YES

We show the termination of the TRS R:

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

and R consists of:

r1: f(x,c(y)) -> f(x,s(f(y,y)))
r2: f(s(x),y) -> f(x,s(c(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),y) -> f(x,s(c(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:
      
        c > f#
      
      argument filter:
    
        pi(f#) = 2
        pi(c) = [1]
    
    2. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        f#_A(x1,x2) = ((1,0,0),(1,1,0),(1,1,1)) x2
        c_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (1,1,1)
    

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

and R consists of:

r1: f(x,c(y)) -> f(x,s(f(y,y)))
r2: f(s(x),y) -> f(x,s(c(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:
      
        c > s > f#
      
      argument filter:
    
        pi(f#) = [1]
        pi(s) = 1
        pi(c) = [1]
    
    2. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        f#_A(x1,x2) = ((0,0,0),(1,0,0),(0,0,0)) x1
        s_A(x1) = ((1,0,0),(0,0,0),(1,1,0)) x1 + (1,1,0)
        c_A(x1) = ((1,0,0),(1,0,0),(0,0,0)) x1 + (1,1,0)
    

The next rules are strictly ordered:

  p1

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