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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          a#_A(x1) = ((1,0),(0,0)) x1 + (0,4)
          b_A(x1) = ((1,0),(0,0)) x1 + (0,1)
          b#_A(x1) = ((1,0),(0,0)) x1 + (0,3)
          a_A(x1) = x1 + (0,5)
          c_A(x1) = x1
          c#_A(x1) = x1 + (0,1)
          u_A(x1) = ((1,0),(0,0)) x1 + (1,1)
          v_A(x1) = x1 + (1,1)
          w_A(x1) = x1 + (1,1)
  
    precedence:
    
      b = c = w > a# = a > b# > u = v > c#
    
    partial status:
  
      pi(a#) = []
      pi(b) = []
      pi(b#) = []
      pi(a) = [1]
      pi(c) = []
      pi(c#) = [1]
      pi(u) = []
      pi(v) = [1]
      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: b#(c(x)) -> b#(x)
p2: b#(c(x)) -> b#(b(x))
p3: b#(c(x)) -> c#(b(b(x)))
p4: c#(a(x)) -> c#(x)
p5: c#(a(x)) -> c#(c(x))
p6: c#(a(x)) -> a#(c(c(x)))
p7: a#(b(x)) -> a#(x)
p8: 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}
  {p4, p5}
  {p7, p8}


-- 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          b#_A(x1) = x1
          c_A(x1) = x1 + (0,2)
          b_A(x1) = x1
          a_A(x1) = ((1,0),(0,0)) x1 + (0,1)
          u_A(x1) = ((1,0),(1,1)) x1 + (1,1)
          w_A(x1) = x1 + (1,0)
          v_A(x1) = x1 + (1,1)
  
    precedence:
    
      c = b = a = u > w = v > b#
    
    partial status:
  
      pi(b#) = []
      pi(c) = []
      pi(b) = []
      pi(a) = []
      pi(u) = []
      pi(w) = []
      pi(v) = [1]

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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          b#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          c_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      b# = c
    
    partial status:
  
      pi(b#) = []
      pi(c) = []

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: 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          c#_A(x1) = x1 + (0,2)
          a_A(x1) = x1 + (0,2)
          c_A(x1) = x1
          b_A(x1) = ((1,0),(0,0)) x1 + (0,3)
          v_A(x1) = x1 + (1,1)
          u_A(x1) = x1 + (1,1)
          w_A(x1) = x1 + (1,1)
  
    precedence:
    
      c# = a = c = b = v = u = w
    
    partial status:
  
      pi(c#) = []
      pi(a) = []
      pi(c) = []
      pi(b) = []
      pi(v) = []
      pi(u) = []
      pi(w) = []

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#(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#(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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          c#_A(x1) = x1 + (0,1)
          a_A(x1) = x1 + (0,2)
          c_A(x1) = x1
          b_A(x1) = ((1,0),(0,0)) x1
          v_A(x1) = x1 + (1,1)
          u_A(x1) = ((1,0),(1,1)) x1 + (1,1)
          w_A(x1) = x1 + (1,1)
  
    precedence:
    
      a = c = b > c# > v = u = w
    
    partial status:
  
      pi(c#) = [1]
      pi(a) = []
      pi(c) = []
      pi(b) = []
      pi(v) = [1]
      pi(u) = [1]
      pi(w) = []

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: 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          a#_A(x1) = x1
          b_A(x1) = x1 + (0,1)
          a_A(x1) = x1
          c_A(x1) = ((1,0),(0,0)) x1 + (0,3)
          w_A(x1) = ((1,0),(1,1)) x1 + (1,1)
          v_A(x1) = ((1,0),(1,1)) x1 + (1,1)
          u_A(x1) = x1 + (1,1)
  
    precedence:
    
      c = w = v > a > b = u > a#
    
    partial status:
  
      pi(a#) = []
      pi(b) = [1]
      pi(a) = []
      pi(c) = []
      pi(w) = []
      pi(v) = []
      pi(u) = [1]

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: 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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          a#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          b_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      a# = b
    
    partial status:
  
      pi(a#) = []
      pi(b) = []

The next rules are strictly ordered:

  p1

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