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

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

We add the following weak dependency pairs:

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

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^#(g(x), s(0()), y) -> c_1(f^#(g(s(0())), y, g(x)))
  , g^#(s(x)) -> c_2(g^#(x))
  , g^#(0()) -> c_3() }
Strict Trs:
  { f(g(x), s(0()), y) -> f(g(s(0())), y, g(x))
  , g(s(x)) -> s(g(x))
  , g(0()) -> 0() }
Obligation:
  runtime complexity
Answer:
  YES(O(1),O(n^1))

We replace rewrite rules by usable rules:

  Strict Usable Rules:
    { g(s(x)) -> s(g(x))
    , g(0()) -> 0() }

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

Strict DPs:
  { f^#(g(x), s(0()), y) -> c_1(f^#(g(s(0())), y, g(x)))
  , g^#(s(x)) -> c_2(g^#(x))
  , g^#(0()) -> c_3() }
Strict Trs:
  { g(s(x)) -> s(g(x))
  , g(0()) -> 0() }
Obligation:
  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(s) = {1}, Uargs(f^#) = {1, 2, 3}, Uargs(c_1) = {1},
  Uargs(c_2) = {1}

TcT has computed the following constructor-restricted matrix
interpretation.

            [g](x1) = [0 1] x1 + [0]                      
                      [0 1]      [0]                      
                                                          
            [s](x1) = [1 0] x1 + [0]                      
                      [0 1]      [1]                      
                                                          
                [0] = [0]                                 
                      [1]                                 
                                                          
  [f^#](x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [0]
                      [0 0]      [0 0]      [0 0]      [0]
                                                          
          [c_1](x1) = [1 0] x1 + [0]                      
                      [0 1]      [0]                      
                                                          
          [g^#](x1) = [0]                                 
                      [0]                                 
                                                          
          [c_2](x1) = [1 0] x1 + [0]                      
                      [0 1]      [0]                      
                                                          
              [c_3] = [0]                                 
                      [0]                                 

The order satisfies the following ordering constraints:

               [g(s(x))] =  [0 1] x + [1]                 
                            [0 1]     [1]                 
                         >  [0 1] x + [0]                 
                            [0 1]     [1]                 
                         =  [s(g(x))]                     
                                                          
                [g(0())] =  [1]                           
                            [1]                           
                         >  [0]                           
                            [1]                           
                         =  [0()]                         
                                                          
  [f^#(g(x), s(0()), y)] =  [0 1] x + [1 0] y + [0]       
                            [0 0]     [0 0]     [0]       
                         ?  [0 1] x + [1 0] y + [2]       
                            [0 0]     [0 0]     [0]       
                         =  [c_1(f^#(g(s(0())), y, g(x)))]
                                                          
             [g^#(s(x))] =  [0]                           
                            [0]                           
                         >= [0]                           
                            [0]                           
                         =  [c_2(g^#(x))]                 
                                                          
              [g^#(0())] =  [0]                           
                            [0]                           
                         >= [0]                           
                            [0]                           
                         =  [c_3()]                       
                                                          

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),O(n^1)).

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

Consider the dependency graph:

  1: f^#(g(x), s(0()), y) -> c_1(f^#(g(s(0())), y, g(x)))
     -->_1 f^#(g(x), s(0()), y) -> c_1(f^#(g(s(0())), y, g(x))) :1
  
  2: g^#(s(x)) -> c_2(g^#(x))
     -->_1 g^#(0()) -> c_3() :3
     -->_1 g^#(s(x)) -> c_2(g^#(x)) :2
  
  3: g^#(0()) -> c_3()
  

Only the nodes {2,3} are reachable from nodes {2,3} that start
derivation from marked basic terms. The nodes not reachable are
removed from the problem.

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

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

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

  DPs:
    { 1: g^#(s(x)) -> c_2(g^#(x))
    , 2: g^#(0()) -> c_3() }

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

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

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

{ g^#(0()) -> c_3() }

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

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

We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.

DPs:
  { 1: g^#(s(x)) -> c_2(g^#(x)) }

Sub-proof:
----------
  The following argument positions are usable:
    Uargs(c_2) = {1}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
      [s](x1) = [1] x1 + [2]
                            
    [g^#](x1) = [4] x1 + [0]
                            
    [c_2](x1) = [1] x1 + [1]
  
  The order satisfies the following ordering constraints:
  
    [g^#(s(x))] = [4] x + [8]  
                > [4] x + [1]  
                = [c_2(g^#(x))]
                               

The strictly oriented rules are moved into the weak component.

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

Weak DPs: { g^#(s(x)) -> c_2(g^#(x)) }
Obligation:
  runtime complexity
Answer:
  YES(O(1),O(1))

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

{ g^#(s(x)) -> c_2(g^#(x)) }

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

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

Empty rules are trivially bounded

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