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 [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

NOT(or(x, y)) → NOT(not(not(y)))
NOT(and(x, y)) → NOT(not(not(x)))
NOT(and(x, y)) → NOT(y)
NOT(or(x, y)) → NOT(not(not(x)))
NOT(or(x, y)) → NOT(x)
NOT(and(x, y)) → NOT(not(not(y)))
NOT(or(x, y)) → NOT(not(y))
NOT(and(x, y)) → NOT(x)
NOT(or(x, y)) → NOT(not(x))
NOT(and(x, y)) → NOT(not(y))
NOT(and(x, y)) → NOT(not(x))
NOT(or(x, y)) → 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(or(x, y)) → NOT(not(not(y)))
NOT(and(x, y)) → NOT(not(not(x)))
NOT(and(x, y)) → NOT(y)
NOT(or(x, y)) → NOT(not(not(x)))
NOT(or(x, y)) → NOT(x)
NOT(and(x, y)) → NOT(not(not(y)))
NOT(or(x, y)) → NOT(not(y))
NOT(and(x, y)) → NOT(x)
NOT(or(x, y)) → NOT(not(x))
NOT(and(x, y)) → NOT(not(y))
NOT(and(x, y)) → NOT(not(x))
NOT(or(x, y)) → 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 [15].


The following pairs can be oriented strictly and are deleted.


NOT(or(x, y)) → NOT(not(not(y)))
NOT(and(x, y)) → NOT(not(not(x)))
NOT(and(x, y)) → NOT(y)
NOT(or(x, y)) → NOT(not(not(x)))
NOT(or(x, y)) → NOT(x)
NOT(and(x, y)) → NOT(not(not(y)))
NOT(or(x, y)) → NOT(not(y))
NOT(and(x, y)) → NOT(x)
NOT(or(x, y)) → NOT(not(x))
NOT(and(x, y)) → NOT(not(y))
NOT(and(x, y)) → NOT(not(x))
NOT(or(x, y)) → NOT(y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(NOT(x1)) = (1/4)x_1   
POL(not(x1)) = x_1   
POL(or(x1, x2)) = 1/4 + (4)x_1 + (2)x_2   
POL(and(x1, x2)) = 1/4 + (4)x_1 + (2)x_2   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

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))))



↳ 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.