YES
We show the termination of the TRS R:

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


-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2}
  {p3}


-- Reduction pair.

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

p1: f#(x,c(y)) -> f#(y,y)

and R consists of:

r1: f(x,c(y)) -> f(x,s(f(y,y)))
r2: f(s(x),y) -> f(x,s(c(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,x2) = x2
          c_A(x1) = x1 + 3
  
    base order 2:
  
      precedence:
      
        c > f#
      
  
    partial status:
  
      pi(f#) = [2]
      pi(c) = []

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the monotone 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,x2) = x1 + x2 + 1
          s_A(x1) = x1 + 1
          c_A(x1) = x1
  
    base order 2:
  
      precedence:
      
        s = c > f#
      
  
    partial status:
  
      pi(f#) = [1, 2]
      pi(s) = [1]
      pi(c) = [1]

The following rules are removed:

  p1

Then no dependency pair remains.