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:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      and#_A(x1,x2,x3) = x1 + ((1,0),(1,1)) x2 + ((1,0),(1,1)) x3
      not_A(x1) = ((1,0),(1,1)) x1 + (1,1)
      band_A(x1,x2) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (1,5)

The next rules are strictly ordered:

  p1

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