YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: f#(a(),a()) -> f#(a(),b())
p2: f#(a(),b()) -> f#(s(a()),c())
p3: f#(s(X),c()) -> f#(X,c())
p4: f#(c(),c()) -> f#(a(),a())

and R consists of:

r1: f(a(),a()) -> f(a(),b())
r2: f(a(),b()) -> f(s(a()),c())
r3: f(s(X),c()) -> f(X,c())
r4: f(c(),c()) -> f(a(),a())

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

p1: f#(a(),a()) -> f#(a(),b())
p2: f#(a(),b()) -> f#(s(a()),c())
p3: f#(s(X),c()) -> f#(X,c())
p4: f#(c(),c()) -> f#(a(),a())

and R consists of:

r1: f(a(),a()) -> f(a(),b())
r2: f(a(),b()) -> f(s(a()),c())
r3: f(s(X),c()) -> f(X,c())
r4: f(c(),c()) -> f(a(),a())

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1,x2) = ((1,0),(0,0)) x1 + (3,5)
            a_A() = (1,4)
            b_A() = (1,3)
            s_A(x1) = ((1,0),(0,0)) x1 + (0,1)
            c_A() = (2,2)
    
      precedence:
      
        f# = c > a > b > s
      
      partial status:
    
        pi(f#) = []
        pi(a) = []
        pi(b) = []
        pi(s) = []
        pi(c) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1,x2) = (3,3)
            a_A() = (4,4)
            b_A() = (2,1)
            s_A(x1) = (1,2)
            c_A() = (5,5)
    
      precedence:
      
        c > f# = a = b = s
      
      partial status:
    
        pi(f#) = []
        pi(a) = []
        pi(b) = []
        pi(s) = []
        pi(c) = []
    

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

p1: f#(a(),a()) -> f#(a(),b())
p2: f#(a(),b()) -> f#(s(a()),c())
p3: f#(s(X),c()) -> f#(X,c())

and R consists of:

r1: f(a(),a()) -> f(a(),b())
r2: f(a(),b()) -> f(s(a()),c())
r3: f(s(X),c()) -> f(X,c())
r4: f(c(),c()) -> f(a(),a())

The estimated dependency graph contains the following SCCs:

  {p3}


-- Reduction pair.

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

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

and R consists of:

r1: f(a(),a()) -> f(a(),b())
r2: f(a(),b()) -> f(s(a()),c())
r3: f(s(X),c()) -> f(X,c())
r4: f(c(),c()) -> f(a(),a())

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1,x2) = ((1,0),(1,1)) x1 + x2
            s_A(x1) = x1 + (2,2)
            c_A() = (1,1)
    
      precedence:
      
        f# = s = c
      
      partial status:
    
        pi(f#) = [1]
        pi(s) = [1]
        pi(c) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1,x2) = (0,0)
            s_A(x1) = (2,2)
            c_A() = (1,1)
    
      precedence:
      
        f# > s > c
      
      partial status:
    
        pi(f#) = []
        pi(s) = []
        pi(c) = []
    

The next rules are strictly ordered:

  p1

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