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. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        f#_A(x1,x2) = ((1,0,0),(0,1,0),(1,1,1)) x1 + ((1,0,0),(0,1,0),(1,1,1)) x2
        c_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (2,2,1)
        s_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (1,1,1)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > c > f#
      
      argument filter:
    
        pi(f#) = 2
        pi(c) = 1
        pi(s) = 1
    

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. matrix interpretations:
    
      carrier: N^3
      order: lexicographic order
      interpretations:
        f#_A(x1,x2) = ((1,0,0),(0,1,0),(1,1,1)) x1 + ((1,0,0),(0,1,0),(1,1,1)) x2
        s_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (2,1,1)
        a_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (1,2,2)
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        s > f# > a
      
      argument filter:
    
        pi(f#) = 1
        pi(s) = [1]
        pi(a) = 1
    

The next rules are strictly ordered:

  p1

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