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)) -> p(minus(X, Y))
  , p(s(X)) -> X
  , div(0(), s(Y)) -> 0()
  , div(s(X), s(Y)) -> s(div(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(p^#(minus(X, Y)))
  , p^#(s(X)) -> c_3()
  , div^#(0(), s(Y)) -> c_4()
  , div^#(s(X), s(Y)) -> c_5(div^#(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(p^#(minus(X, Y)))
  , p^#(s(X)) -> c_3()
  , div^#(0(), s(Y)) -> c_4()
  , div^#(s(X), s(Y)) -> c_5(div^#(minus(X, Y), s(Y))) }
Strict Trs:
  { minus(X, 0()) -> X
  , minus(s(X), s(Y)) -> p(minus(X, Y))
  , p(s(X)) -> X
  , div(0(), s(Y)) -> 0()
  , div(s(X), s(Y)) -> s(div(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)) -> p(minus(X, Y))
    , p(s(X)) -> X }

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(p^#(minus(X, Y)))
  , p^#(s(X)) -> c_3()
  , div^#(0(), s(Y)) -> c_4()
  , div^#(s(X), s(Y)) -> c_5(div^#(minus(X, Y), s(Y))) }
Strict Trs:
  { minus(X, 0()) -> X
  , minus(s(X), s(Y)) -> p(minus(X, Y))
  , p(s(X)) -> X }
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(p) = {1}, Uargs(c_2) = {1}, Uargs(p^#) = {1},
  Uargs(div^#) = {1}, Uargs(c_5) = {1}

TcT has computed the following constructor-restricted matrix
interpretation.

    [minus](x1, x2) = [1 0] x1 + [2]           
                      [2 2]      [0]           
                                               
                [0] = [0]                      
                      [0]                      
                                               
            [s](x1) = [1 2] x1 + [2]           
                      [0 0]      [0]           
                                               
            [p](x1) = [1 0] x1 + [0]           
                      [2 0]      [0]           
                                               
  [minus^#](x1, x2) = [2 0] x1 + [0 0] x2 + [2]
                      [1 2]      [1 1]      [2]
                                               
              [c_1] = [1]                      
                      [1]                      
                                               
          [c_2](x1) = [1 0] x1 + [1]           
                      [0 1]      [2]           
                                               
          [p^#](x1) = [2 0] x1 + [2]           
                      [0 0]      [2]           
                                               
              [c_3] = [1]                      
                      [1]                      
                                               
    [div^#](x1, x2) = [1 0] x1 + [0]           
                      [0 0]      [0]           
                                               
              [c_4] = [1]                      
                      [1]                      
                                               
          [c_5](x1) = [1 0] x1 + [2]           
                      [0 1]      [2]           

The following symbols are considered usable

  {minus, p, minus^#, p^#, div^#}

The order satisfies the following ordering constraints:

        [minus(X, 0())] = [1 0] X + [2]                  
                          [2 2]     [0]                  
                        > [1 0] X + [0]                  
                          [0 1]     [0]                  
                        = [X]                            
                                                         
    [minus(s(X), s(Y))] = [1 2] X + [4]                  
                          [2 4]     [4]                  
                        > [1 0] X + [2]                  
                          [2 0]     [4]                  
                        = [p(minus(X, Y))]               
                                                         
              [p(s(X))] = [1 2] X + [2]                  
                          [2 4]     [4]                  
                        > [1 0] X + [0]                  
                          [0 1]     [0]                  
                        = [X]                            
                                                         
      [minus^#(X, 0())] = [2 0] X + [2]                  
                          [1 2]     [2]                  
                        > [1]                            
                          [1]                            
                        = [c_1()]                        
                                                         
  [minus^#(s(X), s(Y))] = [2 4] X + [0 0] Y + [6]        
                          [1 2]     [1 2]     [6]        
                        ? [2 0] X + [7]                  
                          [0 0]     [4]                  
                        = [c_2(p^#(minus(X, Y)))]        
                                                         
            [p^#(s(X))] = [2 4] X + [6]                  
                          [0 0]     [2]                  
                        > [1]                            
                          [1]                            
                        = [c_3()]                        
                                                         
     [div^#(0(), s(Y))] = [0]                            
                          [0]                            
                        ? [1]                            
                          [1]                            
                        = [c_4()]                        
                                                         
