Termination w.r.t. Q proof of /tmp/tmpIAV-1X/4.21.trs

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:

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.

↳ QTRS

  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

AND(x, or(y, z)) → AND(x, y)
AND(x, or(y, z)) → OR(and(x, y), and(x, z))
AND(x, or(y, z)) → AND(x, z)
OR(x, or(y, y)) → OR(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.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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.
We use the reduction pair processor [15].

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.
none
Used ordering: Polynomial interpretation [25,35]:

POL(or(x₁, x₂)) = 1 + (4)x_1
POL(OR(x₁, x₂)) = (2)x_2

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

AND(x, or(y, z)) → AND(x, y)
AND(x, or(y, z)) → AND(x, z)
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.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(AND(x₁, x₂)) = (4)x_2
POL(or(x₁, x₂)) = 3 + x_1 + x_2
POL(and(x₁, x₂)) = 4 + (4)x_1 + (4)x_2

The value of delta used in the strict ordering is 12.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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