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

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:
  { f(0()) -> s(0())
  , f(s(0())) -> s(s(0()))
  , f(s(0())) -> *(s(s(0())), f(0())) }
Weak Trs:
  { f(+(x, y)) -> *(f(x), f(y))
  , f(+(x, s(0()))) -> +(s(s(0())), f(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(0()) -> s(0())
  , f(s(0())) -> s(s(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).
  
        [f](x1) = [1] x1 + [1]         
                                       
            [0] = [0]                  
                                       
        [s](x1) = [1] x1 + [0]         
                                       
    [*](x1, x2) = [1] x1 + [1] x2 + [0]
                                       
    [+](x1, x2) = [1] x1 + [1] x2 + [3]
  
  This order satisfies the following ordering constraints:
  
             [f(0())] =  [1]                   
                      >  [0]                   
                      =  [s(0())]              
                                               
          [f(s(0()))] =  [1]                   
                      >  [0]                   
                      =  [s(s(0()))]           
                                               
          [f(s(0()))] =  [1]                   
                      >= [1]                   
                      =  [*(s(s(0())), f(0()))]
                                               
         [f(+(x, y))] =  [1] x + [1] y + [4]   
                      >  [1] x + [1] y + [2]   
                      =  [*(f(x), f(y))]       
                                               
    [f(+(x, s(0())))] =  [1] x + [4]           
                      >= [1] x + [4]           
                      =  [+(s(s(0())), f(x))]  
                                               

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: { f(s(0())) -> *(s(s(0())), f(0())) }
Weak Trs:
  { f(0()) -> s(0())
  , f(s(0())) -> s(s(0()))
  , f(+(x, y)) -> *(f(x), f(y))
  , f(+(x, s(0()))) -> +(s(s(0())), f(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(0())) -> *(s(s(0())), f(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).
  
        [f](x1) = [3] x1 + [1]         
                                       
            [0] = [0]                  
                                       
        [s](x1) = [1] x1 + [1]         
                                       
    [*](x1, x2) = [1] x1 + [1] x2 + [0]
                                       
    [+](x1, x2) = [1] x1 + [1] x2 + [1]
  
  This order satisfies the following ordering constraints:
  
             [f(0())] =  [1]                   
                      >= [1]                   
                      =  [s(0())]              
                                               
          [f(s(0()))] =  [4]                   
                      >  [2]                   
                      =  [s(s(0()))]           
                                               
          [f(s(0()))] =  [4]                   
                      >  [3]                   
                      =  [*(s(s(0())), f(0()))]
                                               
         [f(+(x, y))] =  [3] x + [3] y + [4]   
                      >  [3] x + [3] y + [2]   
                      =  [*(f(x), f(y))]       
                                               
    [f(+(x, s(0())))] =  [3] x + [7]           
                      >  [3] x + [4]           
                      =  [+(s(s(0())), f(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(0()) -> s(0())
  , f(s(0())) -> s(s(0()))
  , f(s(0())) -> *(s(s(0())), f(0()))
  , f(+(x, y)) -> *(f(x), f(y))
  , f(+(x, s(0()))) -> +(s(s(0())), f(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))