YES

We show the termination of the TRS R:

  and(not(not(x)),y,not(z)) -> and(y,band(x,z),x)

-- SCC decomposition.

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

p1: and#(not(not(x)),y,not(z)) -> and#(y,band(x,z),x)

and R consists of:

r1: and(not(not(x)),y,not(z)) -> and(y,band(x,z),x)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: and#(not(not(x)),y,not(z)) -> and#(y,band(x,z),x)

and R consists of:

r1: and(not(not(x)),y,not(z)) -> and(y,band(x,z),x)

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:
        
          and#_A(x1,x2,x3) = max{x1 - 1, x2 + 1, x3 - 1}
          not_A(x1) = x1 + 5
          band_A(x1,x2) = max{x1 - 1, x2 + 2}
    
      precedence:
      
        and# = not = band
      
      partial status:
    
        pi(and#) = []
        pi(not) = []
        pi(band) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          and#_A(x1,x2,x3) = 0
          not_A(x1) = max{5, x1}
          band_A(x1,x2) = 7
    
      precedence:
      
        and# = not = band
      
      partial status:
    
        pi(and#) = []
        pi(not) = [1]
        pi(band) = []
    

The next rules are strictly ordered:

  p1

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