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: standard order
        interpretations:
          f#_A(x1,x2) = ((0,1),(0,0)) x2 + (5,7)
          a_A() = (1,1)
          f_A(x1,x2) = x1 + ((0,0),(0,1)) x2 + (2,1)
          b_A() = (2,5)
  
    precedence:
    
      f# = f > a = 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(),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(),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: standard order
        interpretations:
          f#_A(x1,x2) = ((0,1),(0,0)) x2 + (3,12)
          a_A() = (6,1)
          f_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(1,1)) x2 + (2,1)
          b_A() = (1,10)
  
    precedence:
    
      f# = a = f = b
    
    partial status:
  
      pi(f#) = []
      pi(a) = []
      pi(f) = []
      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(),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(),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: standard order
        interpretations:
          f#_A(x1,x2) = ((0,1),(0,0)) x2 + (3,2)
          a_A() = (2,3)
          f_A(x1,x2) = ((0,0),(0,1)) x2 + (1,1)
          b_A() = (4,3)
  
    precedence:
    
      f# = b > a > f
    
    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: standard order
        interpretations:
          f#_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (8,0)
          b_A() = (13,1)
          f_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,0),(0,0)) x2 + (1,11)
          a_A() = (7,10)
  
    precedence:
    
      f# > b = f = a
    
    partial status:
  
      pi(f#) = []
      pi(b) = []
      pi(f) = []
      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: standard order
        interpretations:
          f#_A(x1,x2) = ((1,0),(0,0)) x2 + (12,7)
          b_A() = (10,1)
          f_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,0),(1,0)) x2 + (1,8)
          a_A() = (11,6)
  
    precedence:
    
      f# > a > b > f
    
    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: standard order
        interpretations:
          f#_A(x1,x2) = ((0,1),(0,0)) x2 + (3,2)
          b_A() = (2,3)
          f_A(x1,x2) = ((0,0),(0,1)) x2 + (1,1)
          a_A() = (4,3)
  
    precedence:
    
      f# = a > b > f
    
    partial status:
  
      pi(f#) = []
      pi(b) = []
      pi(f) = []
      pi(a) = []

The next rules are strictly ordered:

  p1

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