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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        a#_A(x1) = x1
        b_A(x1) = x1
        b#_A(x1) = x1
        a_A(x1) = x1
        c_A(x1) = x1
        c#_A(x1) = x1
        u_A(x1) = x1 + 1
        v_A(x1) = x1 + 1
        w_A(x1) = x1 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        a#_A(x1) = 2
        b_A(x1) = x1
        b#_A(x1) = x1
        a_A(x1) = 1
        c_A(x1) = x1 + 4
        c#_A(x1) = 3
        u_A(x1) = 0
        v_A(x1) = 1
        w_A(x1) = 0
    

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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        c#_A(x1) = x1
        a_A(x1) = x1
        c_A(x1) = x1
        b_A(x1) = x1
        v_A(x1) = x1 + 1
        u_A(x1) = x1 + 1
        w_A(x1) = x1 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        c#_A(x1) = x1
        a_A(x1) = x1 + 1
        c_A(x1) = x1
        b_A(x1) = 1
        v_A(x1) = 0
        u_A(x1) = 0
        w_A(x1) = 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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        a#_A(x1) = x1
        b_A(x1) = x1
        a_A(x1) = x1
        c_A(x1) = x1
        w_A(x1) = x1 + 1
        v_A(x1) = x1 + 1
        u_A(x1) = x1 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        a#_A(x1) = x1
        b_A(x1) = x1 + 1
        a_A(x1) = x1
        c_A(x1) = 1
        w_A(x1) = 0
        v_A(x1) = 0
        u_A(x1) = 1
    

The next rules are strictly ordered:

  p1, p2

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