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, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        f#_A(x1,x2) = max{x1 + 5, x2}
        c_A(x1) = x1 + 5
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          f#_A(x1,x2) = max{x1, x2 - 3}
          c_A(x1) = x1 + 3
          f#!_A(x1,x2) = max{0, x1 - 5}
          c!_A(x1) = max{2, x1 - 4}
  
    partial status:
  
      pi(f#) = [1, 2]
      pi(c) = [1]

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 reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        f#_A(x1,x2) = x2 + 5
        s_A(x1) = max{0, x1 - 12}
        c_A(x1) = x1 + 8
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          f#_A(x1,x2) = max{0, x1 - 21, x2 - 16}
          s_A(x1) = x1 + 6
          c_A(x1) = max{9, x1 - 34}
          f#!_A(x1,x2) = max{2, x1 - 3}
          s!_A(x1) = max{5, x1 - 8}
          c!_A(x1) = max{4, x1 - 30}
  
    partial status:
  
      pi(f#) = []
      pi(s) = []
      pi(c) = []

The following rules are removed:

  p1

Then no dependency pair remains.