Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/SK90SLASH2.31.trs

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(true) -> false
not₁(false) -> true
odd₁(0) -> false
odd₁(s₁(x)) -> not₁(odd₁(x))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
+₂(s₁(x), y) -> s₁(+₂(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

not₁(true) -> false
not₁(false) -> true
odd₁(0) -> false
odd₁(s₁(x)) -> not₁(odd₁(x))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
+₂(s₁(x), y) -> s₁(+₂(x, y))

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:

ODD₁(s₁(x)) -> NOT₁(odd₁(x))
+¹₂(x, s₁(y)) -> +¹₂(x, y)
+¹₂(s₁(x), y) -> +¹₂(x, y)
ODD₁(s₁(x)) -> ODD₁(x)

The TRS R consists of the following rules:

not₁(true) -> false
not₁(false) -> true
odd₁(0) -> false
odd₁(s₁(x)) -> not₁(odd₁(x))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
+₂(s₁(x), y) -> s₁(+₂(x, y))

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:

ODD₁(s₁(x)) -> NOT₁(odd₁(x))
+¹₂(x, s₁(y)) -> +¹₂(x, y)
+¹₂(s₁(x), y) -> +¹₂(x, y)
ODD₁(s₁(x)) -> ODD₁(x)

The TRS R consists of the following rules:

not₁(true) -> false
not₁(false) -> true
odd₁(0) -> false
odd₁(s₁(x)) -> not₁(odd₁(x))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
+₂(s₁(x), y) -> s₁(+₂(x, y))

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:

+¹₂(x, s₁(y)) -> +¹₂(x, y)
+¹₂(s₁(x), y) -> +¹₂(x, y)

The TRS R consists of the following rules:

not₁(true) -> false
not₁(false) -> true
odd₁(0) -> false
odd₁(s₁(x)) -> not₁(odd₁(x))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
+₂(s₁(x), y) -> s₁(+₂(x, y))

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.

+¹₂(x, s₁(y)) -> +¹₂(x, y)

The remaining pairs can at least by weakly be oriented.

+¹₂(s₁(x), y) -> +¹₂(x, y)

Used ordering: Combined order from the following AFS and order.
+¹₂(x1, x2) = +¹₁(x2)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

[+^1₁, s_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

+¹₂(s₁(x), y) -> +¹₂(x, y)

The TRS R consists of the following rules:

not₁(true) -> false
not₁(false) -> true
odd₁(0) -> false
odd₁(s₁(x)) -> not₁(odd₁(x))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
+₂(s₁(x), y) -> s₁(+₂(x, y))

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.

+¹₂(s₁(x), y) -> +¹₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
+¹₂(x1, x2) = +¹₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

[+^1₁, s_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

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

not₁(true) -> false
not₁(false) -> true
odd₁(0) -> false
odd₁(s₁(x)) -> not₁(odd₁(x))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
+₂(s₁(x), y) -> s₁(+₂(x, y))

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:

ODD₁(s₁(x)) -> ODD₁(x)

The TRS R consists of the following rules:

not₁(true) -> false
not₁(false) -> true
odd₁(0) -> false
odd₁(s₁(x)) -> not₁(odd₁(x))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
+₂(s₁(x), y) -> s₁(+₂(x, y))

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.

ODD₁(s₁(x)) -> ODD₁(x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ODD₁(x1) = ODD₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > ODD₁

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:

not₁(true) -> false
not₁(false) -> true
odd₁(0) -> false
odd₁(s₁(x)) -> not₁(odd₁(x))
+₂(x, 0) -> x
+₂(x, s₁(y)) -> s₁(+₂(x, y))
+₂(s₁(x), y) -> s₁(+₂(x, y))

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.