YES

We show the termination of the TRS R:

  f(a(),f(b(),x)) -> f(a(),f(a(),f(a(),x)))
  f(b(),f(a(),x)) -> f(b(),f(b(),f(b(),x)))

-- SCC decomposition.

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

p1: f#(a(),f(b(),x)) -> f#(a(),f(a(),f(a(),x)))
p2: f#(a(),f(b(),x)) -> f#(a(),f(a(),x))
p3: f#(a(),f(b(),x)) -> f#(a(),x)
p4: f#(b(),f(a(),x)) -> f#(b(),f(b(),f(b(),x)))
p5: f#(b(),f(a(),x)) -> f#(b(),f(b(),x))
p6: f#(b(),f(a(),x)) -> f#(b(),x)

and R consists of:

r1: f(a(),f(b(),x)) -> f(a(),f(a(),f(a(),x)))
r2: f(b(),f(a(),x)) -> f(b(),f(b(),f(b(),x)))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}
  {p4, p5, p6}


-- Reduction pair.

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

p1: f#(a(),f(b(),x)) -> f#(a(),f(a(),f(a(),x)))
p2: f#(a(),f(b(),x)) -> f#(a(),x)
p3: f#(a(),f(b(),x)) -> f#(a(),f(a(),x))

and R consists of:

r1: f(a(),f(b(),x)) -> f(a(),f(a(),f(a(),x)))
r2: f(b(),f(a(),x)) -> f(b(),f(b(),f(b(),x)))

The set of usable rules consists of

  r1

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          f#_A(x1,x2) = x2
          a_A() = (1,8)
          f_A(x1,x2) = ((0,0),(1,0)) x1 + x2 + (0,2)
          b_A() = (8,7)
  
    precedence:
    
      f# = a = f = b
    
    partial status:
  
      pi(f#) = [2]
      pi(a) = []
      pi(f) = [2]
      pi(b) = []

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: f#(a(),f(b(),x)) -> f#(a(),f(a(),f(a(),x)))
p2: f#(a(),f(b(),x)) -> f#(a(),f(a(),x))

and R consists of:

r1: f(a(),f(b(),x)) -> f(a(),f(a(),f(a(),x)))
r2: f(b(),f(a(),x)) -> f(b(),f(b(),f(b(),x)))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

r1: f(a(),f(b(),x)) -> f(a(),f(a(),f(a(),x)))
r2: f(b(),f(a(),x)) -> f(b(),f(b(),f(b(),x)))

The set of usable rules consists of

  r1

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          f#_A(x1,x2) = x2 + (1,2)
          a_A() = (3,8)
          f_A(x1,x2) = ((0,0),(1,0)) x1 + (1,1)
          b_A() = (4,9)
  
    precedence:
    
      f# = a = f = b
    
    partial status:
  
      pi(f#) = []
      pi(a) = []
      pi(f) = []
      pi(b) = []

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: f#(a(),f(b(),x)) -> f#(a(),f(a(),x))

and R consists of:

r1: f(a(),f(b(),x)) -> f(a(),f(a(),f(a(),x)))
r2: f(b(),f(a(),x)) -> f(b(),f(b(),f(b(),x)))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: f#(a(),f(b(),x)) -> f#(a(),f(a(),x))

and R consists of:

r1: f(a(),f(b(),x)) -> f(a(),f(a(),f(a(),x)))
r2: f(b(),f(a(),x)) -> f(b(),f(b(),f(b(),x)))

The set of usable rules consists of

  r1

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          f#_A(x1,x2) = ((1,0),(0,0)) x2 + (2,4)
          a_A() = (2,3)
          f_A(x1,x2) = ((1,0),(0,0)) x1 + (2,1)
          b_A() = (5,2)
  
    precedence:
    
      a = f > f# = b
    
    partial status:
  
      pi(f#) = []
      pi(a) = []
      pi(f) = []
      pi(b) = []

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: f#(b(),f(a(),x)) -> f#(b(),f(b(),f(b(),x)))
p2: f#(b(),f(a(),x)) -> f#(b(),x)
p3: f#(b(),f(a(),x)) -> f#(b(),f(b(),x))

and R consists of:

r1: f(a(),f(b(),x)) -> f(a(),f(a(),f(a(),x)))
r2: f(b(),f(a(),x)) -> f(b(),f(b(),f(b(),x)))

The set of usable rules consists of

  r2

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          f#_A(x1,x2) = ((1,0),(1,0)) x2 + (6,3)
          b_A() = (0,1)
          f_A(x1,x2) = x1 + ((1,0),(1,0)) x2 + (2,2)
          a_A() = (3,0)
  
    precedence:
    
      f# = b = f = a
    
    partial status:
  
      pi(f#) = []
      pi(b) = []
      pi(f) = [1]
      pi(a) = []

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: f(a(),f(b(),x)) -> f(a(),f(a(),f(a(),x)))
r2: f(b(),f(a(),x)) -> f(b(),f(b(),f(b(),x)))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

r1: f(a(),f(b(),x)) -> f(a(),f(a(),f(a(),x)))
r2: f(b(),f(a(),x)) -> f(b(),f(b(),f(b(),x)))

The set of usable rules consists of

  r2

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          f#_A(x1,x2) = ((1,0),(0,0)) x2 + (2,4)
          b_A() = (1,1)
          f_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,3)
          a_A() = (7,2)
  
    precedence:
    
      f# = b = f = a
    
    partial status:
  
      pi(f#) = []
      pi(b) = []
      pi(f) = []
      pi(a) = []

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: f#(b(),f(a(),x)) -> f#(b(),x)

and R consists of:

r1: f(a(),f(b(),x)) -> f(a(),f(a(),f(a(),x)))
r2: f(b(),f(a(),x)) -> f(b(),f(b(),f(b(),x)))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: f#(b(),f(a(),x)) -> f#(b(),x)

and R consists of:

r1: f(a(),f(b(),x)) -> f(a(),f(a(),f(a(),x)))
r2: f(b(),f(a(),x)) -> f(b(),f(b(),f(b(),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:
          f#_A(x1,x2) = ((0,0),(1,0)) x2 + (2,4)
          b_A() = (1,2)
          f_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (1,2)
          a_A() = (3,1)
  
    precedence:
    
      f# = b = f = a
    
    partial status:
  
      pi(f#) = []
      pi(b) = []
      pi(f) = [1]
      pi(a) = []

The next rules are strictly ordered:

  p1

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