YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    c#_A(x1) = max{0, x1 - 12}
    a_A(x1,x2) = max{13, x1 + 1, x2 + 10}
    c_A(x1) = max{17, x1 + 4}
    |0|_A = 1

The next rules are strictly ordered:

  p1, p2, p3

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