Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 MRRProof (⇔)
↳4 QDP
↳5 PisEmptyProof (⇔)
↳6 TRUE

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, 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(x)))
NOT(or(x, y)) → NOT(not(x))
NOT(or(x, y)) → NOT(x)
NOT(or(x, y)) → NOT(not(not(y)))
NOT(or(x, y)) → NOT(not(y))
NOT(or(x, y)) → NOT(y)
NOT(and(x, y)) → NOT(not(not(x)))
NOT(and(x, y)) → NOT(not(x))
NOT(and(x, y)) → NOT(x)
NOT(and(x, y)) → NOT(not(not(y)))
NOT(and(x, y)) → NOT(not(y))
NOT(and(x, y)) → NOT(y)


Used ordering: Polynomial interpretation [POLO]:

POL(NOT(x1)) = x1   
POL(and(x1, x2)) = 1 + x1 + x2   
POL(not(x1)) = x1   
POL(or(x1, x2)) = 1 + x1 + x2   

(4) Obligation:

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.

(5) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(6) TRUE