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:

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

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^1
        order: lexicographic order
        interpretations:
          f#_A(x1,x2) = x1 + 3
          s_A(x1) = x1 + 1
          c_A() = 2
  
    precedence:
    
      f# = s = c
    
    partial status:
  
      pi(f#) = [1]
      pi(s) = [1]
      pi(c) = []

The next rules are strictly ordered:

  p1

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