YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: not#(and(x,y)) -> not#(x)
p2: not#(and(x,y)) -> not#(y)
p3: not#(or(x,y)) -> and#(not(x),not(y))
p4: not#(or(x,y)) -> not#(x)
p5: not#(or(x,y)) -> not#(y)
p6: and#(x,or(y,z)) -> and#(x,y)
p7: and#(x,or(y,z)) -> and#(x,z)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p4, p5}
  {p6, p7}


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

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

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

The next rules are strictly ordered:

  p2, p3
  r1, r2, r3

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

  (no rules)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

  (no rules)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      not#_A(x1) = x1
      and_A(x1,x2) = x1 + x2 + 1

The next rules are strictly ordered:

  p1, p2

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

-- Reduction pair.

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

p1: and#(x,or(y,z)) -> and#(x,y)
p2: and#(x,or(y,z)) -> and#(x,z)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      and#_A(x1,x2) = x2
      or_A(x1,x2) = x1 + x2 + 1

The next rules are strictly ordered:

  p1, p2

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