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 the following innermost 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 the following constructor-restricted matrix
interpretation.

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

The following symbols are considered usable

  {minus, minus^#, quot^#}

The order satisfies the following ordering constraints:

        [minus(x, 0())] = [1 1] x + [2]                   
                          [1 1]     [0]                   
                        > [1 0] x + [0]                   
                          [0 1]     [0]                   
                        = [x]                             
                                                          
    [minus(s(x), s(y))] = [1 1] x + [4]                   
                          [1 1]     [2]                   
                        > [1 1] x + [2]                   
                          [1 1]     [0]                   
                        = [minus(x, y)]                   
                                                          
      [minus^#(x, 0())] = [0 0] x + [2]                   
                          [1 1]     [2]                   
                        > [1]                             
                          [1]                             
                        = [c_1()]                         
                                                          
  [minus^#(s(x), s(y))] = [0 0] x + [0 0] y + [2]         
                          [1 1]     [1 1]     [6]         
                        ? [0 0] x + [0 0] y + [3]         
                          [1 1]     [1 1]     [3]         
                        = [c_2(minus^#(x, y))]            
                                                          
    [quot^#(0(), s(y))] = [0]                             
                          [0]                             
                        ? [1]                             
                          [0]                             
                        = [c_3()]                         
                                                          
   [quot^#(s(x), s(y))] = [1 1] x + [0]                   
                          [0 0]     [0]                   
                        ? [1 1] x + [4]                   
                          [0 0]     [2]                   
                        = [c_4(quot^#(minus(x, y), s(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(1),O(n^1)).

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

We estimate the number of application of {2} by applications of
Pre({2}) = {3}. Here rules are labeled as follows:

  DPs:
    { 1: minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
    , 2: quot^#(0(), s(y)) -> c_3()
    , 3: quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y)))
    , 4: minus^#(x, 0()) -> c_1() }

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

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

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

Strict DPs:
  { minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
  , 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(1),O(n^1))

We use the processor 'Small Polynomial Path Order (PS,1-bounded)'
to orient following rules strictly.

DPs:
  { 1: minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
  , 2: quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) }
Trs: { minus(s(x), s(y)) -> minus(x, y) }

Sub-proof:
----------
  The input was oriented with the instance of 'Small Polynomial Path
  Order (PS,1-bounded)' as induced by the safe mapping
  
   safe(minus) = {}, safe(0) = {}, safe(s) = {1}, safe(quot) = {},
   safe(minus^#) = {2}, safe(c_1) = {}, safe(c_2) = {},
   safe(quot^#) = {2}, safe(c_3) = {}, safe(c_4) = {}
  
  and precedence
  
   minus > minus^#, minus ~ quot^# .
  
  Following symbols are considered recursive:
  
   {minus^#, quot^#}
  
  The recursion depth is 1.
  
  Further, following argument filtering is employed:
  
   pi(minus) = 1, pi(0) = [], pi(s) = [1], pi(quot) = [],
   pi(minus^#) = [1], pi(c_1) = [], pi(c_2) = [1], pi(quot^#) = [1],
   pi(c_3) = [], pi(c_4) = [1]
  
  Usable defined function symbols are a subset of:
  
   {minus, minus^#, quot^#}
  
  For your convenience, here are the satisfied ordering constraints:
  
    pi(minus^#(s(x), s(y))) =  minus^#(s(; x);)                  
                            >  c_2(minus^#(x;);)                 
                            =  pi(c_2(minus^#(x, y)))            
                                                                 
     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))                   
                                                                 

The strictly oriented rules are moved into the weak component.

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

Weak DPs:
  { minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
  , 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(1),O(1))

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

{ minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
, quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) }

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

Weak Trs:
  { minus(x, 0()) -> x
  , minus(s(x), s(y)) -> minus(x, y) }
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)).

Rules: Empty
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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