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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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:

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

The TRS R consists of the following rules:

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

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:

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

The TRS R consists of the following rules:

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

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:

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

The TRS R consists of the following rules:

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

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:

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

The TRS R consists of the following rules:

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

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)
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)

Lexicographic Path Order [19].
Precedence:
[AND1, or2]


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:

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

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:

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

The TRS R consists of the following rules:

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

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.


NOT(and(x, y)) → NOT(x)
NOT(and(x, y)) → NOT(y)
NOT(or(x, y)) → NOT(x)
NOT(or(x, y)) → NOT(y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
NOT(x1)  =  NOT(x1)
and(x1, x2)  =  and(x1, x2)
or(x1, x2)  =  or(x1, x2)

Lexicographic Path Order [19].
Precedence:
[NOT1, and2]


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:

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

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.