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.

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 contains 1 SCC with 8 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

AND2(xor2(x, y), z) -> AND2(y, z)
AND2(xor2(x, y), z) -> AND2(x, z)
Used argument filtering: AND2(x1, x2)  =  x1
xor2(x1, x2)  =  xor2(x1, x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ 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.