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

  r1

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              g#_A(x1) = x1
              c_A(x1,x2) = x2 + 1
              s_A(x1) = x1 + 2
              g_A(x1) = x1
              g#!_A(x1) = x1 + 2
              c!_A(x1,x2) = x1 + 1
              s!_A(x1) = x1 + 4
              g!_A(x1) = 5
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              g#_A(x1) = x1
              c_A(x1,x2) = 1
              s_A(x1) = x1 + 2
              g_A(x1) = x1
              g#!_A(x1) = 0
              c!_A(x1,x2) = x1 + 3
              s!_A(x1) = 1
              g!_A(x1) = 2
        
  
    partial status:
  
      pi(g#) = []
      pi(c) = [2]
      pi(s) = [1]
      pi(g) = []

The following rules are removed:

  p1

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

  r1

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              f#_A(x1) = x1
              c_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 1
              g_A(x1) = x1
              f#!_A(x1) = 4
              c!_A(x1,x2) = x1 + x2 + 2
              s!_A(x1) = 1
              g!_A(x1) = 0
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              f#_A(x1) = x1
              c_A(x1,x2) = x1 + x2
              s_A(x1) = x1 + 1
              g_A(x1) = x1
              f#!_A(x1) = 0
              c!_A(x1,x2) = x1 + 3
              s!_A(x1) = x1 + 1
              g!_A(x1) = 0
        
  
    partial status:
  
      pi(f#) = [1]
      pi(c) = []
      pi(s) = []
      pi(g) = []

The following rules are removed:

  p1

Then no dependency pair remains.