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
      ↳ RuleRemovalProof

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.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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)


Used ordering: POLO with Polynomial interpretation [25]:

POL(NOT(x1)) = x1   
POL(and(x1, x2)) = 2 + x1 + x2   
POL(not(x1)) = x1   
POL(or(x1, x2)) = 2 + x1 + x2   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ RuleRemovalProof
        ↳ 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.