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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 5
          c_A(x1,x2) = max{0, x1 - 2}
          s_A(x1) = x1 + 3
    
      precedence:
      
        f# = c = s
      
      partial status:
    
        pi(f#) = []
        pi(c) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 3
          c_A(x1,x2) = 0
          s_A(x1) = max{6, x1 + 3}
    
      precedence:
      
        f# = c = s
      
      partial status:
    
        pi(f#) = []
        pi(c) = []
        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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = max{13, x1 + 11}
          s_A(x1) = max{3, x1}
          f_A(x1) = max{12, x1 + 11}
          c_A(x1,x2) = max{x1 + 8, x2 + 4}
    
      precedence:
      
        g# = s = f = c
      
      partial status:
    
        pi(g#) = [1]
        pi(s) = [1]
        pi(f) = [1]
        pi(c) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = max{12, x1 + 10}
          s_A(x1) = x1 + 5
          f_A(x1) = x1 + 6
          c_A(x1,x2) = max{x1, x2 + 4}
    
      precedence:
      
        g# = s = f = c
      
      partial status:
    
        pi(g#) = []
        pi(s) = [1]
        pi(f) = []
        pi(c) = [1, 2]
    

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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = max{5, x1 + 1}
          c_A(x1,x2) = max{3, x2 - 2}
          s_A(x1) = max{9, x1 + 8}
    
      precedence:
      
        c > g# = s
      
      partial status:
    
        pi(g#) = [1]
        pi(c) = []
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = x1 + 3
          c_A(x1,x2) = 0
          s_A(x1) = max{6, x1 + 3}
    
      precedence:
      
        g# = c = s
      
      partial status:
    
        pi(g#) = []
        pi(c) = []
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1

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