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.

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 contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

AND2(x, or2(y, z)) -> AND2(x, z)
AND2(x, or2(y, z)) -> AND2(x, y)
Used argument filtering: AND2(x1, x2)  =  x2
or2(x1, x2)  =  or2(x1, x2)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ 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
            ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

NOT1(and2(x, y)) -> NOT1(x)
NOT1(and2(x, y)) -> NOT1(y)
NOT1(or2(x, y)) -> NOT1(x)
NOT1(or2(x, y)) -> NOT1(y)
Used argument filtering: NOT1(x1)  =  x1
and2(x1, x2)  =  and2(x1, x2)
or2(x1, x2)  =  or2(x1, x2)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ 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.