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:

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

The next rules are strictly ordered:

  p5, p6, p7

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: b#(c(x)) -> c#(b(b(x)))
p5: a#(b(x)) -> a#(x)
p6: 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:

  {p5, p6}
  {p2, p3}


-- Reduction pair.

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

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

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      a#_A(x1) = ((1,0,0),(0,1,0),(0,1,1)) x1
      b_A(x1) = ((1,0,0),(0,1,0),(0,1,0)) x1 + (0,1,1)
      a_A(x1) = ((1,0,0),(0,1,0),(0,1,0)) x1 + (0,0,1)
      c_A(x1) = ((1,0,0),(0,0,0),(0,1,0)) x1 + (0,1,3)
      w_A(x1) = ((1,0,0),(1,0,0),(0,0,0)) x1 + (1,1,0)
      v_A(x1) = x1 + (1,0,0)
      u_A(x1) = x1 + (1,1,0)

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: 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:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      b#_A(x1) = ((1,0,0),(0,0,0),(0,1,0)) x1
      c_A(x1) = x1 + (0,1,1)
      b_A(x1) = x1 + (0,0,2)
      a_A(x1) = x1 + (0,1,3)
      u_A(x1) = x1 + (1,1,0)
      w_A(x1) = ((1,0,0),(1,0,0),(0,0,0)) x1 + (1,1,0)
      v_A(x1) = x1 + (1,1,0)

The next rules are strictly ordered:

  p1, p2

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