YES

We show the termination of the TRS R:

  *(x,*(minus(y),y)) -> *(minus(*(y,y)),x)

-- SCC decomposition.

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

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

and R consists of:

r1: *(x,*(minus(y),y)) -> *(minus(*(y,y)),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: *#(x,*(minus(y),y)) -> *#(minus(*(y,y)),x)
p2: *#(x,*(minus(y),y)) -> *#(y,y)

and R consists of:

r1: *(x,*(minus(y),y)) -> *(minus(*(y,y)),x)

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:
        
          *#_A(x1,x2) = max{6, x1 - 2, x2 - 3}
          *_A(x1,x2) = max{x1 + 12, x2 + 12}
          minus_A(x1) = max{1, x1 - 2}
    
      precedence:
      
        *# = * = minus
      
      partial status:
    
        pi(*#) = []
        pi(*) = []
        pi(minus) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          *#_A(x1,x2) = 14
          *_A(x1,x2) = 12
          minus_A(x1) = 13
    
      precedence:
      
        *# > minus > *
      
      partial status:
    
        pi(*#) = []
        pi(*) = []
        pi(minus) = []
    

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,*(minus(y),y)) -> *#(minus(*(y,y)),x)

and R consists of:

r1: *(x,*(minus(y),y)) -> *(minus(*(y,y)),x)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: *(x,*(minus(y),y)) -> *(minus(*(y,y)),x)

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:
        
          *#_A(x1,x2) = max{x1 + 5, x2 - 4}
          *_A(x1,x2) = max{x1 + 17, x2 + 16}
          minus_A(x1) = max{1, x1 - 11}
    
      precedence:
      
        *# = * = minus
      
      partial status:
    
        pi(*#) = []
        pi(*) = []
        pi(minus) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          *#_A(x1,x2) = 0
          *_A(x1,x2) = 8
          minus_A(x1) = 13
    
      precedence:
      
        *# = minus > *
      
      partial status:
    
        pi(*#) = []
        pi(*) = []
        pi(minus) = []
    

The next rules are strictly ordered:

  p1

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