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 [1,13] we result in the following initial DP problem:
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, z)
AND(x, or(y, z)) → AND(x, y)
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
      ↳ EdgeDeletionProof

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, z)
AND(x, or(y, z)) → AND(x, y)
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.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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, z)
AND(x, or(y, z)) → AND(x, y)
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.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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 [13].


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: Combined order from the following AFS and order.
OR(x1, x2)  =  OR(x2)
or(x1, x2)  =  or(x1)

Recursive Path Order [2].
Precedence:
or1 > OR1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


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.
none
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(x2)

Recursive Path Order [2].
Precedence:
or2 > AND1
and1 > AND1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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.