YES

We show the termination of the TRS R:

  not(not(x)) -> x
  not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
  not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

-- SCC decomposition.

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

p1: not#(or(x,y)) -> not#(not(not(x)))
p2: not#(or(x,y)) -> not#(not(x))
p3: not#(or(x,y)) -> not#(x)
p4: not#(or(x,y)) -> not#(not(not(y)))
p5: not#(or(x,y)) -> not#(not(y))
p6: not#(or(x,y)) -> not#(y)
p7: not#(and(x,y)) -> not#(not(not(x)))
p8: not#(and(x,y)) -> not#(not(x))
p9: not#(and(x,y)) -> not#(x)
p10: not#(and(x,y)) -> not#(not(not(y)))
p11: not#(and(x,y)) -> not#(not(y))
p12: not#(and(x,y)) -> not#(y)

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12}


-- Reduction pair.

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

p1: not#(or(x,y)) -> not#(not(not(x)))
p2: not#(and(x,y)) -> not#(y)
p3: not#(and(x,y)) -> not#(not(y))
p4: not#(and(x,y)) -> not#(not(not(y)))
p5: not#(and(x,y)) -> not#(x)
p6: not#(and(x,y)) -> not#(not(x))
p7: not#(and(x,y)) -> not#(not(not(x)))
p8: not#(or(x,y)) -> not#(y)
p9: not#(or(x,y)) -> not#(not(y))
p10: not#(or(x,y)) -> not#(not(not(y)))
p11: not#(or(x,y)) -> not#(x)
p12: not#(or(x,y)) -> not#(not(x))

and R consists of:

r1: not(not(x)) -> x
r2: not(or(x,y)) -> and(not(not(not(x))),not(not(not(y))))
r3: not(and(x,y)) -> or(not(not(not(x))),not(not(not(y))))

The set of usable rules consists of

  r1, r2, r3

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      not#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(1,0,0,0)) x1
      or_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(0,1,0,0),(0,1,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(0,1,0,0),(0,0,1,0)) x2 + (3,0,0,1)
      not_A(x1) = ((1,0,0,0),(1,1,0,0),(0,1,0,0),(0,0,1,0)) x1 + (0,1,1,1)
      and_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(0,0,0,0),(0,1,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(1,1,0,0),(0,0,0,0)) x2 + (3,3,1,1)

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12

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