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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

not1(and2(x, y)) -> or2(not1(x), not1(y))
not1(or2(x, y)) -> and2(not1(x), not1(y))
and2(x, or2(y, z)) -> 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)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(AND2(x1, x2)) = 2·x2   
POL(or2(x1, x2)) = 1 + x1 + 2·x2   

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(and2(x, y)) -> or2(not1(x), not1(y))
not1(or2(x, y)) -> and2(not1(x), not1(y))
and2(x, or2(y, z)) -> 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(and2(x, y)) -> or2(not1(x), not1(y))
not1(or2(x, y)) -> and2(not1(x), not1(y))
and2(x, or2(y, z)) -> 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)
The remaining pairs can at least be oriented weakly.

NOT1(or2(x, y)) -> NOT1(x)
NOT1(or2(x, y)) -> NOT1(y)
Used ordering: Polynomial interpretation [21]:

POL(NOT1(x1)) = x1   
POL(and2(x1, x2)) = 1 + 2·x1 + 3·x2   
POL(or2(x1, x2)) = 3·x1 + 3·x2   

The following usable rules [14] were oriented: none



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

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

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

The TRS R consists of the following rules:

not1(and2(x, y)) -> or2(not1(x), not1(y))
not1(or2(x, y)) -> and2(not1(x), not1(y))
and2(x, or2(y, z)) -> 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(or2(x, y)) -> NOT1(x)
NOT1(or2(x, y)) -> NOT1(y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(NOT1(x1)) = 2·x1   
POL(or2(x1, x2)) = 1 + 2·x1 + x2   

The following usable rules [14] were oriented: none



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

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

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