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

The next rules are strictly ordered:

  p1

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