YES

We show the termination of the TRS R:

  f(c(X,s(Y))) -> f(c(s(X),Y))
  g(c(s(X),Y)) -> f(c(X,s(Y)))

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(c(X,s(Y))) -> f#(c(s(X),Y))
p2: g#(c(s(X),Y)) -> f#(c(X,s(Y)))

and R consists of:

r1: f(c(X,s(Y))) -> f(c(s(X),Y))
r2: g(c(s(X),Y)) -> f(c(X,s(Y)))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(c(X,s(Y))) -> f#(c(s(X),Y))

and R consists of:

r1: f(c(X,s(Y))) -> f(c(s(X),Y))
r2: g(c(s(X),Y)) -> f(c(X,s(Y)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1) = x1 + (3,1)
            c_A(x1,x2) = x2 + (1,1)
            s_A(x1) = x1 + (2,3)
    
      precedence:
      
        f# = c = s
      
      partial status:
    
        pi(f#) = []
        pi(c) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1) = x1
            c_A(x1,x2) = x2
            s_A(x1) = x1 + (1,1)
    
      precedence:
      
        f# > c = s
      
      partial status:
    
        pi(f#) = [1]
        pi(c) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.