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)  =  x1
xor(x1, x2)  =  xor(x1, x2)

Lexicographic path order with status [19].
Quasi-Precedence:
trivial

Status:
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:

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.