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:
  { average(x, s(s(s(y)))) -> s(average(s(x), y))
  , average(s(x), y) -> average(x, s(y))
  , average(0(), s(s(0()))) -> s(0())
  , average(0(), s(0())) -> 0()
  , average(0(), 0()) -> 0() }
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:
  { average(0(), s(s(0()))) -> s(0())
  , average(0(), s(0())) -> 0()
  , average(0(), 0()) -> 0() }

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).
  
    [average](x1, x2) = [1] x1 + [3] x2 + [0]
                                             
              [s](x1) = [1] x1 + [0]         
                                             
                  [0] = [1]                  
  
  This order satisfies the following ordering constraints:
  
     [average(x, s(s(s(y))))] =  [1] x + [3] y + [0]  
                              >= [1] x + [3] y + [0]  
                              =  [s(average(s(x), y))]
                                                      
           [average(s(x), y)] =  [1] x + [3] y + [0]  
                              >= [1] x + [3] y + [0]  
                              =  [average(x, s(y))]   
                                                      
    [average(0(), s(s(0())))] =  [4]                  
                              >  [1]                  
                              =  [s(0())]             
                                                      
       [average(0(), s(0()))] =  [4]                  
                              >  [1]                  
                              =  [0()]                
                                                      
          [average(0(), 0())] =  [4]                  
                              >  [1]                  
                              =  [0()]                
                                                      

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:
  { average(x, s(s(s(y)))) -> s(average(s(x), y))
  , average(s(x), y) -> average(x, s(y)) }
Weak Trs:
  { average(0(), s(s(0()))) -> s(0())
  , average(0(), s(0())) -> 0()
  , average(0(), 0()) -> 0() }
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:
  { average(x, s(s(s(y)))) -> s(average(s(x), y))
  , average(s(x), y) -> average(x, 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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
    [average](x1, x2) = [3] x1 + [2] x2 + [1]
                                             
              [s](x1) = [1] x1 + [1]         
                                             
                  [0] = [0]                  
  
  This order satisfies the following ordering constraints:
  
     [average(x, s(s(s(y))))] = [3] x + [2] y + [7]  
                              > [3] x + [2] y + [5]  
                              = [s(average(s(x), y))]
                                                     
           [average(s(x), y)] = [3] x + [2] y + [4]  
                              > [3] x + [2] y + [3]  
                              = [average(x, s(y))]   
                                                     
    [average(0(), s(s(0())))] = [5]                  
                              > [1]                  
                              = [s(0())]             
                                                     
       [average(0(), s(0()))] = [3]                  
                              > [0]                  
                              = [0()]                
                                                     
          [average(0(), 0())] = [1]                  
                              > [0]                  
                              = [0()]                
                                                     

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

Empty rules are trivially bounded

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