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

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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            c#_A(x1) = ((1,0),(0,0)) x1 + (1,6)
            c_A(x1) = ((0,1),(1,0)) x1 + (7,2)
            b_A(x1) = ((0,0),(0,1)) x1 + (4,3)
            a#_A(x1,x2) = x1 + ((0,1),(0,0)) x2 + (5,5)
            |0|_A() = (4,1)
            a_A(x1,x2) = x2 + (9,9)
    
      precedence:
      
        a > c > b > a# > c# = |0|
      
      partial status:
    
        pi(c#) = []
        pi(c) = []
        pi(b) = []
        pi(a#) = [1]
        pi(|0|) = []
        pi(a) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            c#_A(x1) = (0,0)
            c_A(x1) = (1,0)
            b_A(x1) = (0,0)
            a#_A(x1,x2) = ((0,1),(0,0)) x1 + (1,1)
            |0|_A() = (2,1)
            a_A(x1,x2) = (2,1)
    
      precedence:
      
        a > |0| > c > b > c# = a#
      
      partial status:
    
        pi(c#) = []
        pi(c) = []
        pi(b) = []
        pi(a#) = []
        pi(|0|) = []
        pi(a) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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:

  (no SCCs)