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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        c#_A(x1) = ((0,1),(0,0)) x1
        c_A(x1) = ((0,0),(1,1)) x1 + (1,1)
        b_A(x1) = x1 + (1,0)
        a#_A(x1,x2) = x1 + ((1,1),(0,0)) x2
        |0|_A() = (1,1)
        a_A(x1,x2) = x1 + ((0,0),(1,1)) x2 + (1,3)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        c#_A(x1) = (1,0)
        c_A(x1) = (2,2)
        b_A(x1) = (1,2)
        a#_A(x1,x2) = x1
        |0|_A() = (1,1)
        a_A(x1,x2) = x1 + (0,1)
    

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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        c#_A(x1) = ((0,1),(0,0)) x1
        c_A(x1) = ((0,1),(1,0)) x1 + (1,4)
        b_A(x1) = x1 + (2,1)
        a#_A(x1,x2) = x2 + (5,0)
        |0|_A() = (1,1)
        a_A(x1,x2) = x2 + (5,5)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        c#_A(x1) = (0,0)
        c_A(x1) = (2,1)
        b_A(x1) = (1,1)
        a#_A(x1,x2) = (1,1)
        |0|_A() = (1,1)
        a_A(x1,x2) = (3,1)
    

The next rules are strictly ordered:

  p1, p2

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