YES
We show the termination of the TRS R:

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


-- 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: f#(c(s(x),s(y))) -> g#(c(x,y))
p3: g#(c(x,s(y))) -> g#(c(s(x),y))
p4: g#(c(s(x),s(y))) -> f#(c(x,y))

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

p1: f#(c(s(x),y)) -> f#(c(x,s(y)))
p2: f#(c(s(x),s(y))) -> g#(c(x,y))
p3: g#(c(s(x),s(y))) -> f#(c(x,y))
p4: 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: f(c(s(x),s(y))) -> g(c(x,y))
r3: g(c(x,s(y))) -> g(c(s(x),y))
r4: g(c(s(x),s(y))) -> f(c(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          f#_A(x1) = x1 + 2
          c_A(x1,x2) = x1 + x2 + 2
          s_A(x1) = x1 + 1
          g#_A(x1) = x1
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          f#_A(x1) = 9
          c_A(x1,x2) = 0
          s_A(x1) = max{10, x1 + 1}
          g#_A(x1) = max{9, x1 + 8}
          f#!_A(x1) = 3
          c!_A(x1,x2) = 4
          s!_A(x1) = max{6, x1 + 1}
          g#!_A(x1) = 2
  
    partial status:
  
      pi(f#) = []
      pi(c) = [2]
      pi(s) = []
      pi(g#) = [1]

The following rules are removed:

  p2, p3, p4



-- SCC decomposition.

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

The estimated dependency graph contains the following SCCs:

  {p1}


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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          f#_A(x1) = x1
          c_A(x1,x2) = 3
          s_A(x1) = 2
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          f#_A(x1) = 0
          c_A(x1,x2) = max{0, x1 - 17}
          s_A(x1) = max{16, x1 + 11}
          f#!_A(x1) = x1 + 9
          c!_A(x1,x2) = max{1, x1 - 1}
          s!_A(x1) = 0
  
    partial status:
  
      pi(f#) = [1]
      pi(c) = []
      pi(s) = []

The following rules are removed:

  p1

Then no dependency pair remains.