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:

not1(x) -> xor2(x, true)
or2(x, y) -> xor2(and2(x, y), xor2(x, y))
implies2(x, y) -> xor2(and2(x, y), xor2(x, true))
and2(x, true) -> x
and2(x, false) -> false
and2(x, x) -> x
xor2(x, false) -> x
xor2(x, x) -> false
and2(xor2(x, y), z) -> xor2(and2(x, z), and2(y, z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

not1(x) -> xor2(x, true)
or2(x, y) -> xor2(and2(x, y), xor2(x, y))
implies2(x, y) -> xor2(and2(x, y), xor2(x, true))
and2(x, true) -> x
and2(x, false) -> false
and2(x, x) -> x
xor2(x, false) -> x
xor2(x, x) -> false
and2(xor2(x, y), z) -> xor2(and2(x, z), and2(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:

IMPLIES2(x, y) -> XOR2(x, true)
IMPLIES2(x, y) -> AND2(x, y)
OR2(x, y) -> XOR2(x, y)
OR2(x, y) -> AND2(x, y)
AND2(xor2(x, y), z) -> AND2(y, z)
AND2(xor2(x, y), z) -> AND2(x, z)
NOT1(x) -> XOR2(x, true)
AND2(xor2(x, y), z) -> XOR2(and2(x, z), and2(y, z))
OR2(x, y) -> XOR2(and2(x, y), xor2(x, y))
IMPLIES2(x, y) -> XOR2(and2(x, y), xor2(x, true))

The TRS R consists of the following rules:

not1(x) -> xor2(x, true)
or2(x, y) -> xor2(and2(x, y), xor2(x, y))
implies2(x, y) -> xor2(and2(x, y), xor2(x, true))
and2(x, true) -> x
and2(x, false) -> false
and2(x, x) -> x
xor2(x, false) -> x
xor2(x, x) -> false
and2(xor2(x, y), z) -> xor2(and2(x, z), and2(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

IMPLIES2(x, y) -> XOR2(x, true)
IMPLIES2(x, y) -> AND2(x, y)
OR2(x, y) -> XOR2(x, y)
OR2(x, y) -> AND2(x, y)
AND2(xor2(x, y), z) -> AND2(y, z)
AND2(xor2(x, y), z) -> AND2(x, z)
NOT1(x) -> XOR2(x, true)
AND2(xor2(x, y), z) -> XOR2(and2(x, z), and2(y, z))
OR2(x, y) -> XOR2(and2(x, y), xor2(x, y))
IMPLIES2(x, y) -> XOR2(and2(x, y), xor2(x, true))

The TRS R consists of the following rules:

not1(x) -> xor2(x, true)
or2(x, y) -> xor2(and2(x, y), xor2(x, y))
implies2(x, y) -> xor2(and2(x, y), xor2(x, true))
and2(x, true) -> x
and2(x, false) -> false
and2(x, x) -> x
xor2(x, false) -> x
xor2(x, x) -> false
and2(xor2(x, y), z) -> xor2(and2(x, z), and2(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
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

AND2(xor2(x, y), z) -> AND2(y, z)
AND2(xor2(x, y), z) -> AND2(x, z)

The TRS R consists of the following rules:

not1(x) -> xor2(x, true)
or2(x, y) -> xor2(and2(x, y), xor2(x, y))
implies2(x, y) -> xor2(and2(x, y), xor2(x, true))
and2(x, true) -> x
and2(x, false) -> false
and2(x, x) -> x
xor2(x, false) -> x
xor2(x, x) -> false
and2(xor2(x, y), z) -> xor2(and2(x, z), and2(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.


AND2(xor2(x, y), z) -> AND2(y, z)
AND2(xor2(x, y), z) -> AND2(x, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( AND2(x1, x2) ) = x1 + 3x2 + 2


POL( xor2(x1, x2) ) = 2x1 + 2x2 + 3



The following usable rules [14] were oriented: none



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

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

not1(x) -> xor2(x, true)
or2(x, y) -> xor2(and2(x, y), xor2(x, y))
implies2(x, y) -> xor2(and2(x, y), xor2(x, true))
and2(x, true) -> x
and2(x, false) -> false
and2(x, x) -> x
xor2(x, false) -> x
xor2(x, x) -> false
and2(xor2(x, y), z) -> xor2(and2(x, z), and2(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.