YES(O(1),O(n^1))

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

Strict Trs:
  { minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y)
  , quot(0(), s(y)) -> 0()
  , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We add following weak dependency pairs:

Strict DPs:
  { minus^#(x, 0()) -> c_1()
  , minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
  , quot^#(0(), s(y)) -> c_3()
  , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(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^1)).

Strict DPs:
  { minus^#(x, 0()) -> c_1()
  , minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
  , quot^#(0(), s(y)) -> c_3()
  , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) }
Strict Trs:
  { minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y)
  , quot(0(), s(y)) -> 0()
  , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We replace rewrite rules by usable rules:

  Strict Usable Rules:
    { minus(x, 0()) -> x
    , minus(s(x), s(y)) -> minus(x, y) }

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

Strict DPs:
  { minus^#(x, 0()) -> c_1()
  , minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
  , quot^#(0(), s(y)) -> c_3()
  , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) }
Strict Trs:
  { minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y) }
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(c_2) = {1}, Uargs(quot^#) = {1}, Uargs(c_4) = {1}

TcT has computed following constructor-restricted matrix
interpretation.

    [minus](x1, x2) = [1] x1 + [1]         
                                           
                [0] = [1]                  
                                           
            [s](x1) = [1] x1 + [2]         
                                           
  [minus^#](x1, x2) = [1] x1 + [1] x2 + [2]
                                           
              [c_1] = [2]                  
                                           
          [c_2](x1) = [1] x1 + [1]         
                                           
   [quot^#](x1, x2) = [1] x1 + [1] x2 + [1]
                                           
              [c_3] = [1]                  
                                           
          [c_4](x1) = [1] x1 + [2]         

This order satisfies following ordering constraints:

      [minus(x, 0())] = [1] x + [1]  
                      > [1] x + [0]  
                      = [x]          
                                     
  [minus(s(x), s(y))] = [1] x + [3]  
                      > [1] x + [1]  
                      = [minus(x, y)]
                                     

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(n^1)).

Strict DPs:
  { quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) }
Weak DPs:
  { minus^#(x, 0()) -> c_1()
  , minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
  , quot^#(0(), s(y)) -> c_3() }
Weak Trs:
  { minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, 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^#(x, 0()) -> c_1()
, minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
, quot^#(0(), s(y)) -> c_3() }

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

Strict DPs:
  { quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) }
Weak Trs:
  { minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The following argument positions are usable:
  Uargs(c_4) = {1}

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

   [minus](x1, x2) = [1] x1 + [0]         
                                          
               [0] = [3]                  
                                          
           [s](x1) = [1] x1 + [2]         
                                          
  [quot^#](x1, x2) = [1] x1 + [1] x2 + [0]
                                          
         [c_4](x1) = [1] x1 + [1]         

This order satisfies following ordering constraints:

       [minus(x, 0())] =  [1] x + [0]                     
                       >= [1] x + [0]                     
                       =  [x]                             
                                                          
   [minus(s(x), s(y))] =  [1] x + [2]                     
                       >  [1] x + [0]                     
                       =  [minus(x, y)]                   
                                                          
  [quot^#(s(x), s(y))] =  [1] x + [1] y + [4]             
                       >  [1] x + [1] y + [3]             
                       =  [c_4(quot^#(minus(x, y), s(y)))]
                                                          

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