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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a#_A(x1) = x1 + 1
          b_A(x1) = x1
          b#_A(x1) = x1 + 1
          a_A(x1) = x1
          c_A(x1) = x1
          c#_A(x1) = x1 + 1
          u_A(x1) = x1 + 1
          v_A(x1) = x1 + 1
          w_A(x1) = x1 + 1
    
      precedence:
      
        a# = b = b# = a = c = c# = u = v = w
      
      partial status:
    
        pi(a#) = []
        pi(b) = []
        pi(b#) = []
        pi(a) = []
        pi(c) = []
        pi(c#) = []
        pi(u) = []
        pi(v) = []
        pi(w) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a#_A(x1) = x1 + 2
          b_A(x1) = x1 + 15
          b#_A(x1) = 7
          a_A(x1) = x1
          c_A(x1) = 0
          c#_A(x1) = 7
          u_A(x1) = x1 + 1
          v_A(x1) = x1 + 1
          w_A(x1) = 0
    
      precedence:
      
        b# > c# > c = u = v = w > a > b > a#
      
      partial status:
    
        pi(a#) = [1]
        pi(b) = [1]
        pi(b#) = []
        pi(a) = []
        pi(c) = []
        pi(c#) = []
        pi(u) = []
        pi(v) = []
        pi(w) = []
    

The next rules are strictly ordered:

  p4, p8, p9

We remove them from the problem.

-- SCC decomposition.

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: c#(a(x)) -> c#(x)
p5: c#(a(x)) -> c#(c(x))
p6: c#(a(x)) -> a#(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 estimated dependency graph contains the following SCCs:

  {p4, p5}
  {p2, p3}


-- Reduction pair.

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

p1: c#(a(x)) -> c#(c(x))
p2: 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 set of usable rules consists of

  r1, r2, r3, r7, r8, r9

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          c#_A(x1) = x1 + 2
          a_A(x1) = x1
          c_A(x1) = x1
          b_A(x1) = max{0, x1 - 1}
          v_A(x1) = max{3, x1 + 2}
          u_A(x1) = x1 + 1
          w_A(x1) = x1 + 1
    
      precedence:
      
        c# = a = c = b = v = u = w
      
      partial status:
    
        pi(c#) = []
        pi(a) = []
        pi(c) = []
        pi(b) = []
        pi(v) = []
        pi(u) = []
        pi(w) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          c#_A(x1) = max{0, x1 - 2}
          a_A(x1) = max{7, x1 + 3}
          c_A(x1) = x1
          b_A(x1) = 9
          v_A(x1) = 0
          u_A(x1) = x1 + 1
          w_A(x1) = x1 + 1
    
      precedence:
      
        b > c = w > a > v > c# = u
      
      partial status:
    
        pi(c#) = []
        pi(a) = []
        pi(c) = []
        pi(b) = []
        pi(v) = []
        pi(u) = []
        pi(w) = []
    

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: 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}


-- Reduction pair.

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

p1: 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 set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          c#_A(x1) = max{2, x1 + 1}
          a_A(x1) = max{1, x1}
    
      precedence:
      
        c# = a
      
      partial status:
    
        pi(c#) = [1]
        pi(a) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          c#_A(x1) = x1 + 2
          a_A(x1) = x1 + 1
    
      precedence:
      
        c# = a
      
      partial status:
    
        pi(c#) = [1]
        pi(a) = [1]
    

The next rules are strictly ordered:

  p1

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: b#(c(x)) -> b#(x)
p2: b#(c(x)) -> b#(b(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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          b#_A(x1) = x1
          c_A(x1) = x1
          b_A(x1) = x1
          a_A(x1) = max{0, x1 - 1}
          u_A(x1) = max{3, x1 + 2}
          w_A(x1) = x1 + 1
          v_A(x1) = x1 + 1
    
      precedence:
      
        b# = c = b = a = u = w = v
      
      partial status:
    
        pi(b#) = []
        pi(c) = []
        pi(b) = []
        pi(a) = []
        pi(u) = []
        pi(w) = []
        pi(v) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          b#_A(x1) = max{4, x1 + 2}
          c_A(x1) = max{4, x1 + 3}
          b_A(x1) = x1
          a_A(x1) = 0
          u_A(x1) = 0
          w_A(x1) = max{0, x1 - 2}
          v_A(x1) = x1
    
      precedence:
      
        u > a = w = v > b > c > b#
      
      partial status:
    
        pi(b#) = []
        pi(c) = []
        pi(b) = []
        pi(a) = []
        pi(u) = []
        pi(w) = []
        pi(v) = []
    

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: b#(c(x)) -> b#(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}


-- Reduction pair.

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

p1: b#(c(x)) -> b#(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

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          b#_A(x1) = max{4, x1 + 3}
          c_A(x1) = max{3, x1 + 2}
    
      precedence:
      
        b# = c
      
      partial status:
    
        pi(b#) = [1]
        pi(c) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          b#_A(x1) = max{1, x1 - 1}
          c_A(x1) = x1
    
      precedence:
      
        b# = c
      
      partial status:
    
        pi(b#) = []
        pi(c) = [1]
    

The next rules are strictly ordered:

  p1

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