YES

We show the termination of the TRS R:

  a(b(x)) -> b(a(a(x)))
  b(c(x)) -> c(b(b(x)))
  c(a(x)) -> a(c(c(x)))
  u(a(x)) -> x
  v(b(x)) -> x
  w(c(x)) -> x
  a(u(x)) -> x
  b(v(x)) -> x
  c(w(x)) -> x

-- SCC decomposition.

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

p1: a#(b(x)) -> b#(a(a(x)))
p2: a#(b(x)) -> a#(a(x))
p3: a#(b(x)) -> a#(x)
p4: b#(c(x)) -> c#(b(b(x)))
p5: b#(c(x)) -> b#(b(x))
p6: b#(c(x)) -> b#(x)
p7: c#(a(x)) -> a#(c(c(x)))
p8: c#(a(x)) -> c#(c(x))
p9: c#(a(x)) -> c#(x)

and R consists of:

r1: a(b(x)) -> b(a(a(x)))
r2: b(c(x)) -> c(b(b(x)))
r3: c(a(x)) -> a(c(c(x)))
r4: u(a(x)) -> x
r5: v(b(x)) -> x
r6: w(c(x)) -> x
r7: a(u(x)) -> x
r8: b(v(x)) -> x
r9: c(w(x)) -> x

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9}


-- Reduction pair.

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

p1: a#(b(x)) -> b#(a(a(x)))
p2: b#(c(x)) -> b#(x)
p3: b#(c(x)) -> b#(b(x))
p4: b#(c(x)) -> c#(b(b(x)))
p5: c#(a(x)) -> c#(x)
p6: c#(a(x)) -> c#(c(x))
p7: c#(a(x)) -> a#(c(c(x)))
p8: a#(b(x)) -> a#(x)
p9: a#(b(x)) -> a#(a(x))

and R consists of:

r1: a(b(x)) -> b(a(a(x)))
r2: b(c(x)) -> c(b(b(x)))
r3: c(a(x)) -> a(c(c(x)))
r4: u(a(x)) -> x
r5: v(b(x)) -> x
r6: w(c(x)) -> x
r7: a(u(x)) -> x
r8: b(v(x)) -> x
r9: c(w(x)) -> x

The set of usable rules consists of

  r1, r2, r3, r7, r8, r9

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        w > v > u > a# > c# > a > b > b# > c
      
      argument filter:
    
        pi(a#) = 1
        pi(b) = 1
        pi(b#) = 1
        pi(a) = 1
        pi(c) = 1
        pi(c#) = 1
        pi(u) = [1]
        pi(v) = [1]
        pi(w) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        w > v > u > b > c > c# > a# > b# > a
      
      argument filter:
    
        pi(a#) = []
        pi(b) = 1
        pi(b#) = [1]
        pi(a) = []
        pi(c) = [1]
        pi(c#) = []
        pi(u) = [1]
        pi(v) = [1]
        pi(w) = 1
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        w > v > u > c# > a > a# > b# > b > c
      
      argument filter:
    
        pi(a#) = []
        pi(b) = 1
        pi(b#) = []
        pi(a) = []
        pi(c) = [1]
        pi(c#) = []
        pi(u) = 1
        pi(v) = 1
        pi(w) = 1
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p7

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: a(b(x)) -> b(a(a(x)))
r2: b(c(x)) -> c(b(b(x)))
r3: c(a(x)) -> a(c(c(x)))
r4: u(a(x)) -> x
r5: v(b(x)) -> x
r6: w(c(x)) -> x
r7: a(u(x)) -> x
r8: b(v(x)) -> x
r9: c(w(x)) -> x

The estimated dependency graph contains the following SCCs:

  {p1, p2}
  {p3, p4}


-- Reduction pair.

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

p1: c#(a(x)) -> c#(x)
p2: c#(a(x)) -> c#(c(x))

and R consists of:

r1: a(b(x)) -> b(a(a(x)))
r2: b(c(x)) -> c(b(b(x)))
r3: c(a(x)) -> a(c(c(x)))
r4: u(a(x)) -> x
r5: v(b(x)) -> x
r6: w(c(x)) -> x
r7: a(u(x)) -> x
r8: b(v(x)) -> x
r9: c(w(x)) -> x

The set of usable rules consists of

  r1, r2, r3, r7, r8, r9

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        w > u > v > a > c > b > c#
      
      argument filter:
    
        pi(c#) = 1
        pi(a) = 1
        pi(c) = 1
        pi(b) = 1
        pi(v) = 1
        pi(u) = [1]
        pi(w) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        w > u > v > c > b > c# > a
      
      argument filter:
    
        pi(c#) = [1]
        pi(a) = 1
        pi(c) = 1
        pi(b) = 1
        pi(v) = [1]
        pi(u) = [1]
        pi(w) = [1]
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        u > w > v > c# > a > b > c
      
      argument filter:
    
        pi(c#) = [1]
        pi(a) = [1]
        pi(c) = 1
        pi(b) = []
        pi(v) = 1
        pi(u) = 1
        pi(w) = 1
    

The next rules are strictly ordered:

  p1, p2

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

-- Reduction pair.

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

p1: a#(b(x)) -> a#(x)
p2: a#(b(x)) -> a#(a(x))

and R consists of:

r1: a(b(x)) -> b(a(a(x)))
r2: b(c(x)) -> c(b(b(x)))
r3: c(a(x)) -> a(c(c(x)))
r4: u(a(x)) -> x
r5: v(b(x)) -> x
r6: w(c(x)) -> x
r7: a(u(x)) -> x
r8: b(v(x)) -> x
r9: c(w(x)) -> x

The set of usable rules consists of

  r1, r2, r3, r7, r8, r9

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        u > v > w > b > a > c > a#
      
      argument filter:
    
        pi(a#) = 1
        pi(b) = 1
        pi(a) = 1
        pi(c) = 1
        pi(w) = 1
        pi(v) = [1]
        pi(u) = 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        u > v > w > a > c > b > a#
      
      argument filter:
    
        pi(a#) = 1
        pi(b) = 1
        pi(a) = 1
        pi(c) = 1
        pi(w) = [1]
        pi(v) = [1]
        pi(u) = [1]
    
    3. lexicographic path order with precedence:
    
      precedence:
      
        u > v > w > b > c > a > a#
      
      argument filter:
    
        pi(a#) = 1
        pi(b) = [1]
        pi(a) = 1
        pi(c) = []
        pi(w) = 1
        pi(v) = 1
        pi(u) = 1
    

The next rules are strictly ordered:

  p1, p2

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