YES

We show the termination of the TRS R:

  f(x,f(a(),y)) -> f(a(),f(f(f(a(),x),h(a())),y))

-- SCC decomposition.

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

p1: f#(x,f(a(),y)) -> f#(a(),f(f(f(a(),x),h(a())),y))
p2: f#(x,f(a(),y)) -> f#(f(f(a(),x),h(a())),y)
p3: f#(x,f(a(),y)) -> f#(f(a(),x),h(a()))
p4: f#(x,f(a(),y)) -> f#(a(),x)

and R consists of:

r1: f(x,f(a(),y)) -> f(a(),f(f(f(a(),x),h(a())),y))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p4}


-- Reduction pair.

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

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

and R consists of:

r1: f(x,f(a(),y)) -> f(a(),f(f(f(a(),x),h(a())),y))

The set of usable rules consists of

  r1

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            f#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (3,2)
            f_A(x1,x2) = ((1,0),(1,0)) x2 + (2,0)
            a_A() = (0,1)
            h_A(x1) = (0,2)
    
      precedence:
      
        f# = f = a = h
      
      partial status:
    
        pi(f#) = []
        pi(f) = []
        pi(a) = []
        pi(h) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            f#_A(x1,x2) = (0,0)
            f_A(x1,x2) = (1,1)
            a_A() = (2,2)
            h_A(x1) = (0,0)
    
      precedence:
      
        f# = f = a = h
      
      partial status:
    
        pi(f#) = []
        pi(f) = []
        pi(a) = []
        pi(h) = []
    

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#(x,f(a(),y)) -> f#(a(),f(f(f(a(),x),h(a())),y))
p2: f#(x,f(a(),y)) -> f#(f(f(a(),x),h(a())),y)

and R consists of:

r1: f(x,f(a(),y)) -> f(a(),f(f(f(a(),x),h(a())),y))

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#(x,f(a(),y)) -> f#(a(),f(f(f(a(),x),h(a())),y))
p2: f#(x,f(a(),y)) -> f#(f(f(a(),x),h(a())),y)

and R consists of:

r1: f(x,f(a(),y)) -> f(a(),f(f(f(a(),x),h(a())),y))

The set of usable rules consists of

  r1

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{x1 + 23, x2 + 83}
          f_A(x1,x2) = max{22, x2 + 15}
          a_A = 16
          h_A(x1) = max{59, x1 + 41}
    
      precedence:
      
        f# = f > a > h
      
      partial status:
    
        pi(f#) = []
        pi(f) = []
        pi(a) = []
        pi(h) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = 0
          f_A(x1,x2) = 13
          a_A = 10
          h_A(x1) = max{25, x1 + 15}
    
      precedence:
      
        f# = f = a = h
      
      partial status:
    
        pi(f#) = []
        pi(f) = []
        pi(a) = []
        pi(h) = []
    

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#(x,f(a(),y)) -> f#(a(),f(f(f(a(),x),h(a())),y))

and R consists of:

r1: f(x,f(a(),y)) -> f(a(),f(f(f(a(),x),h(a())),y))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: f#(x,f(a(),y)) -> f#(a(),f(f(f(a(),x),h(a())),y))

and R consists of:

r1: f(x,f(a(),y)) -> f(a(),f(f(f(a(),x),h(a())),y))

The set of usable rules consists of

  r1

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            f#_A(x1,x2) = x2 + (1,3)
            f_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,0),(0,0)) x2 + (5,0)
            a_A() = (4,2)
            h_A(x1) = ((1,0),(0,0)) x1 + (0,3)
    
      precedence:
      
        f# = f = a = h
      
      partial status:
    
        pi(f#) = []
        pi(f) = []
        pi(a) = []
        pi(h) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            f#_A(x1,x2) = ((1,1),(0,0)) x2 + (1,2)
            f_A(x1,x2) = (2,1)
            a_A() = (2,1)
            h_A(x1) = (1,0)
    
      precedence:
      
        a > f > f# = h
      
      partial status:
    
        pi(f#) = []
        pi(f) = []
        pi(a) = []
        pi(h) = []
    

The next rules are strictly ordered:

  p1

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