YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

r1: g(c(x,s(y))) -> g(c(s(x),y))
r2: f(c(s(x),y)) -> f(c(x,s(y)))
r3: f(f(x)) -> f(d(f(x)))
r4: f(x) -> x

The estimated dependency graph contains the following SCCs:

  {p1}
  {p2}


-- 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: g(c(x,s(y))) -> g(c(s(x),y))
r2: f(c(s(x),y)) -> f(c(x,s(y)))
r3: f(f(x)) -> f(d(f(x)))
r4: f(x) -> x

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      g#_A(x1) = ((0,0,0),(0,0,0),(1,0,0)) x1
      c_A(x1,x2) = ((1,0,0),(1,0,0),(1,1,0)) x2 + (1,1,1)
      s_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#(c(s(x),y)) -> f#(c(x,s(y)))

and R consists of:

r1: g(c(x,s(y))) -> g(c(s(x),y))
r2: f(c(s(x),y)) -> f(c(x,s(y)))
r3: f(f(x)) -> f(d(f(x)))
r4: f(x) -> x

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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