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

(0) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
NOT(x1)  =  x1
and(x1, x2)  =  and(x1, x2)
or(x1, x2)  =  or(x1, x2)
not(x1)  =  not(x1)

Recursive path order with status [RPO].
Precedence:

not₁ > and₂
not₁ > or₂

Status:

trivial

The following usable rules [FROCOS05] were oriented:

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

(6) Obligation:

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

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

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

(7) PisEmptyProof (EQUIVALENT transformation)

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

(8) TRUE