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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          c#_A(x1) = ((1,0),(0,0)) x1
          c_A(x1) = ((1,0),(1,0)) x1 + (2,1)
          b_A(x1) = ((0,1),(0,0)) x1 + (1,2)
          a#_A(x1,x2) = ((1,0),(0,0)) x2 + (2,0)
          |0|_A() = (1,0)
          a_A(x1,x2) = x1 + ((1,0),(1,0)) x2 + (4,3)
  
    precedence:
    
      a > c# = c = b = a# = |0|
    
    partial status:
  
      pi(c#) = []
      pi(c) = []
      pi(b) = []
      pi(a#) = []
      pi(|0|) = []
      pi(a) = [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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          c#_A(x1) = ((0,1),(0,0)) x1 + (5,11)
          c_A(x1) = ((0,1),(1,0)) x1 + (1,5)
          b_A(x1) = ((1,0),(0,0)) x1 + (3,1)
          a#_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (1,1)
          |0|_A() = (11,10)
          a_A(x1,x2) = ((1,0),(1,1)) x2 + (6,6)
  
    precedence:
    
      c = b = a > c# = a# = |0|
    
    partial status:
  
      pi(c#) = []
      pi(c) = []
      pi(b) = []
      pi(a#) = []
      pi(|0|) = []
      pi(a) = [2]

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))

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:

  (no SCCs)