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:

not1(not1(x)) -> x
not1(or2(x, y)) -> and2(not1(x), not1(y))
not1(and2(x, y)) -> or2(not1(x), not1(y))
and2(x, or2(y, z)) -> or2(and2(x, y), and2(x, z))
and2(or2(y, z), x) -> or2(and2(x, y), and2(x, z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

not1(not1(x)) -> x
not1(or2(x, y)) -> and2(not1(x), not1(y))
not1(and2(x, y)) -> or2(not1(x), not1(y))
and2(x, or2(y, z)) -> or2(and2(x, y), and2(x, z))
and2(or2(y, z), x) -> or2(and2(x, y), and2(x, z))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

NOT1(and2(x, y)) -> NOT1(x)
NOT1(and2(x, y)) -> NOT1(y)
NOT1(or2(x, y)) -> NOT1(x)
NOT1(or2(x, y)) -> NOT1(y)
AND2(x, or2(y, z)) -> AND2(x, z)
NOT1(or2(x, y)) -> AND2(not1(x), not1(y))
AND2(x, or2(y, z)) -> AND2(x, y)
AND2(or2(y, z), x) -> AND2(x, z)
AND2(or2(y, z), x) -> AND2(x, y)

The TRS R consists of the following rules:

not1(not1(x)) -> x
not1(or2(x, y)) -> and2(not1(x), not1(y))
not1(and2(x, y)) -> or2(not1(x), not1(y))
and2(x, or2(y, z)) -> or2(and2(x, y), and2(x, z))
and2(or2(y, z), x) -> or2(and2(x, y), and2(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:

NOT1(and2(x, y)) -> NOT1(x)
NOT1(and2(x, y)) -> NOT1(y)
NOT1(or2(x, y)) -> NOT1(x)
NOT1(or2(x, y)) -> NOT1(y)
AND2(x, or2(y, z)) -> AND2(x, z)
NOT1(or2(x, y)) -> AND2(not1(x), not1(y))
AND2(x, or2(y, z)) -> AND2(x, y)
AND2(or2(y, z), x) -> AND2(x, z)
AND2(or2(y, z), x) -> AND2(x, y)

The TRS R consists of the following rules:

not1(not1(x)) -> x
not1(or2(x, y)) -> and2(not1(x), not1(y))
not1(and2(x, y)) -> or2(not1(x), not1(y))
and2(x, or2(y, z)) -> or2(and2(x, y), and2(x, z))
and2(or2(y, z), x) -> or2(and2(x, y), and2(x, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

AND2(x, or2(y, z)) -> AND2(x, z)
AND2(x, or2(y, z)) -> AND2(x, y)
AND2(or2(y, z), x) -> AND2(x, z)
AND2(or2(y, z), x) -> AND2(x, y)

The TRS R consists of the following rules:

not1(not1(x)) -> x
not1(or2(x, y)) -> and2(not1(x), not1(y))
not1(and2(x, y)) -> or2(not1(x), not1(y))
and2(x, or2(y, z)) -> or2(and2(x, y), and2(x, z))
and2(or2(y, z), x) -> or2(and2(x, y), and2(x, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


AND2(x, or2(y, z)) -> AND2(x, z)
AND2(x, or2(y, z)) -> AND2(x, y)
AND2(or2(y, z), x) -> AND2(x, z)
AND2(or2(y, z), x) -> AND2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( AND2(x1, x2) ) = x1 + x2 + 3


POL( or2(x1, x2) ) = 2x1 + x2 + 1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

not1(not1(x)) -> x
not1(or2(x, y)) -> and2(not1(x), not1(y))
not1(and2(x, y)) -> or2(not1(x), not1(y))
and2(x, or2(y, z)) -> or2(and2(x, y), and2(x, z))
and2(or2(y, z), x) -> or2(and2(x, y), and2(x, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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

NOT1(and2(x, y)) -> NOT1(x)
NOT1(and2(x, y)) -> NOT1(y)
NOT1(or2(x, y)) -> NOT1(x)
NOT1(or2(x, y)) -> NOT1(y)

The TRS R consists of the following rules:

not1(not1(x)) -> x
not1(or2(x, y)) -> and2(not1(x), not1(y))
not1(and2(x, y)) -> or2(not1(x), not1(y))
and2(x, or2(y, z)) -> or2(and2(x, y), and2(x, z))
and2(or2(y, z), x) -> or2(and2(x, y), and2(x, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


NOT1(and2(x, y)) -> NOT1(x)
NOT1(and2(x, y)) -> NOT1(y)
NOT1(or2(x, y)) -> NOT1(x)
NOT1(or2(x, y)) -> NOT1(y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( and2(x1, x2) ) = 3x1 + 3x2 + 1


POL( NOT1(x1) ) = 2x1 + 2


POL( or2(x1, x2) ) = x1 + 3x2 + 1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

not1(not1(x)) -> x
not1(or2(x, y)) -> and2(not1(x), not1(y))
not1(and2(x, y)) -> or2(not1(x), not1(y))
and2(x, or2(y, z)) -> or2(and2(x, y), and2(x, z))
and2(or2(y, z), x) -> or2(and2(x, y), and2(x, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.