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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            and#_A(x1,x2,x3) = ((0,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (2,1)
            not_A(x1) = ((1,0),(0,0)) x1 + (3,1)
            band_A(x1,x2) = (1,0)
    
      precedence:
      
        not = band > and#
      
      partial status:
    
        pi(and#) = []
        pi(not) = []
        pi(band) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            and#_A(x1,x2,x3) = (0,0)
            not_A(x1) = (1,1)
            band_A(x1,x2) = (0,0)
    
      precedence:
      
        and# = not = band
      
      partial status:
    
        pi(and#) = []
        pi(not) = []
        pi(band) = []
    

The next rules are strictly ordered:

  p1

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