YES(O(1),O(n^2))

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict Trs: { *(x, +(y, z)) -> +(*(x, y), *(x, z)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We use the processor 'custom shape polynomial interpretation' to
orient following rules strictly.

Trs: { *(x, +(y, z)) -> +(*(x, y), *(x, z)) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  TcT has computed the following constructor-restricted polynomial
  interpretation.
  [*](x1, x2) = 3*x1 + 3*x1*x2 + 3*x2 + x2^2
                                            
  [+](x1, x2) = 1 + x1 + x2                 
                                            
  
  This order satisfies the following ordering constraints.
  
    [*(x, +(y, z))] = 6*x + 3*x*y + 3*x*z + 4 + 5*y + 5*z + y^2 + y*z + z*y + z^2
                    > 1 + 6*x + 3*x*y + 3*y + y^2 + 3*x*z + 3*z + z^2            
                    = [+(*(x, y), *(x, z))]                                      
                                                                                 

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: { *(x, +(y, z)) -> +(*(x, y), *(x, z)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

Hurray, we answered YES(O(1),O(n^2))