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(m, n, s(r)) -> p(m, r, n)
  , p(m, s(n), 0()) -> p(0(), n, m)
  , p(m, 0(), 0()) -> m }
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(m, 0(), 0()) -> m }

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, x2, x3) = [3] x1 + [2] x2 + [2] x3 + [1]
                                                    
            [s](x1) = [1] x1 + [0]                  
                                                    
                [0] = [0]                           
  
  This order satisfies the following ordering constraints:
  
      [p(m, n, s(r))] =  [3] m + [2] n + [2] r + [1]
                      >= [3] m + [2] n + [2] r + [1]
                      =  [p(m, r, n)]               
                                                    
    [p(m, s(n), 0())] =  [3] m + [2] n + [1]        
                      >= [2] m + [2] n + [1]        
                      =  [p(0(), n, m)]             
                                                    
     [p(m, 0(), 0())] =  [3] m + [1]                
                      >  [1] m + [0]                
                      =  [m]                        
                                                    

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(m, n, s(r)) -> p(m, r, n)
  , p(m, s(n), 0()) -> p(0(), n, m) }
Weak Trs: { p(m, 0(), 0()) -> m }
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(m, n, s(r)) -> p(m, r, n)
  , p(m, s(n), 0()) -> p(0(), n, m) }

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, x2, x3) = [2] x1 + [2] x2 + [2] x3 + [0]
                                                    
            [s](x1) = [1] x1 + [2]                  
                                                    
                [0] = [0]                           
  
  This order satisfies the following ordering constraints:
  
      [p(m, n, s(r))] =  [2] m + [2] n + [2] r + [4]
                      >  [2] m + [2] n + [2] r + [0]
                      =  [p(m, r, n)]               
                                                    
    [p(m, s(n), 0())] =  [2] m + [2] n + [4]        
                      >  [2] m + [2] n + [0]        
                      =  [p(0(), n, m)]             
                                                    
     [p(m, 0(), 0())] =  [2] m + [0]                
                      >= [1] m + [0]                
                      =  [m]                        
                                                    

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(m, n, s(r)) -> p(m, r, n)
  , p(m, s(n), 0()) -> p(0(), n, m)
  , p(m, 0(), 0()) -> m }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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