    [div^#(s(X), s(Y))] = [1 2] X + [2]                  
                          [0 0]     [0]                  
                        ? [1 0] X + [4]                  
                          [0 0]     [2]                  
                        = [c_5(div^#(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(p^#(minus(X, Y)))
  , div^#(0(), s(Y)) -> c_4()
  , div^#(s(X), s(Y)) -> c_5(div^#(minus(X, Y), s(Y))) }
Weak DPs:
  { minus^#(X, 0()) -> c_1()
  , p^#(s(X)) -> c_3() }
Weak Trs:
  { minus(X, 0()) -> X
  , minus(s(X), s(Y)) -> p(minus(X, Y))
  , p(s(X)) -> X }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

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

  DPs:
    { 1: minus^#(s(X), s(Y)) -> c_2(p^#(minus(X, Y)))
    , 2: div^#(0(), s(Y)) -> c_4()
    , 3: div^#(s(X), s(Y)) -> c_5(div^#(minus(X, Y), s(Y)))
    , 4: minus^#(X, 0()) -> c_1()
    , 5: p^#(s(X)) -> c_3() }

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

Strict DPs: { div^#(s(X), s(Y)) -> c_5(div^#(minus(X, Y), s(Y))) }
Weak DPs:
  { minus^#(X, 0()) -> c_1()
  , minus^#(s(X), s(Y)) -> c_2(p^#(minus(X, Y)))
  , p^#(s(X)) -> c_3()
  , div^#(0(), s(Y)) -> c_4() }
Weak Trs:
  { minus(X, 0()) -> X
  , minus(s(X), s(Y)) -> p(minus(X, Y))
  , p(s(X)) -> X }
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()
, minus^#(s(X), s(Y)) -> c_2(p^#(minus(X, Y)))
, p^#(s(X)) -> c_3()
, div^#(0(), s(Y)) -> c_4() }

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

Strict DPs: { div^#(s(X), s(Y)) -> c_5(div^#(minus(X, Y), s(Y))) }
Weak Trs:
  { minus(X, 0()) -> X
  , minus(s(X), s(Y)) -> p(minus(X, Y))
  , p(s(X)) -> X }
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: div^#(s(X), s(Y)) -> c_5(div^#(minus(X, Y), s(Y))) }
Trs:
  { minus(s(X), s(Y)) -> p(minus(X, Y))
  , p(s(X)) -> X }

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(p) = {},
   safe(div) = {}, safe(minus^#) = {}, safe(c_1) = {}, safe(c_2) = {},
   safe(p^#) = {}, safe(c_3) = {}, safe(div^#) = {}, safe(c_4) = {},
   safe(c_5) = {}
  
  and precedence
  
   minus > p, minus ~ div^# .
  
  Following symbols are considered recursive:
  
   {p, div^#}
  
  The recursion depth is 1.
  
  Further, following argument filtering is employed:
  
   pi(minus) = 1, pi(0) = [], pi(s) = [1], pi(p) = 1, pi(div) = [],
   pi(minus^#) = [], pi(c_1) = [], pi(c_2) = [], pi(p^#) = [],
   pi(c_3) = [], pi(div^#) = [1, 2], pi(c_4) = [], pi(c_5) = [1]
  
  Usable defined function symbols are a subset of:
  
   {minus, p, minus^#, p^#, div^#}
  
  For your convenience, here are the satisfied ordering constraints:
  
    pi(div^#(s(X), s(Y))) =  div^#(s(; X),  s(; Y);)          
                          >  c_5(div^#(X,  s(; Y););)         
                          =  pi(c_5(div^#(minus(X, Y), s(Y))))
                                                              
        pi(minus(X, 0())) =  X                                
                          >= X                                
                          =  pi(X)                            
                                                              
    pi(minus(s(X), s(Y))) =  s(; X)                           
                          >  X                                
                          =  pi(p(minus(X, Y)))               
                                                              
              pi(p(s(X))) =  s(; X)                           
                          >  X                                
                          =  pi(X)                            
                                                              

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: { div^#(s(X), s(Y)) -> c_5(div^#(minus(X, Y), s(Y))) }
Weak Trs:
  { minus(X, 0()) -> X
  , minus(s(X), s(Y)) -> p(minus(X, Y))
  , p(s(X)) -> X }
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.

{ div^#(s(X), s(Y)) -> c_5(div^#(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)) -> p(minus(X, Y))
  , p(s(X)) -> X }
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))