YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

r1: *(x,+(y,z)) -> +(*(x,y),*(x,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{0, x2 - 2}
          +_A(x1,x2) = max{x1 + 3, x2 + 1}
    
      precedence:
      
        *# = +
      
      partial status:
    
        pi(*#) = []
        pi(+) = [2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          *#_A(x1,x2) = 0
          +_A(x1,x2) = x2
    
      precedence:
      
        *# = +
      
      partial status:
    
        pi(*#) = []
        pi(+) = [2]
    

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

and R consists of:

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

and R consists of:

r1: *(x,+(y,z)) -> +(*(x,y),*(x,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 + 1}
          +_A(x1,x2) = max{x1, x2}
    
      precedence:
      
        *# = +
      
      partial status:
    
        pi(*#) = [2]
        pi(+) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          *#_A(x1,x2) = x2 + 1
          +_A(x1,x2) = max{x1, x2}
    
      precedence:
      
        *# = +
      
      partial status:
    
        pi(*#) = [2]
        pi(+) = [1, 2]
    

The next rules are strictly ordered:

  p1

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