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:

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

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: 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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      g#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,0,0),(1,1,0,0)) x1
      s_A(x1) = ((1,0,0,0),(0,0,0,0),(0,1,0,0),(1,1,1,0)) x1 + (1,2,1,1)
      f_A(x1) = ((1,0,0,0),(1,1,0,0),(0,0,0,0),(0,0,0,0)) x1 + (1,1,0,1)
      c_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + x2 + (0,1,1,1)

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: 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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      g#_A(x1) = ((1,0,0,0),(1,1,0,0),(1,0,1,0),(1,0,1,1)) x1
      c_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(0,1,0,0),(1,1,1,0),(0,0,1,1)) x2 + (0,1,1,1)
      s_A(x1) = ((1,0,0,0),(0,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (1,0,0,0)

The next rules are strictly ordered:

  p1

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