YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

r1: f(x,f(a(),y)) -> f(a(),f(f(a(),h(f(a(),x))),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(a(),h(f(a(),x))),y))
p2: f#(x,f(a(),y)) -> f#(a(),x)
p3: f#(x,f(a(),y)) -> f#(f(a(),h(f(a(),x))),y)

and R consists of:

r1: f(x,f(a(),y)) -> f(a(),f(f(a(),h(f(a(),x))),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: lexicographic order
          interpretations:
            f#_A(x1,x2) = x1 + x2 + (1,2)
            f_A(x1,x2) = x2 + (0,1)
            a_A() = (0,0)
            h_A(x1) = (0,0)
    
      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: lexicographic order
          interpretations:
            f#_A(x1,x2) = (3,3)
            f_A(x1,x2) = (0,0)
            a_A() = (2,2)
            h_A(x1) = (1,1)
    
      precedence:
      
        h > f# = a > f
      
      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(a(),h(f(a(),x))),y))
p2: f#(x,f(a(),y)) -> f#(f(a(),h(f(a(),x))),y)

and R consists of:

r1: f(x,f(a(),y)) -> f(a(),f(f(a(),h(f(a(),x))),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(a(),h(f(a(),x))),y))
p2: f#(x,f(a(),y)) -> f#(f(a(),h(f(a(),x))),y)

and R consists of:

r1: f(x,f(a(),y)) -> f(a(),f(f(a(),h(f(a(),x))),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 + 11, x2 + 25}
          f_A(x1,x2) = max{12, x1, x2 + 7}
          a_A = 0
          h_A(x1) = 1
    
      precedence:
      
        f# > f = a = h
      
      partial status:
    
        pi(f#) = [1]
        pi(f) = [1]
        pi(a) = []
        pi(h) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = 0
          f_A(x1,x2) = max{6, x1 - 4}
          a_A = 2
          h_A(x1) = 19
    
      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(a(),h(f(a(),x))),y))

and R consists of:

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

and R consists of:

r1: f(x,f(a(),y)) -> f(a(),f(f(a(),h(f(a(),x))),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: lexicographic order
          interpretations:
            f#_A(x1,x2) = x2
            f_A(x1,x2) = ((0,0),(1,0)) x1 + x2
            a_A() = (1,2)
            h_A(x1) = (0,0)
    
      precedence:
      
        f > f# > a = h
      
      partial status:
    
        pi(f#) = [2]
        pi(f) = []
        pi(a) = []
        pi(h) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1,x2) = x2
            f_A(x1,x2) = (1,2)
            a_A() = (2,3)
            h_A(x1) = (1,1)
    
      precedence:
      
        f# = f = a = h
      
      partial status:
    
        pi(f#) = [2]
        pi(f) = []
        pi(a) = []
        pi(h) = []
    

The next rules are strictly ordered:

  p1

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