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 reduction pair:

  max/plus interpretations on natural numbers:
  
    c#_A(x1) = max{0, x1 - 70}
    c_A(x1) = x1 + 16
    a_A(x1,x2) = max{x1 - 8, x2 + 20}
    |0|_A = 42
    b_A(x1,x2) = max{x1 + 3, x2 + 40}

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- 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(a(a(y,|0|()),x))) -> c#(y)
p3: c#(c(c(a(x,y)))) -> c#(y)
p4: c#(c(c(a(x,y)))) -> c#(c(y))
p5: 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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- 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(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(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 set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    c#_A(x1) = x1
    c_A(x1) = max{9, x1 + 5}
    a_A(x1,x2) = max{15, x1 - 2, x2 + 5}
    |0|_A = 7
    b_A(x1,x2) = max{x1 - 2, x2 + 13}

The next rules are strictly ordered:

  p2, p3, p4, p5

We remove them from the problem.

-- 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))))

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}


-- 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))))

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 reduction pair:

  max/plus interpretations on natural numbers:
  
    c#_A(x1) = max{0, x1 - 5}
    c_A(x1) = max{6, x1 + 5}
    a_A(x1,x2) = max{8, x1, x2 + 7}
    b_A(x1,x2) = max{15, x1 + 2, x2 - 4}
    |0|_A = 1

The next rules are strictly ordered:

  p1

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