Termination w.r.t. Q proof of TRS/Cimeboolean

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:

xor(x, F) → x
xor(x, neg(x)) → F
and(x, T) → x
and(x, F) → F
and(x, x) → x
and(xor(x, y), z) → xor(and(x, z), and(y, z))
xor(x, x) → F
impl(x, y) → xor(and(x, y), xor(x, T))
or(x, y) → xor(and(x, y), xor(x, y))
equiv(x, y) → xor(x, xor(y, T))
neg(x) → xor(x, T)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

xor(x, F) → x
xor(x, neg(x)) → F
and(x, T) → x
and(x, F) → F
and(x, x) → x
and(xor(x, y), z) → xor(and(x, z), and(y, z))
xor(x, x) → F
impl(x, y) → xor(and(x, y), xor(x, T))
or(x, y) → xor(and(x, y), xor(x, y))
equiv(x, y) → xor(x, xor(y, T))
neg(x) → xor(x, T)

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:

IMPL(x, y) → XOR(and(x, y), xor(x, T))
OR(x, y) → XOR(x, y)
OR(x, y) → AND(x, y)
IMPL(x, y) → AND(x, y)
IMPL(x, y) → XOR(x, T)
AND(xor(x, y), z) → AND(y, z)
AND(xor(x, y), z) → AND(x, z)
EQUIV(x, y) → XOR(y, T)
OR(x, y) → XOR(and(x, y), xor(x, y))
AND(xor(x, y), z) → XOR(and(x, z), and(y, z))
NEG(x) → XOR(x, T)
EQUIV(x, y) → XOR(x, xor(y, T))

The TRS R consists of the following rules:

xor(x, F) → x
xor(x, neg(x)) → F
and(x, T) → x
and(x, F) → F
and(x, x) → x
and(xor(x, y), z) → xor(and(x, z), and(y, z))
xor(x, x) → F
impl(x, y) → xor(and(x, y), xor(x, T))
or(x, y) → xor(and(x, y), xor(x, y))
equiv(x, y) → xor(x, xor(y, T))
neg(x) → xor(x, T)

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:

IMPL(x, y) → XOR(and(x, y), xor(x, T))
OR(x, y) → XOR(x, y)
OR(x, y) → AND(x, y)
IMPL(x, y) → AND(x, y)
IMPL(x, y) → XOR(x, T)
AND(xor(x, y), z) → AND(y, z)
AND(xor(x, y), z) → AND(x, z)
EQUIV(x, y) → XOR(y, T)
OR(x, y) → XOR(and(x, y), xor(x, y))
AND(xor(x, y), z) → XOR(and(x, z), and(y, z))
NEG(x) → XOR(x, T)
EQUIV(x, y) → XOR(x, xor(y, T))

The TRS R consists of the following rules:

xor(x, F) → x
xor(x, neg(x)) → F
and(x, T) → x
and(x, F) → F
and(x, x) → x
and(xor(x, y), z) → xor(and(x, z), and(y, z))
xor(x, x) → F
impl(x, y) → xor(and(x, y), xor(x, T))
or(x, y) → xor(and(x, y), xor(x, y))
equiv(x, y) → xor(x, xor(y, T))
neg(x) → xor(x, T)

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:

IMPL(x, y) → XOR(and(x, y), xor(x, T))
AND(xor(x, y), z) → AND(y, z)
AND(xor(x, y), z) → AND(x, z)
AND(xor(x, y), z) → XOR(and(x, z), and(y, z))
NEG(x) → XOR(x, T)
OR(x, y) → AND(x, y)
OR(x, y) → XOR(x, y)
IMPL(x, y) → AND(x, y)
IMPL(x, y) → XOR(x, T)
EQUIV(x, y) → XOR(y, T)
OR(x, y) → XOR(and(x, y), xor(x, y))
EQUIV(x, y) → XOR(x, xor(y, T))

The TRS R consists of the following rules:

xor(x, F) → x
xor(x, neg(x)) → F
and(x, T) → x
and(x, F) → F
and(x, x) → x
and(xor(x, y), z) → xor(and(x, z), and(y, z))
xor(x, x) → F
impl(x, y) → xor(and(x, y), xor(x, T))
or(x, y) → xor(and(x, y), xor(x, y))
equiv(x, y) → xor(x, xor(y, T))
neg(x) → xor(x, T)

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 10 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

xor(x, F) → x
xor(x, neg(x)) → F
and(x, T) → x
and(x, F) → F
and(x, x) → x
and(xor(x, y), z) → xor(and(x, z), and(y, z))
xor(x, x) → F
impl(x, y) → xor(and(x, y), xor(x, T))
or(x, y) → xor(and(x, y), xor(x, y))
equiv(x, y) → xor(x, xor(y, T))
neg(x) → xor(x, T)

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
AND(x1, x2) = AND(x1)
xor(x1, x2) = xor(x1, x2)

Lexicographic path order with status [19].
Precedence:

xor₂ > AND₁

Status:

AND₁: [1]
xor₂: multiset

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:

xor(x, F) → x
xor(x, neg(x)) → F
and(x, T) → x
and(x, F) → F
and(x, x) → x
and(xor(x, y), z) → xor(and(x, z), and(y, z))
xor(x, x) → F
impl(x, y) → xor(and(x, y), xor(x, T))
or(x, y) → xor(and(x, y), xor(x, y))
equiv(x, y) → xor(x, xor(y, T))
neg(x) → xor(x, T)

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.