Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

not(not(x)) → x
not(or(x, y)) → and(not(x), not(y))
not(and(x, y)) → or(not(x), not(y))
and(x, or(y, z)) → or(and(x, y), and(x, z))
and(or(y, z), x) → or(and(x, y), and(x, z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

not(not(x)) → x
not(or(x, y)) → and(not(x), not(y))
not(and(x, y)) → or(not(x), not(y))
and(x, or(y, z)) → or(and(x, y), and(x, z))
and(or(y, z), x) → or(and(x, y), and(x, z))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

AND(x, or(y, z)) → AND(x, y)
NOT(and(x, y)) → NOT(y)
NOT(or(x, y)) → NOT(x)
AND(or(y, z), x) → AND(x, y)
NOT(and(x, y)) → NOT(x)
NOT(or(x, y)) → AND(not(x), not(y))
AND(or(y, z), x) → AND(x, z)
AND(x, or(y, z)) → AND(x, z)
NOT(or(x, y)) → NOT(y)

The TRS R consists of the following rules:

not(not(x)) → x
not(or(x, y)) → and(not(x), not(y))
not(and(x, y)) → or(not(x), not(y))
and(x, or(y, z)) → or(and(x, y), and(x, z))
and(or(y, z), x) → or(and(x, y), and(x, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

AND(x, or(y, z)) → AND(x, y)
NOT(and(x, y)) → NOT(y)
NOT(or(x, y)) → NOT(x)
AND(or(y, z), x) → AND(x, y)
NOT(and(x, y)) → NOT(x)
NOT(or(x, y)) → AND(not(x), not(y))
AND(or(y, z), x) → AND(x, z)
AND(x, or(y, z)) → AND(x, z)
NOT(or(x, y)) → NOT(y)

The TRS R consists of the following rules:

not(not(x)) → x
not(or(x, y)) → and(not(x), not(y))
not(and(x, y)) → or(not(x), not(y))
and(x, or(y, z)) → or(and(x, y), and(x, z))
and(or(y, z), x) → or(and(x, y), and(x, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

AND(x, or(y, z)) → AND(x, y)
AND(or(y, z), x) → AND(x, y)
AND(or(y, z), x) → AND(x, z)
AND(x, or(y, z)) → AND(x, z)

The TRS R consists of the following rules:

not(not(x)) → x
not(or(x, y)) → and(not(x), not(y))
not(and(x, y)) → or(not(x), not(y))
and(x, or(y, z)) → or(and(x, y), and(x, z))
and(or(y, z), x) → or(and(x, y), and(x, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QDPSizeChangeProof
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

AND(x, or(y, z)) → AND(x, y)
AND(or(y, z), x) → AND(x, y)
AND(or(y, z), x) → AND(x, z)
AND(x, or(y, z)) → AND(x, z)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof

Q DP problem:
The TRS P consists of the following rules:

NOT(and(x, y)) → NOT(y)
NOT(or(x, y)) → NOT(x)
NOT(and(x, y)) → NOT(x)
NOT(or(x, y)) → NOT(y)

The TRS R consists of the following rules:

not(not(x)) → x
not(or(x, y)) → and(not(x), not(y))
not(and(x, y)) → or(not(x), not(y))
and(x, or(y, z)) → or(and(x, y), and(x, z))
and(or(y, z), x) → or(and(x, y), and(x, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QDPSizeChangeProof

Q DP problem:
The TRS P consists of the following rules:

NOT(and(x, y)) → NOT(y)
NOT(or(x, y)) → NOT(x)
NOT(and(x, y)) → NOT(x)
NOT(or(x, y)) → NOT(y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: