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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          f#_A(x1) = ((0,0),(1,0)) x1 + (4,3)
          c_A(x1,x2) = x1 + (1,1)
          s_A(x1) = x1 + (2,2)
  
    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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          g#_A(x1) = ((0,0),(1,0)) x1 + (1,1)
          s_A(x1) = x1 + (3,0)
          f_A(x1) = (1,1)
          c_A(x1,x2) = x1 + (2,0)
  
    precedence:
    
      g# = s = f = c
    
    partial status:
  
      pi(g#) = []
      pi(s) = [1]
      pi(f) = []
      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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          g#_A(x1) = x1 + (1,2)
          c_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (1,2)
          s_A(x1) = ((1,0),(1,1)) x1 + (3,4)
  
    precedence:
    
      g# = c = s
    
    partial status:
  
      pi(g#) = [1]
      pi(c) = [2]
      pi(s) = []

The next rules are strictly ordered:

  p1

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