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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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:

AND2(x, or2(y, z)) -> OR2(and2(x, y), and2(x, z))
AND2(x, or2(y, z)) -> AND2(x, z)
AND2(x, or2(y, z)) -> AND2(x, y)
OR2(x, or2(y, y)) -> OR2(x, y)
AND2(x, and2(y, y)) -> AND2(x, y)

The TRS R consists of the following rules:

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

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:

AND2(x, or2(y, z)) -> OR2(and2(x, y), and2(x, z))
AND2(x, or2(y, z)) -> AND2(x, z)
AND2(x, or2(y, z)) -> AND2(x, y)
OR2(x, or2(y, y)) -> OR2(x, y)
AND2(x, and2(y, y)) -> AND2(x, y)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

OR2(x, or2(y, y)) -> OR2(x, y)

The TRS R consists of the following rules:

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

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.


OR2(x, or2(y, y)) -> OR2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(OR2(x1, x2)) = 3·x1·x2 + 3·x2   
POL(or2(x1, x2)) = 1 + 3·x1   

The following usable rules [14] were oriented: none



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

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

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

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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

AND2(x, or2(y, z)) -> AND2(x, z)
AND2(x, or2(y, z)) -> AND2(x, y)
AND2(x, and2(y, y)) -> AND2(x, y)

The TRS R consists of the following rules:

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

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(x, or2(y, z)) -> AND2(x, z)
AND2(x, or2(y, z)) -> AND2(x, y)
AND2(x, and2(y, y)) -> AND2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(AND2(x1, x2)) = 3·x1·x2 + 3·x2   
POL(and2(x1, x2)) = 3 + 3·x1 + 3·x2   
POL(or2(x1, x2)) = 3 + x1 + x2   

The following usable rules [14] were oriented: none



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

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

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

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.