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(x) → xor(x, true)
or(x, y) → xor(and(x, y), xor(x, y))
implies(x, y) → xor(and(x, y), xor(x, true))
and(x, true) → x
and(x, false) → false
and(x, x) → x
xor(x, false) → x
xor(x, x) → false
and(xor(x, y), z) → xor(and(x, z), and(y, z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

not(x) → xor(x, true)
or(x, y) → xor(and(x, y), xor(x, y))
implies(x, y) → xor(and(x, y), xor(x, true))
and(x, true) → x
and(x, false) → false
and(x, x) → x
xor(x, false) → x
xor(x, x) → false
and(xor(x, y), z) → xor(and(x, z), and(y, 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:

IMPLIES(x, y) → XOR(x, true)
IMPLIES(x, y) → AND(x, y)
OR(x, y) → XOR(x, y)
OR(x, y) → AND(x, y)
AND(xor(x, y), z) → AND(y, z)
AND(xor(x, y), z) → AND(x, z)
NOT(x) → XOR(x, true)
AND(xor(x, y), z) → XOR(and(x, z), and(y, z))
OR(x, y) → XOR(and(x, y), xor(x, y))
IMPLIES(x, y) → XOR(and(x, y), xor(x, true))

The TRS R consists of the following rules:

not(x) → xor(x, true)
or(x, y) → xor(and(x, y), xor(x, y))
implies(x, y) → xor(and(x, y), xor(x, true))
and(x, true) → x
and(x, false) → false
and(x, x) → x
xor(x, false) → x
xor(x, x) → false
and(xor(x, y), z) → xor(and(x, z), and(y, 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:

IMPLIES(x, y) → XOR(x, true)
IMPLIES(x, y) → AND(x, y)
OR(x, y) → XOR(x, y)
OR(x, y) → AND(x, y)
AND(xor(x, y), z) → AND(y, z)
AND(xor(x, y), z) → AND(x, z)
NOT(x) → XOR(x, true)
AND(xor(x, y), z) → XOR(and(x, z), and(y, z))
OR(x, y) → XOR(and(x, y), xor(x, y))
IMPLIES(x, y) → XOR(and(x, y), xor(x, true))

The TRS R consists of the following rules:

not(x) → xor(x, true)
or(x, y) → xor(and(x, y), xor(x, y))
implies(x, y) → xor(and(x, y), xor(x, true))
and(x, true) → x
and(x, false) → false
and(x, x) → x
xor(x, false) → x
xor(x, x) → false
and(xor(x, y), z) → xor(and(x, z), and(y, 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:

OR(x, y) → AND(x, y)
OR(x, y) → XOR(x, y)
IMPLIES(x, y) → AND(x, y)
IMPLIES(x, y) → XOR(x, true)
AND(xor(x, y), z) → AND(y, z)
AND(xor(x, y), z) → AND(x, z)
NOT(x) → XOR(x, true)
OR(x, y) → XOR(and(x, y), xor(x, y))
AND(xor(x, y), z) → XOR(and(x, z), and(y, z))
IMPLIES(x, y) → XOR(and(x, y), xor(x, true))

The TRS R consists of the following rules:

not(x) → xor(x, true)
or(x, y) → xor(and(x, y), xor(x, y))
implies(x, y) → xor(and(x, y), xor(x, true))
and(x, true) → x
and(x, false) → false
and(x, x) → x
xor(x, false) → x
xor(x, x) → false
and(xor(x, y), z) → xor(and(x, z), and(y, z))

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 8 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:

not(x) → xor(x, true)
or(x, y) → xor(and(x, y), xor(x, y))
implies(x, y) → xor(and(x, y), xor(x, true))
and(x, true) → x
and(x, false) → false
and(x, x) → x
xor(x, false) → x
xor(x, x) → false
and(xor(x, y), z) → xor(and(x, z), and(y, 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(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)

Recursive Path Order [2].
Precedence:
xor2 > AND1

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:

not(x) → xor(x, true)
or(x, y) → xor(and(x, y), xor(x, y))
implies(x, y) → xor(and(x, y), xor(x, true))
and(x, true) → x
and(x, false) → false
and(x, x) → x
xor(x, false) → x
xor(x, x) → false
and(xor(x, y), z) → xor(and(x, z), and(y, 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.