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

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

Trs: { minus(x, 0()) -> x }

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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
            [p](x1) = [1] x1 + [0]         
                                           
                [0] = [0]                  
                                           
            [s](x1) = [1] x1 + [0]         
                                           
    [minus](x1, x2) = [1] x1 + [1] x2 + [1]
  
  This order satisfies the following ordering constraints:
  
            [p(0())] =  [0]                
                     >= [0]                
                     =  [0()]              
                                           
           [p(s(x))] =  [1] x + [0]        
                     >= [1] x + [0]        
                     =  [x]                
                                           
     [minus(x, 0())] =  [1] x + [1]        
                     >  [1] x + [0]        
                     =  [x]                
                                           
    [minus(x, s(y))] =  [1] x + [1] y + [1]
                     >= [1] x + [1] y + [1]
                     =  [minus(p(x), y)]   
                                           

We return to the main proof.

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

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

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

Trs:
  { p(0()) -> 0()
  , p(s(x)) -> x }

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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
            [p](x1) = [1] x1 + [3]         
                                           
                [0] = [3]                  
                                           
            [s](x1) = [1] x1 + [3]         
                                           
    [minus](x1, x2) = [1] x1 + [1] x2 + [3]
  
  This order satisfies the following ordering constraints:
  
            [p(0())] =  [6]                
                     >  [3]                
                     =  [0()]              
                                           
           [p(s(x))] =  [1] x + [6]        
                     >  [1] x + [0]        
                     =  [x]                
                                           
     [minus(x, 0())] =  [1] x + [6]        
                     >  [1] x + [0]        
                     =  [x]                
                                           
    [minus(x, s(y))] =  [1] x + [1] y + [6]
                     >= [1] x + [1] y + [6]
                     =  [minus(p(x), y)]   
                                           

We return to the main proof.

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

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

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

Trs: { minus(x, s(y)) -> minus(p(x), 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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
            [p](x1) = [1] x1 + [1]         
                                           
                [0] = [1]                  
                                           
            [s](x1) = [1] x1 + [1]         
                                           
    [minus](x1, x2) = [2] x1 + [3] x2 + [3]
  
  This order satisfies the following ordering constraints:
  
            [p(0())] = [2]                
                     > [1]                
                     = [0()]              
                                          
           [p(s(x))] = [1] x + [2]        
                     > [1] x + [0]        
                     = [x]                
                                          
     [minus(x, 0())] = [2] x + [6]        
                     > [1] x + [0]        
                     = [x]                
                                          
    [minus(x, s(y))] = [2] x + [3] y + [6]
                     > [2] x + [3] y + [5]
                     = [minus(p(x), 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:
  { p(0()) -> 0()
  , p(s(x)) -> x
  , minus(x, 0()) -> x
  , minus(x, s(y)) -> minus(p(x), y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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