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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{x1 + 3, x2 + 6}
          c_A(x1) = x1 + 3
          s_A(x1) = max{2, x1 + 1}
    
      precedence:
      
        f# = c = s
      
      partial status:
    
        pi(f#) = [1, 2]
        pi(c) = [1]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{x1 + 1, x2 + 1}
          c_A(x1) = x1
          s_A(x1) = x1 + 2
    
      precedence:
      
        f# = c = s
      
      partial status:
    
        pi(f#) = []
        pi(c) = [1]
        pi(s) = []
    

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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{13, x1 + 7, x2 + 4}
          s_A(x1) = x1 + 6
          a_A(x1) = x1 + 3
    
      precedence:
      
        f# = s = a
      
      partial status:
    
        pi(f#) = [1, 2]
        pi(s) = [1]
        pi(a) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{x1 + 2, x2 + 3}
          s_A(x1) = x1 + 1
          a_A(x1) = max{4, x1 + 1}
    
      precedence:
      
        f# = s = a
      
      partial status:
    
        pi(f#) = [1]
        pi(s) = [1]
        pi(a) = []
    

The next rules are strictly ordered:

  p1

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