YES

We show the termination of the TRS R:

  +(*(x,y),*(a(),y)) -> *(+(x,a()),y)
  *(*(x,y),z) -> *(x,*(y,z))

-- SCC decomposition.

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

p1: +#(*(x,y),*(a(),y)) -> *#(+(x,a()),y)
p2: +#(*(x,y),*(a(),y)) -> +#(x,a())
p3: *#(*(x,y),z) -> *#(x,*(y,z))
p4: *#(*(x,y),z) -> *#(y,z)

and R consists of:

r1: +(*(x,y),*(a(),y)) -> *(+(x,a()),y)
r2: *(*(x,y),z) -> *(x,*(y,z))

The estimated dependency graph contains the following SCCs:

  {p3, p4}


-- Reduction pair.

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

p1: *#(*(x,y),z) -> *#(x,*(y,z))
p2: *#(*(x,y),z) -> *#(y,z)

and R consists of:

r1: +(*(x,y),*(a(),y)) -> *(+(x,a()),y)
r2: *(*(x,y),z) -> *(x,*(y,z))

The set of usable rules consists of

  r2

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          *#_A(x1,x2) = x1 + 4
          *_A(x1,x2) = max{x1 + 3, x2}
    
      precedence:
      
        *# = *
      
      partial status:
    
        pi(*#) = []
        pi(*) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          *#_A(x1,x2) = 0
          *_A(x1,x2) = 2
    
      precedence:
      
        *# = *
      
      partial status:
    
        pi(*#) = []
        pi(*) = []
    

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: *#(*(x,y),z) -> *#(y,z)

and R consists of:

r1: +(*(x,y),*(a(),y)) -> *(+(x,a()),y)
r2: *(*(x,y),z) -> *(x,*(y,z))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: *#(*(x,y),z) -> *#(y,z)

and R consists of:

r1: +(*(x,y),*(a(),y)) -> *(+(x,a()),y)
r2: *(*(x,y),z) -> *(x,*(y,z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          *#_A(x1,x2) = max{x1 + 1, x2}
          *_A(x1,x2) = max{x1 - 1, x2}
    
      precedence:
      
        *# = *
      
      partial status:
    
        pi(*#) = [1, 2]
        pi(*) = [2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          *#_A(x1,x2) = max{x1 - 1, x2 + 1}
          *_A(x1,x2) = x2
    
      precedence:
      
        *# = *
      
      partial status:
    
        pi(*#) = []
        pi(*) = [2]
    

The next rules are strictly ordered:

  p1

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