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 input was oriented with the instance of 'Small Polynomial Path
Order (PS)' as induced by the safe mapping

 safe(minus) = {}, safe(0) = {}, safe(s) = {1}, safe(quot^#) = {2},
 safe(c_4) = {}

and precedence

 minus ~ quot^# .

Following symbols are considered recursive:

 {quot^#}

The recursion depth is 1.

Further, following argument filtering is employed:

 pi(minus) = 1, pi(0) = [], pi(s) = [1], pi(quot^#) = [1],
 pi(c_4) = [1]

Usable defined function symbols are a subset of:

 {minus, quot^#}

For your convenience, here are the satisfied ordering constraints:

  pi(quot^#(s(x), s(y))) =  quot^#(s(; x);)                   
                         >  c_4(quot^#(x;);)                  
                         =  pi(c_4(quot^#(minus(x, y), s(y))))
                                                              
       pi(minus(x, 0())) =  x                                 
                         >= x                                 
                         =  pi(x)                             
                                                              
   pi(minus(s(x), s(y))) =  s(; x)                            
                         >  x                                 
                         =  pi(minus(x, y))                   
                                                              

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