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:
  { add(0(), x) -> x
  , add(s(x), y) -> s(add(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: { add(0(), 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).
  
    [add](x1, x2) = [2] x1 + [3] x2 + [0]
                                         
              [0] = [2]                  
                                         
          [s](x1) = [1] x1 + [0]         
  
  This order satisfies the following ordering constraints:
  
     [add(0(), x)] =  [3] x + [4]        
                   >  [1] x + [0]        
                   =  [x]                
                                         
    [add(s(x), y)] =  [2] x + [3] y + [0]
                   >= [2] x + [3] y + [0]
                   =  [s(add(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: { add(s(x), y) -> s(add(x, y)) }
Weak Trs: { add(0(), x) -> 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: { add(s(x), y) -> s(add(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).
  
    [add](x1, x2) = [2] x1 + [3] x2 + [0]
                                         
              [0] = [3]                  
                                         
          [s](x1) = [1] x1 + [2]         
  
  This order satisfies the following ordering constraints:
  
     [add(0(), x)] = [3] x + [6]        
                   > [1] x + [0]        
                   = [x]                
                                        
    [add(s(x), y)] = [2] x + [3] y + [4]
                   > [2] x + [3] y + [2]
                   = [s(add(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:
  { add(0(), x) -> x
  , add(s(x), y) -> s(add(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))