YES(?,O(n^1))

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

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

We add following dependency tuples:

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

and mark the set of starting terms.

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

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

We estimate the number of application of {1,2} by applications of
Pre({1,2}) = {3}. Here rules are labeled as follows:

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

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

Strict DPs:
  { +^#(s(x), s(y)) -> c_3(+^#(s(x), +(y, 0())), +^#(y, 0())) }
Weak DPs:
  { +^#(0(), y) -> c_1()
  , +^#(s(x), 0()) -> c_2() }
Weak Trs:
  { +(0(), y) -> y
  , +(s(x), 0()) -> s(x)
  , +(s(x), s(y)) -> s(+(s(x), +(y, 0()))) }
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()
, +^#(s(x), 0()) -> c_2() }

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

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

Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:

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

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

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

The following argument positions are usable:
  Uargs(c_1) = {1}

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

    [+](x1, x2) = [1] x1 + [2] x2 + [0]
                                       
            [0] = [0]                  
                                       
        [s](x1) = [1] x1 + [2]         
                                       
  [+^#](x1, x2) = [2] x2 + [0]         
                                       
      [c_1](x1) = [1] x1 + [3]         

This order satisfies following ordering constraints:

        [+(0(), y)] =  [2] y + [0]                
                    >= [1] y + [0]                
                    =  [y]                        
                                                  
     [+(s(x), 0())] =  [1] x + [2]                
                    >= [1] x + [2]                
                    =  [s(x)]                     
                                                  
    [+(s(x), s(y))] =  [2] y + [1] x + [6]        
                    >  [2] y + [1] x + [4]        
                    =  [s(+(s(x), +(y, 0())))]    
                                                  
  [+^#(s(x), s(y))] =  [2] y + [4]                
                    >  [2] y + [3]                
                    =  [c_1(+^#(s(x), +(y, 0())))]
                                                  

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