YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p3, p4}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

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

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: standard order
    interpretations:
      c#_A(x1) = x1
      c_A(x1) = ((0,1),(1,0)) x1 + (3,1)
      b_A(x1) = x1 + (1,2)
      a#_A(x1,x2) = ((0,1),(0,0)) x2 + (4,0)
      |0|_A() = (1,1)
      a_A(x1,x2) = ((1,1),(0,1)) x2 + (4,4)

The next rules are strictly ordered:

  p1, p2

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