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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        and#_A(x1,x2,x3) = max{0, x1 - 9, x2 - 8, x3 - 4}
        not_A(x1) = max{10, x1 + 5}
        band_A(x1,x2) = max{7, x1 + 2, x2 + 2}
  
    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.