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:

and(x, or(y, z)) → or(and(x, y), and(x, z))
and(x, and(y, y)) → and(x, y)
or(or(x, y), and(y, z)) → or(x, y)
or(x, and(x, y)) → x
or(true, y) → true
or(x, false) → x
or(x, x) → x
or(x, or(y, y)) → or(x, y)
and(x, true) → x
and(false, y) → false
and(x, x) → x

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:

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

The TRS R consists of the following rules:

and(x, or(y, z)) → or(and(x, y), and(x, z))
and(x, and(y, y)) → and(x, y)
or(or(x, y), and(y, z)) → or(x, y)
or(x, and(x, y)) → x
or(true, y) → true
or(x, false) → x
or(x, x) → x
or(x, or(y, y)) → or(x, y)
and(x, true) → x
and(false, y) → false
and(x, x) → x

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:

OR(x, or(y, y)) → OR(x, y)

The TRS R consists of the following rules:

and(x, or(y, z)) → or(and(x, y), and(x, z))
and(x, and(y, y)) → and(x, y)
or(or(x, y), and(y, z)) → or(x, y)
or(x, and(x, y)) → x
or(true, y) → true
or(x, false) → x
or(x, x) → x
or(x, or(y, y)) → or(x, y)
and(x, true) → x
and(false, y) → false
and(x, x) → x

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.

OR(x, or(y, y)) → OR(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
OR(x1, x2)  =  OR(x1, x2)
or(x1, x2)  =  or(x1)
and(x1, x2)  =  and(x1, x2)
true  =  true
false  =  false

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

OR₂ > or₁
[and₂, false] > or₁
true > or₁

Status:

OR₂: [2,1]
or₁: multiset
and₂: [1,2]
true: multiset
false: multiset

The following usable rules [FROCOS05] were oriented:

and(x, or(y, z)) → or(and(x, y), and(x, z))
and(x, and(y, y)) → and(x, y)
or(or(x, y), and(y, z)) → or(x, y)
or(x, and(x, y)) → x
or(true, y) → true
or(x, false) → x
or(x, x) → x
or(x, or(y, y)) → or(x, y)
and(x, true) → x
and(false, y) → false
and(x, x) → x

(7) Obligation:

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

and(x, or(y, z)) → or(and(x, y), and(x, z))
and(x, and(y, y)) → and(x, y)
or(or(x, y), and(y, z)) → or(x, y)
or(x, and(x, y)) → x
or(true, y) → true
or(x, false) → x
or(x, x) → x
or(x, or(y, y)) → or(x, y)
and(x, true) → x
and(false, y) → false
and(x, x) → x

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:

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

The TRS R consists of the following rules:

and(x, or(y, z)) → or(and(x, y), and(x, z))
and(x, and(y, y)) → and(x, y)
or(or(x, y), and(y, z)) → or(x, y)
or(x, and(x, y)) → x
or(true, y) → true
or(x, false) → x
or(x, x) → x
or(x, or(y, y)) → or(x, y)
and(x, true) → x
and(false, y) → false
and(x, x) → x

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.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
AND(x1, x2)  =  AND(x2)
or(x1, x2)  =  or(x1, x2)
and(x1, x2)  =  and(x1, x2)
true  =  true
false  =  false

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

and₂ > [AND₁, or₂] > true
and₂ > false

Status:

AND₁: multiset
or₂: multiset
and₂: multiset
true: multiset
false: multiset

The following usable rules [FROCOS05] were oriented:

and(x, or(y, z)) → or(and(x, y), and(x, z))
and(x, and(y, y)) → and(x, y)
or(or(x, y), and(y, z)) → or(x, y)
or(x, and(x, y)) → x
or(true, y) → true
or(x, false) → x
or(x, x) → x
or(x, or(y, y)) → or(x, y)
and(x, true) → x
and(false, y) → false
and(x, x) → x

(12) Obligation:

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

and(x, or(y, z)) → or(and(x, y), and(x, z))
and(x, and(y, y)) → and(x, y)
or(or(x, y), and(y, z)) → or(x, y)
or(x, and(x, y)) → x
or(true, y) → true
or(x, false) → x
or(x, x) → x
or(x, or(y, y)) → or(x, y)
and(x, true) → x
and(false, y) → false
and(x, x) → x

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