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:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      not#_A(x1) = x1
      and_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,0)) x2 + (1,1)
      or_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2 + (1,1)

The next rules are strictly ordered:

  p1, p2, p3, p4

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