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:

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

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

The TRS R consists of the following rules:

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

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 2 SCCs with 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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

AND(x, or(y, z)) → AND(x, z)
AND(x, or(y, z)) → AND(x, y)
AND(or(y, z), x) → AND(x, y)
AND(or(y, z), x) → AND(x, z)

The TRS R consists of the following rules:

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

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

AND(x, or(y, z)) → AND(x, z)
AND(x, or(y, z)) → AND(x, y)
AND(or(y, z), x) → AND(x, y)
AND(or(y, z), x) → AND(x, z)

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Quasi-Precedence:

trivial

Status:

AND₂: multiset

The following usable rules [FROCOS05] were oriented: none

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

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

NOT(or(x, y)) → NOT(y)
NOT(or(x, y)) → NOT(x)
NOT(and(x, y)) → NOT(x)
NOT(and(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
or(x1, x2)  =  or(x1, x2)
and(x1, x2)  =  and(x1, x2)

Recursive path order with status [RPO].
Quasi-Precedence:

trivial

Status:

trivial

The following usable rules [FROCOS05] were oriented: none

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

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE