YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: not#(and(x,y)) -> or#(not(x),not(y))
p2: not#(and(x,y)) -> not#(x)
p3: not#(and(x,y)) -> not#(y)
p4: not#(or(x,y)) -> and#(not(x),not(y))
p5: not#(or(x,y)) -> not#(x)
p6: not#(or(x,y)) -> not#(y)

and R consists of:

r1: or(x,x) -> x
r2: and(x,x) -> x
r3: not(not(x)) -> x
r4: not(and(x,y)) -> or(not(x),not(y))
r5: not(or(x,y)) -> and(not(x),not(y))

The estimated dependency graph contains the following SCCs:

  {p2, p3, p5, p6}


-- Reduction pair.

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

p1: not#(and(x,y)) -> not#(x)
p2: not#(or(x,y)) -> not#(y)
p3: not#(or(x,y)) -> not#(x)
p4: not#(and(x,y)) -> not#(y)

and R consists of:

r1: or(x,x) -> x
r2: and(x,x) -> x
r3: not(not(x)) -> x
r4: not(and(x,y)) -> or(not(x),not(y))
r5: not(or(x,y)) -> and(not(x),not(y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        and > not# > or
      
      argument filter:
    
        pi(not#) = [1]
        pi(and) = [1, 2]
        pi(or) = [1, 2]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        or > not# > and
      
      argument filter:
    
        pi(not#) = [1]
        pi(and) = 1
        pi(or) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3, p4

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