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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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)
NOT1(or2(x, y)) -> AND2(not1(x), not1(y))
NOT1(and2(x, y)) -> OR2(not1(x), not1(y))

The TRS R consists of the following rules:

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

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)
NOT1(or2(x, y)) -> AND2(not1(x), not1(y))
NOT1(and2(x, y)) -> OR2(not1(x), not1(y))

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ 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:

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

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 interpretation [21]:

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

The following usable rules [14] were oriented: none



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

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

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

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.