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

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ QDPAfsSolverProof

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

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

The TRS R consists of the following rules:

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

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(not1(x))
NOT1(and2(x, y)) -> NOT1(y)
NOT1(or2(x, y)) -> NOT1(x)
NOT1(or2(x, y)) -> NOT1(not1(not1(x)))
NOT1(or2(x, y)) -> NOT1(y)
NOT1(and2(x, y)) -> NOT1(not1(not1(y)))
NOT1(or2(x, y)) -> NOT1(not1(x))
NOT1(and2(x, y)) -> NOT1(not1(y))
NOT1(or2(x, y)) -> NOT1(not1(y))
NOT1(and2(x, y)) -> NOT1(not1(not1(x)))
NOT1(or2(x, y)) -> NOT1(not1(not1(y)))
Used argument filtering: NOT1(x1)  =  x1
and2(x1, x2)  =  and2(x1, x2)
not1(x1)  =  x1
or2(x1, x2)  =  or2(x1, x2)
Used ordering: Quasi Precedence: [and_2, or_2]


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
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(not1(not1(x))), not1(not1(not1(y))))
not1(and2(x, y)) -> or2(not1(not1(not1(x))), not1(not1(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.