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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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))
NOT(and(x, y)) → OR(not(x), not(y))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

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

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

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)  =  x1
and(x1, x2)  =  and(x1, x2)
or(x1, x2)  =  or(x1, x2)

Recursive Path Order [2].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ PisEmptyProof

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

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

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.