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

We add the following innermost weak dependency pairs:

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

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

We replace rewrite rules by usable rules:

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

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

Strict DPs:
  { f^#(s(0()), g(x)) -> c_1(f^#(x, g(x)))
  , g^#(s(x)) -> c_2(g^#(x)) }
Strict Trs: { g(s(x)) -> g(x) }
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_1) = {1}, Uargs(c_2) = {1}

TcT has computed the following constructor-restricted matrix
interpretation.

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

The following symbols are considered usable

  {g, f^#, g^#}

The order satisfies the following ordering constraints:

            [g(s(x))] = [0 1] x + [1]      
                        [0 0]     [0]      
                      > [0 1] x + [0]      
                        [0 0]     [0]      
                      = [g(x)]             
                                           
  [f^#(s(0()), g(x))] = [2]                
                        [2]                
                      ? [4]                
                        [4]                
                      = [c_1(f^#(x, g(x)))]
                                           
          [g^#(s(x))] = [0 2] x + [4]      
                        [0 2]     [4]      
                      ? [0 2] x + [3]      
                        [1 2]     [3]      
                      = [c_2(g^#(x))]      
                                           

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

Consider the dependency graph:

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

Only the nodes {2} are reachable from nodes {2} 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)) }
Weak Trs: { g(s(x)) -> g(x) }
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: { g^#(s(x)) -> c_2(g^#(x)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'Small Polynomial Path Order (PS,1-bounded)'
to orient following rules strictly.

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

Sub-proof:
----------
  The input was oriented with the instance of 'Small Polynomial Path
  Order (PS,1-bounded)' as induced by the safe mapping
  
   safe(f) = {}, safe(s) = {1}, safe(0) = {}, safe(g) = {},
   safe(f^#) = {}, safe(c_1) = {}, safe(g^#) = {}, safe(c_2) = {}
  
  and precedence
  
   empty .
  
  Following symbols are considered recursive:
  
   {g^#}
  
  The recursion depth is 1.
  
  Further, following argument filtering is employed:
  
   pi(f) = [], pi(s) = [1], pi(0) = [], pi(g) = [], pi(f^#) = [],
   pi(c_1) = [], pi(g^#) = [1], pi(c_2) = [1]
  
  Usable defined function symbols are a subset of:
  
   {f^#, g^#}
  
  For your convenience, here are the satisfied ordering constraints:
  
    pi(g^#(s(x))) = g^#(s(; x);)   
                  > c_2(g^#(x;);)  
                  = pi(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:
  innermost 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:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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