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))
  , *(x, 1()) -> x
  , *(+(x, y), z) -> +(*(x, z), *(y, z))
  , *(1(), y) -> y }
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))
  , *(x, 1()) -> x
  , *(+(x, y), z) -> +(*(x, z), *(y, z))
  , *(1(), y) -> y }

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) = 2*x1 + x1*x2 + 2*x2
                                   
  [+](x1, x2) = 2 + x1 + x2        
                                   
        [1]() = 2                  
                                   
  
  This order satisfies the following ordering constraints.
  
    [*(x, +(y, z))] = 4*x + x*y + x*z + 4 + 2*y + 2*z
                    > 2 + 4*x + x*y + 2*y + x*z + 2*z
                    = [+(*(x, y), *(x, z))]          
                                                     
        [*(x, 1())] = 4*x + 4                        
                    > x                              
                    = [x]                            
                                                     
    [*(+(x, y), z)] = 4 + 2*x + 2*y + 4*z + x*z + y*z
                    > 2 + 2*x + x*z + 4*z + 2*y + y*z
                    = [+(*(x, z), *(y, z))]          
                                                     
        [*(1(), y)] = 4 + 4*y                        
                    > y                              
                    = [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:
  { *(x, +(y, z)) -> +(*(x, y), *(x, z))
  , *(x, 1()) -> x
  , *(+(x, y), z) -> +(*(x, z), *(y, z))
  , *(1(), y) -> y }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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