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:

xor2(x, F) -> x
xor2(x, neg1(x)) -> F
and2(x, T) -> x
and2(x, F) -> F
and2(x, x) -> x
and2(xor2(x, y), z) -> xor2(and2(x, z), and2(y, z))
xor2(x, x) -> F
impl2(x, y) -> xor2(and2(x, y), xor2(x, T))
or2(x, y) -> xor2(and2(x, y), xor2(x, y))
equiv2(x, y) -> xor2(x, xor2(y, T))
neg1(x) -> xor2(x, T)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

xor2(x, F) -> x
xor2(x, neg1(x)) -> F
and2(x, T) -> x
and2(x, F) -> F
and2(x, x) -> x
and2(xor2(x, y), z) -> xor2(and2(x, z), and2(y, z))
xor2(x, x) -> F
impl2(x, y) -> xor2(and2(x, y), xor2(x, T))
or2(x, y) -> xor2(and2(x, y), xor2(x, y))
equiv2(x, y) -> xor2(x, xor2(y, T))
neg1(x) -> xor2(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:

IMPL2(x, y) -> XOR2(and2(x, y), xor2(x, T))
OR2(x, y) -> XOR2(x, y)
OR2(x, y) -> AND2(x, y)
IMPL2(x, y) -> AND2(x, y)
IMPL2(x, y) -> XOR2(x, T)
AND2(xor2(x, y), z) -> AND2(y, z)
AND2(xor2(x, y), z) -> AND2(x, z)
EQUIV2(x, y) -> XOR2(y, T)
OR2(x, y) -> XOR2(and2(x, y), xor2(x, y))
AND2(xor2(x, y), z) -> XOR2(and2(x, z), and2(y, z))
NEG1(x) -> XOR2(x, T)
EQUIV2(x, y) -> XOR2(x, xor2(y, T))

The TRS R consists of the following rules:

xor2(x, F) -> x
xor2(x, neg1(x)) -> F
and2(x, T) -> x
and2(x, F) -> F
and2(x, x) -> x
and2(xor2(x, y), z) -> xor2(and2(x, z), and2(y, z))
xor2(x, x) -> F
impl2(x, y) -> xor2(and2(x, y), xor2(x, T))
or2(x, y) -> xor2(and2(x, y), xor2(x, y))
equiv2(x, y) -> xor2(x, xor2(y, T))
neg1(x) -> xor2(x, T)

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:

IMPL2(x, y) -> XOR2(and2(x, y), xor2(x, T))
OR2(x, y) -> XOR2(x, y)
OR2(x, y) -> AND2(x, y)
IMPL2(x, y) -> AND2(x, y)
IMPL2(x, y) -> XOR2(x, T)
AND2(xor2(x, y), z) -> AND2(y, z)
AND2(xor2(x, y), z) -> AND2(x, z)
EQUIV2(x, y) -> XOR2(y, T)
OR2(x, y) -> XOR2(and2(x, y), xor2(x, y))
AND2(xor2(x, y), z) -> XOR2(and2(x, z), and2(y, z))
NEG1(x) -> XOR2(x, T)
EQUIV2(x, y) -> XOR2(x, xor2(y, T))

The TRS R consists of the following rules:

xor2(x, F) -> x
xor2(x, neg1(x)) -> F
and2(x, T) -> x
and2(x, F) -> F
and2(x, x) -> x
and2(xor2(x, y), z) -> xor2(and2(x, z), and2(y, z))
xor2(x, x) -> F
impl2(x, y) -> xor2(and2(x, y), xor2(x, T))
or2(x, y) -> xor2(and2(x, y), xor2(x, y))
equiv2(x, y) -> xor2(x, xor2(y, T))
neg1(x) -> xor2(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
      ↳ 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:

xor2(x, F) -> x
xor2(x, neg1(x)) -> F
and2(x, T) -> x
and2(x, F) -> F
and2(x, x) -> x
and2(xor2(x, y), z) -> xor2(and2(x, z), and2(y, z))
xor2(x, x) -> F
impl2(x, y) -> xor2(and2(x, y), xor2(x, T))
or2(x, y) -> xor2(and2(x, y), xor2(x, y))
equiv2(x, y) -> xor2(x, xor2(y, T))
neg1(x) -> xor2(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 strictly oriented 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 by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
AND2(x1, x2)  =  AND1(x1)
xor2(x1, x2)  =  xor2(x1, x2)

Lexicographic Path Order [19].
Precedence:
xor2 > AND1


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:

xor2(x, F) -> x
xor2(x, neg1(x)) -> F
and2(x, T) -> x
and2(x, F) -> F
and2(x, x) -> x
and2(xor2(x, y), z) -> xor2(and2(x, z), and2(y, z))
xor2(x, x) -> F
impl2(x, y) -> xor2(and2(x, y), xor2(x, T))
or2(x, y) -> xor2(and2(x, y), xor2(x, y))
equiv2(x, y) -> xor2(x, xor2(y, T))
neg1(x) -> xor2(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.