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

We add following weak dependency pairs:

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

TcT has computed following constructor-restricted matrix
interpretation.

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

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: { f^#(c(X, s(Y))) -> c_1(f^#(c(s(X), Y))) }
Weak DPs: { g^#(c(s(X), Y)) -> c_2(f^#(c(X, s(Y)))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

Consider the dependency graph

  1: f^#(c(X, s(Y))) -> c_1(f^#(c(s(X), Y)))
     -->_1 f^#(c(X, s(Y))) -> c_1(f^#(c(s(X), Y))) :1
  
  2: g^#(c(s(X), Y)) -> c_2(f^#(c(X, s(Y))))
     -->_1 f^#(c(X, s(Y))) -> c_1(f^#(c(s(X), Y))) :1
  

Following roots of the dependency graph are removed, as the
considered set of starting terms is closed under reduction with
respect to these rules (modulo compound contexts).

  { g^#(c(s(X), Y)) -> c_2(f^#(c(X, s(Y)))) }


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

Strict DPs: { f^#(c(X, s(Y))) -> c_1(f^#(c(s(X), Y))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

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

 safe(c) = {1, 2}, safe(s) = {1}, safe(f^#) = {}, safe(c_1) = {}

and precedence

 empty .

Following symbols are considered recursive:

 {f^#}

The recursion depth is 1.

Further, following argument filtering is employed:

 pi(c) = 2, pi(s) = [1], pi(f^#) = [1], pi(c_1) = [1]

Usable defined function symbols are a subset of:

 {f^#}

For your convenience, here are the satisfied ordering constraints:

  pi(f^#(c(X, s(Y)))) = f^#(s(; Y);)            
                      > c_1(f^#(Y;);)           
                      = pi(c_1(f^#(c(s(X), Y))))
                                                

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