YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}
  {p3}
  {p2}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        f#_A(x1) = x1
        c_A(x1,x2) = x1
        s_A(x1) = x1 + 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > c > f#
      
      argument filter:
    
        pi(f#) = [1]
        pi(c) = [1]
        pi(s) = [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: g#(s(f(x))) -> g#(f(x))

and R consists of:

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

The set of usable rules consists of

  r1

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        g#_A(x1) = x1
        s_A(x1) = x1 + 1
        f_A(x1) = x1 + 1
        c_A(x1,x2) = x1 + x2
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > f > c > g#
      
      argument filter:
    
        pi(g#) = 1
        pi(s) = [1]
        pi(f) = 1
        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: g#(c(x,s(y))) -> g#(c(s(x),y))

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        g#_A(x1) = x1
        c_A(x1,x2) = x1 + x2
        s_A(x1) = x1 + 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > c > g#
      
      argument filter:
    
        pi(g#) = 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.