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(), y) -> 0()
  , f(s(x), y) -> f(f(x, y), y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We add following weak dependency pairs:

Strict DPs:
  { f^#(0(), y) -> c_1()
  , f^#(s(x), y) -> c_2(f^#(f(x, y), y)) }

and mark the set of starting terms.

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

Strict DPs:
  { f^#(0(), y) -> c_1()
  , f^#(s(x), y) -> c_2(f^#(f(x, y), y)) }
Strict Trs:
  { f(0(), y) -> 0()
  , f(s(x), y) -> f(f(x, y), y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

The weightgap principle applies (using the following constant
growth matrix-interpretation)

The following argument positions are usable:
  Uargs(f) = {1}, Uargs(f^#) = {1}, Uargs(c_2) = {1}

TcT has computed following constructor-restricted matrix
interpretation.

    [f](x1, x2) = [1] x1 + [1]         
                                       
            [0] = [1]                  
                                       
        [s](x1) = [1] x1 + [2]         
                                       
  [f^#](x1, x2) = [1] x1 + [2] x2 + [1]
                                       
          [c_1] = [1]                  
                                       
      [c_2](x1) = [1] x1 + [2]         

This order satisfies following ordering constraints:

   [f(0(), y)] = [2]            
               > [1]            
               = [0()]          
                                
  [f(s(x), y)] = [1] x + [3]    
               > [1] x + [2]    
               = [f(f(x, y), y)]
                                

Further, it can be verified that all rules not oriented are covered by the weightgap condition.

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

Strict DPs: { f^#(s(x), y) -> c_2(f^#(f(x, y), y)) }
Weak DPs: { f^#(0(), y) -> c_1() }
Weak Trs:
  { f(0(), y) -> 0()
  , f(s(x), y) -> f(f(x, y), y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(1))

We estimate the number of application of {1} by applications of
Pre({1}) = {}. Here rules are labeled as follows:

  DPs:
    { 1: f^#(s(x), y) -> c_2(f^#(f(x, y), y))
    , 2: f^#(0(), y) -> c_1() }

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

Weak DPs:
  { f^#(0(), y) -> c_1()
  , f^#(s(x), y) -> c_2(f^#(f(x, y), y)) }
Weak Trs:
  { f(0(), y) -> 0()
  , f(s(x), y) -> f(f(x, y), y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(1))

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ f^#(0(), y) -> c_1()
, f^#(s(x), y) -> c_2(f^#(f(x, y), y)) }

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

Weak Trs:
  { f(0(), y) -> 0()
  , f(s(x), y) -> f(f(x, y), y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(1))

No rule is usable, rules are removed from the input problem.

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

Rules: Empty
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(1))

The input was oriented with the instance of 'Small Polynomial Path
Order (PS)' as induced by the safe mapping


and precedence

 empty .

Following symbols are considered recursive:

 {}

The recursion depth is 0.

Further, following argument filtering is employed:

 empty

Usable defined function symbols are a subset of:

 {}

For your convenience, here are the satisfied ordering constraints:


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