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

We add following weak dependency pairs:

Strict DPs:
  { +^#(0(), y) -> c_1()
  , +^#(s(x), y) -> c_2(+^#(x, y))
  , -^#(x, 0()) -> c_3()
  , -^#(0(), y) -> c_4()
  , -^#(s(x), s(y)) -> c_5(-^#(x, 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:
  { +^#(0(), y) -> c_1()
  , +^#(s(x), y) -> c_2(+^#(x, y))
  , -^#(x, 0()) -> c_3()
  , -^#(0(), y) -> c_4()
  , -^#(s(x), s(y)) -> c_5(-^#(x, y)) }
Strict Trs:
  { +(0(), y) -> y
  , +(s(x), y) -> s(+(x, y))
  , -(x, 0()) -> x
  , -(0(), y) -> 0()
  , -(s(x), s(y)) -> -(x, y) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^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),O(n^1)).

Strict DPs:
  { +^#(0(), y) -> c_1()
  , +^#(s(x), y) -> c_2(+^#(x, y))
  , -^#(x, 0()) -> c_3()
  , -^#(0(), y) -> c_4()
  , -^#(s(x), s(y)) -> c_5(-^#(x, 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(c_2) = {1}, Uargs(c_5) = {1}

TcT has computed following constructor-restricted matrix
interpretation.

            [0] = [0]         
                              
        [s](x1) = [1] x1 + [0]
                              
  [+^#](x1, x2) = [2] x2 + [0]
                              
          [c_1] = [1]         
                              
      [c_2](x1) = [1] x1 + [1]
                              
  [-^#](x1, x2) = [1]         
                              
          [c_3] = [0]         
                              
          [c_4] = [0]         
                              
      [c_5](x1) = [1] x1 + [1]

This order satisfies following ordering constraints:


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(n^1)).

Strict DPs:
  { +^#(0(), y) -> c_1()
  , +^#(s(x), y) -> c_2(+^#(x, y))
  , -^#(s(x), s(y)) -> c_5(-^#(x, y)) }
Weak DPs:
  { -^#(x, 0()) -> c_3()
  , -^#(0(), y) -> c_4() }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

We estimate the number of application of {1} by applications of
Pre({1}) = {2}. Here rules are labeled as follows:

  DPs:
    { 1: +^#(0(), y) -> c_1()
    , 2: +^#(s(x), y) -> c_2(+^#(x, y))
    , 3: -^#(s(x), s(y)) -> c_5(-^#(x, y))
    , 4: -^#(x, 0()) -> c_3()
    , 5: -^#(0(), y) -> c_4() }

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

Strict DPs:
  { +^#(s(x), y) -> c_2(+^#(x, y))
  , -^#(s(x), s(y)) -> c_5(-^#(x, y)) }
Weak DPs:
  { +^#(0(), y) -> c_1()
  , -^#(x, 0()) -> c_3()
  , -^#(0(), y) -> c_4() }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

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

{ +^#(0(), y) -> c_1()
, -^#(x, 0()) -> c_3()
, -^#(0(), y) -> c_4() }

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

Strict DPs:
  { +^#(s(x), y) -> c_2(+^#(x, y))
  , -^#(s(x), s(y)) -> c_5(-^#(x, y)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The following argument positions are usable:
  Uargs(c_2) = {1}, Uargs(c_5) = {1}

TcT has computed following constructor-based matrix interpretation
satisfying not(EDA).

        [s](x1) = [1] x1 + [2]         
                                       
  [+^#](x1, x2) = [2] x1 + [3] x2 + [0]
                                       
      [c_2](x1) = [1] x1 + [1]         
                                       
  [-^#](x1, x2) = [2] x1 + [0]         
                                       
      [c_5](x1) = [1] x1 + [1]         

This order satisfies following ordering constraints:

     [+^#(s(x), y)] = [3] y + [2] x + [4]
                    > [3] y + [2] x + [1]
                    = [c_2(+^#(x, y))]   
                                         
  [-^#(s(x), s(y))] = [2] x + [4]        
                    > [2] x + [1]        
                    = [c_5(-^#(x, y))]   
                                         

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