YES

We show the termination of the TRS R:

  f(s(X),X) -> f(X,a(X))
  f(X,c(X)) -> f(s(X),X)
  f(X,X) -> c(X)

-- SCC decomposition.

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

p1: f#(s(X),X) -> f#(X,a(X))
p2: f#(X,c(X)) -> f#(s(X),X)

and R consists of:

r1: f(s(X),X) -> f(X,a(X))
r2: f(X,c(X)) -> f(s(X),X)
r3: f(X,X) -> c(X)

The estimated dependency graph contains the following SCCs:

  {p2}
  {p1}


-- Reduction pair.

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

p1: f#(X,c(X)) -> f#(s(X),X)

and R consists of:

r1: f(s(X),X) -> f(X,a(X))
r2: f(X,c(X)) -> f(s(X),X)
r3: f(X,X) -> c(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      f#_A(x1,x2) = max{2, x1 + 1, x2}
      c_A(x1) = max{4, x1 + 3}
      s_A(x1) = max{0, x1 - 1}
    
    2. max/plus interpretations on natural numbers:
    
      f#_A(x1,x2) = 0
      c_A(x1) = 0
      s_A(x1) = 0
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

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

p1: f#(s(X),X) -> f#(X,a(X))

and R consists of:

r1: f(s(X),X) -> f(X,a(X))
r2: f(X,c(X)) -> f(s(X),X)
r3: f(X,X) -> c(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      f#_A(x1,x2) = max{1, x1 - 1, x2}
      s_A(x1) = max{4, x1 + 3}
      a_A(x1) = max{0, x1 - 1}
    
    2. max/plus interpretations on natural numbers:
    
      f#_A(x1,x2) = 0
      s_A(x1) = 0
      a_A(x1) = 0
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.