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:

not(not(x)) → x
not(or(x, y)) → and(not(not(not(x))), not(not(not(y))))
not(and(x, y)) → or(not(not(not(x))), not(not(not(y))))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

not(not(x)) → x
not(or(x, y)) → and(not(not(not(x))), not(not(not(y))))
not(and(x, y)) → or(not(not(not(x))), not(not(not(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:

NOT(and(x, y)) → NOT(x)
NOT(and(x, y)) → NOT(not(x))
NOT(and(x, y)) → NOT(y)
NOT(or(x, y)) → NOT(x)
NOT(or(x, y)) → NOT(not(not(x)))
NOT(or(x, y)) → NOT(y)
NOT(and(x, y)) → NOT(not(not(y)))
NOT(or(x, y)) → NOT(not(x))
NOT(and(x, y)) → NOT(not(y))
NOT(or(x, y)) → NOT(not(y))
NOT(and(x, y)) → NOT(not(not(x)))
NOT(or(x, y)) → NOT(not(not(y)))

The TRS R consists of the following rules:

not(not(x)) → x
not(or(x, y)) → and(not(not(not(x))), not(not(not(y))))
not(and(x, y)) → or(not(not(not(x))), not(not(not(y))))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

NOT(and(x, y)) → NOT(x)
NOT(and(x, y)) → NOT(not(x))
NOT(and(x, y)) → NOT(y)
NOT(or(x, y)) → NOT(x)
NOT(or(x, y)) → NOT(not(not(x)))
NOT(or(x, y)) → NOT(y)
NOT(and(x, y)) → NOT(not(not(y)))
NOT(or(x, y)) → NOT(not(x))
NOT(and(x, y)) → NOT(not(y))
NOT(or(x, y)) → NOT(not(y))
NOT(and(x, y)) → NOT(not(not(x)))
NOT(or(x, y)) → NOT(not(not(y)))

The TRS R consists of the following rules:

not(not(x)) → x
not(or(x, y)) → and(not(not(not(x))), not(not(not(y))))
not(and(x, y)) → or(not(not(not(x))), not(not(not(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.


NOT(and(x, y)) → NOT(x)
NOT(and(x, y)) → NOT(not(x))
NOT(and(x, y)) → NOT(y)
NOT(or(x, y)) → NOT(x)
NOT(or(x, y)) → NOT(not(not(x)))
NOT(or(x, y)) → NOT(y)
NOT(and(x, y)) → NOT(not(not(y)))
NOT(or(x, y)) → NOT(not(x))
NOT(and(x, y)) → NOT(not(y))
NOT(or(x, y)) → NOT(not(y))
NOT(and(x, y)) → NOT(not(not(x)))
NOT(or(x, y)) → NOT(not(not(y)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
NOT(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
not(x1)  =  x1
or(x1, x2)  =  or(x1, x2)

Recursive Path Order [2].
Precedence:
[and2, or2]


The following usable rules [14] were oriented:

not(and(x, y)) → or(not(not(not(x))), not(not(not(y))))
not(not(x)) → x
not(or(x, y)) → and(not(not(not(x))), not(not(not(y))))



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

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

not(not(x)) → x
not(or(x, y)) → and(not(not(not(x))), not(not(not(y))))
not(and(x, y)) → or(not(not(not(x))), not(not(not(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.