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:
  { h(f(x), y) -> f(g(x, y))
  , g(x, y) -> h(x, y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'Small Polynomial Path Order (PS,1-bounded)'
to orient following rules strictly.

Trs: { h(f(x), y) -> f(g(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:
----------
  The input was oriented with the instance of 'Small Polynomial Path
  Order (PS,1-bounded)' as induced by the safe mapping
  
   safe(h) = {2}, safe(f) = {1}, safe(g) = {2}
  
  and precedence
  
   h ~ g .
  
  Following symbols are considered recursive:
  
   {h, g}
  
  The recursion depth is 1.
  
  For your convenience, here are the satisfied ordering constraints:
  
    h(f(; x); y) >  f(; g(x; y))
                                
         g(x; y) >= h(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: { g(x, y) -> h(x, y) }
Weak Trs: { h(f(x), y) -> f(g(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: { g(x, y) -> h(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).
  
    [h](x1, x2) = [2] x1 + [3] x2 + [0]
                                       
        [f](x1) = [1] x1 + [2]         
                                       
    [g](x1, x2) = [2] x1 + [3] x2 + [1]
  
  This order satisfies the following ordering constraints:
  
    [h(f(x), y)] = [2] x + [3] y + [4]
                 > [2] x + [3] y + [3]
                 = [f(g(x, y))]       
                                      
       [g(x, y)] = [2] x + [3] y + [1]
                 > [2] x + [3] y + [0]
                 = [h(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:
  { h(f(x), y) -> f(g(x, y))
  , g(x, y) -> h(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))