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, >MG)
  
    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) = x1 + x2 + 1
          s_A(x1) = x1 + 3
          g#_A(x1) = x1
  
    base order 2:
  
      precedence:
      
        c > s > f# > g#
      
  
    partial status:
  
      pi(f#) = [1]
      pi(c) = [1]
      pi(s) = []
      pi(g#) = []

The following rules are removed:

  p2



-- 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(s(x),s(y))) -> f#(c(x,y))
p3: 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 estimated dependency graph contains the following SCCs:

  {p3}
  {p1}


-- 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: 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, >MG)
  
    base order 1:
  
      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
  
    base order 2:
  
      precedence:
      
        c = s > g#
      
  
    partial status:
  
      pi(g#) = [1]
      pi(c) = [2]
      pi(s) = []

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: 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, >MG)
  
    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) = x1 + 2
          s_A(x1) = x1 + 3
  
    base order 2:
  
      precedence:
      
        c > f# = s
      
  
    partial status:
  
      pi(f#) = [1]
      pi(c) = []
      pi(s) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.