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:
  runtime complexity
Answer:
  YES(O(1),O(n^1))

The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)

The following argument positions are usable:
  Uargs(minus) = {1}, Uargs(s) = {1}, Uargs(p) = {1},
  Uargs(div) = {1}

TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).

  [minus](x1, x2) = [1] x1 + [0]
                                
              [0] = [0]         
                                
          [s](x1) = [1] x1 + [1]
                                
          [p](x1) = [1] x1 + [0]
                                
    [div](x1, x2) = [1] x1 + [0]

The following symbols are considered usable

  {minus, p, div}

The order satisfies the following ordering constraints:

      [minus(X, 0())] =  [1] X + [0]                
                      >= [1] X + [0]                
                      =  [X]                        
                                                    
  [minus(s(X), s(Y))] =  [1] X + [1]                
                      >  [1] X + [0]                
                      =  [p(minus(X, Y))]           
                                                    
            [p(s(X))] =  [1] X + [1]                
                      >  [1] X + [0]                
                      =  [X]                        
                                                    
     [div(0(), s(Y))] =  [0]                        
                      >= [0]                        
                      =  [0()]                      
                                                    
    [div(s(X), s(Y))] =  [1] X + [1]                
                      >= [1] X + [1]                
                      =  [s(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 Trs:
  { minus(X, 0()) -> X
  , div(0(), s(Y)) -> 0()
  , div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y))) }
Weak Trs:
  { minus(s(X), s(Y)) -> p(minus(X, Y))
  , p(s(X)) -> X }
Obligation:
  runtime complexity
Answer:
  YES(O(1),O(n^1))

The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)

The following argument positions are usable:
  Uargs(minus) = {1}, Uargs(s) = {1}, Uargs(p) = {1},
  Uargs(div) = {1}

TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).

  [minus](x1, x2) = [1] x1 + [1]
                                
              [0] = [4]         
                                
          [s](x1) = [1] x1 + [0]
                                
          [p](x1) = [1] x1 + [0]
                                
    [div](x1, x2) = [1] x1 + [0]

The following symbols are considered usable

  {minus, p, div}

The order satisfies the following ordering constraints:

      [minus(X, 0())] =  [1] X + [1]                
                      >  [1] X + [0]                
                      =  [X]                        
                                                    
  [minus(s(X), s(Y))] =  [1] X + [1]                
                      >= [1] X + [1]                
                      =  [p(minus(X, Y))]           
                                                    
            [p(s(X))] =  [1] X + [0]                
                      >= [1] X + [0]                
                      =  [X]                        
                                                    
     [div(0(), s(Y))] =  [4]                        
                      >= [4]                        
                      =  [0()]                      
                                                    
    [div(s(X), s(Y))] =  [1] X + [0]                
                      ?  [1] X + [1]                
                      =  [s(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 Trs:
  { div(0(), s(Y)) -> 0()
  , div(s(X), s(Y)) -> s(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:
  runtime complexity
Answer:
  YES(O(1),O(n^1))

The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)

The following argument positions are usable:
  Uargs(minus) = {1}, Uargs(s) = {1}, Uargs(p) = {1},
  Uargs(div) = {1}

TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).

  [minus](x1, x2) = [1] x1 + [4]         
                                         
              [0] = [4]                  
                                         
          [s](x1) = [1] x1 + [4]         
                                         
          [p](x1) = [1] x1 + [4]         
                                         
    [div](x1, x2) = [1] x1 + [1] x2 + [0]

The following symbols are considered usable

  {minus, p, div}

The order satisfies the following ordering constraints:

      [minus(X, 0())] =  [1] X + [4]                
                      >  [1] X + [0]                
                      =  [X]                        
                                                    
  [minus(s(X), s(Y))] =  [1] X + [8]                
                      >= [1] X + [8]                
                      =  [p(minus(X, Y))]           
                                                    
            [p(s(X))] =  [1] X + [8]                
                      >  [1] X + [0]                
                      =  [X]                        
                                                    
     [div(0(), s(Y))] =  [1] Y + [8]                
                      >  [4]                        
                      =  [0()]                      
                                                    
    [div(s(X), s(Y))] =  [1] X + [1] Y + [8]        
                      ?  [1] X + [1] Y + [12]       
                      =  [s(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 Trs: { div(s(X), s(Y)) -> s(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
  , div(0(), s(Y)) -> 0() }
Obligation:
  runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.

Trs: { div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y))) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  The following argument positions are usable:
    Uargs(minus) = {1}, Uargs(s) = {1}, Uargs(p) = {1},
    Uargs(div) = {1}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [minus](x1, x2) = [1] x1 + [0]
                                  
                [0] = [0]         
                                  
            [s](x1) = [1] x1 + [4]
                                  
            [p](x1) = [1] x1 + [0]
                                  
      [div](x1, x2) = [2] x1 + [0]
  
  The following symbols are considered usable
  
    {minus, p, div}
  
  The order satisfies the following ordering constraints:
  
        [minus(X, 0())] =  [1] X + [0]                
                        >= [1] X + [0]                
                        =  [X]                        
                                                      
    [minus(s(X), s(Y))] =  [1] X + [4]                
                        >  [1] X + [0]                
                        =  [p(minus(X, Y))]           
                                                      
              [p(s(X))] =  [1] X + [4]                
                        >  [1] X + [0]                
                        =  [X]                        
                                                      
       [div(0(), s(Y))] =  [0]                        
                        >= [0]                        
                        =  [0()]                      
                                                      
      [div(s(X), s(Y))] =  [2] X + [8]                
                        >  [2] X + [4]                
                        =  [s(div(minus(X, Y), s(Y)))]
                                                      

We return to the main proof.

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
  , div(0(), s(Y)) -> 0()
  , div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y))) }
Obligation:
  runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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