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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        not#_A(x1) = x1
        or_A(x1,x2) = x1 + x2 + 1
        not_A(x1) = x1
        and_A(x1,x2) = x1 + x2 + 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        not > or > and > not#
      
      argument filter:
    
        pi(not#) = []
        pi(or) = []
        pi(not) = [1]
        pi(and) = []
    

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.