YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2}
  {p3}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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