Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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.

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(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

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.