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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      f#_A(x1,x2) = x1
      a_A() = (1,1,1)
      b_A() = (2,1,1)
      s_A(x1) = ((1,0,0),(0,0,0),(1,0,0)) x1 + (0,1,1)
      c_A() = (2,1,1)

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 monotone reduction pair:

  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),(0,1,1)) x2
      s_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (1,1,1)
      c_A() = (0,0,0)

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4

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