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: { f(s(x), y, y) -> f(y, x, s(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: { f(s(x), y, y) -> f(y, x, s(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).
  
    [f](x1, x2, x3) = [2] x1 + [1] x2 + [1] x3 + [0]
                                                    
            [s](x1) = [1] x1 + [2]                  
  
  This order satisfies the following ordering constraints:
  
    [f(s(x), y, y)] = [2] x + [2] y + [4]
                    > [2] x + [2] y + [2]
                    = [f(y, x, s(x))]    
                                         

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

Empty rules are trivially bounded

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