [0 0 0] x2 + [0]
                     [0 0 0]      [0 0 0]      [0]

This order satisfies following ordering constraints:

         [min(X, 0())] =  [1 0 0]     [0]                                          
                          [0 1 0] X + [0]                                          
                          [0 0 1]     [0]                                          
                       >= [1 0 0]     [0]                                          
                          [0 1 0] X + [0]                                          
                          [0 0 1]     [0]                                          
                       =  [X]                                                      
                                                                                   
     [min(s(X), s(Y))] =  [1 0 1]     [0]                                          
                          [0 1 0] X + [1]                                          
                          [0 1 1]     [1]                                          
                       >= [1 0 0]     [0]                                          
                          [0 1 0] X + [0]                                          
                          [0 0 1]     [0]                                          
                       =  [min(X, Y)]                                              
                                                                                   
     [quot(0(), s(Y))] =  [0]                                                      
                          [0]                                                      
                          [0]                                                      
                       >= [0]                                                      
                          [0]                                                      
                          [0]                                                      
                       =  [0()]                                                    
                                                                                   
    [quot(s(X), s(Y))] =  [1 0 1]     [0]                                          
                          [0 1 0] X + [1]                                          
                          [0 1 1]     [1]                                          
                       >= [1 0 1]     [0]                                          
                          [0 1 0] X + [1]                                          
                          [0 1 1]     [1]                                          
                       =  [s(quot(min(X, Y), s(Y)))]                               
                                                                                   
   [min^#(s(X), s(Y))] =  [0 1 0]     [1]                                          
                          [0 0 0] X + [0]                                          
                          [0 0 0]     [0]                                          
                       >  [0 1 0]     [0]                                          
                          [0 0 0] X + [0]                                          
                          [0 0 0]     [0]                                          
                       =  [c_2(min^#(X, Y))]                                       
                                                                                   
  [quot^#(s(X), s(Y))] =  [0 1 1]     [2]                                          
                          [0 0 0] X + [0]                                          
                          [0 0 0]     [0]                                          
                       >  [0 1 1]     [1]                                          
                          [0 0 0] X + [0]                                          
                          [0 0 0]     [0]                                          
                       =  [c_4(quot^#(min(X, Y), s(Y)), min^#(X, Y))]              
                                                                                   
      [log^#(s(s(X)))] =  [1 2 2]     [3]                                          
                          [0 0 0] X + [0]                                          
                          [0 0 0]     [0]                                          
                       >  [1 1 2]     [2]                                          
                          [0 0 0] X + [0]                                          
                          [0 0 0]     [0]                                          
                       =  [c_6(log^#(s(quot(X, s(s(0()))))), quot^#(X, s(s(0()))))]
                                                                                   

Hurray, we answered YES(?,O(n^3))