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. lexicographic path order with precedence:
    
      precedence:
      
        s > c > f#
      
      argument filter:
    
        pi(f#) = [1]
        pi(c) = [2]
        pi(s) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > c > f#
      
      argument filter:
    
        pi(f#) = 1
        pi(c) = [2]
        pi(s) = [1]
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        s > c > f#
      
      argument filter:
    
        pi(f#) = 1
        pi(c) = [2]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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