YES

We show the termination of the TRS R:

  c(c(c(a(x,y)))) -> b(c(c(c(c(y)))),x)
  c(c(b(c(y),|0|()))) -> a(|0|(),c(c(a(y,|0|()))))
  c(c(a(a(y,|0|()),x))) -> c(y)

-- SCC decomposition.

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

p1: c#(c(c(a(x,y)))) -> c#(c(c(c(y))))
p2: c#(c(c(a(x,y)))) -> c#(c(c(y)))
p3: c#(c(c(a(x,y)))) -> c#(c(y))
p4: c#(c(c(a(x,y)))) -> c#(y)
p5: c#(c(b(c(y),|0|()))) -> c#(c(a(y,|0|())))
p6: c#(c(b(c(y),|0|()))) -> c#(a(y,|0|()))
p7: c#(c(a(a(y,|0|()),x))) -> c#(y)

and R consists of:

r1: c(c(c(a(x,y)))) -> b(c(c(c(c(y)))),x)
r2: c(c(b(c(y),|0|()))) -> a(|0|(),c(c(a(y,|0|()))))
r3: c(c(a(a(y,|0|()),x))) -> c(y)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p7}


-- Reduction pair.

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

p1: c#(c(c(a(x,y)))) -> c#(c(c(c(y))))
p2: c#(c(a(a(y,|0|()),x))) -> c#(y)
p3: c#(c(b(c(y),|0|()))) -> c#(c(a(y,|0|())))
p4: c#(c(c(a(x,y)))) -> c#(y)
p5: c#(c(c(a(x,y)))) -> c#(c(y))
p6: c#(c(c(a(x,y)))) -> c#(c(c(y)))

and R consists of:

r1: c(c(c(a(x,y)))) -> b(c(c(c(c(y)))),x)
r2: c(c(b(c(y),|0|()))) -> a(|0|(),c(c(a(y,|0|()))))
r3: c(c(a(a(y,|0|()),x))) -> c(y)

The set of usable rules consists of

  r1, r2, r3

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      c#_A(x1) = ((1,0),(1,1)) x1
      c_A(x1) = ((1,1),(1,0)) x1 + (1,1)
      a_A(x1,x2) = x1 + ((1,1),(1,0)) x2 + (1,6)
      |0|_A() = (1,1)
      b_A(x1,x2) = x1 + ((1,1),(1,1)) x2 + (10,5)

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6
  r2, r3

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