YES

We show the termination of the TRS R:

  f(s(x),y) -> f(x,s(x))
  f(x,s(y)) -> f(y,x)
  f(c(x),y) -> f(x,s(x))

-- SCC decomposition.

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

p1: f#(s(x),y) -> f#(x,s(x))
p2: f#(x,s(y)) -> f#(y,x)
p3: f#(c(x),y) -> f#(x,s(x))

and R consists of:

r1: f(s(x),y) -> f(x,s(x))
r2: f(x,s(y)) -> f(y,x)
r3: f(c(x),y) -> f(x,s(x))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: f#(s(x),y) -> f#(x,s(x))
p2: f#(c(x),y) -> f#(x,s(x))
p3: f#(x,s(y)) -> f#(y,x)

and R consists of:

r1: f(s(x),y) -> f(x,s(x))
r2: f(x,s(y)) -> f(y,x)
r3: f(c(x),y) -> f(x,s(x))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{x1 + 10, x2 + 8}
          s_A(x1) = max{6, x1 + 3}
          c_A(x1) = x1 + 6
    
      precedence:
      
        f# = s = c
      
      partial status:
    
        pi(f#) = [1, 2]
        pi(s) = [1]
        pi(c) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{x1 + 4, x2 - 2}
          s_A(x1) = max{8, x1 + 7}
          c_A(x1) = x1 + 2
    
      precedence:
      
        f# = s = c
      
      partial status:
    
        pi(f#) = [1]
        pi(s) = [1]
        pi(c) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3

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