Termination w.r.t. Q proof of TRS/SK902.31.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:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

not(true)
not(false)
odd(0)
odd(s(x0))
+(x0, 0)
+(x0, s(x1))
+(s(x0), x1)

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ODD(s(x)) → NOT(odd(x))
+¹(x, s(y)) → +¹(x, y)
+¹(s(x), y) → +¹(x, y)
ODD(s(x)) → ODD(x)

The TRS R consists of the following rules:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

not(true)
not(false)
odd(0)
odd(s(x0))
+(x0, 0)
+(x0, s(x1))
+(s(x0), x1)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

ODD(s(x)) → NOT(odd(x))
+¹(x, s(y)) → +¹(x, y)
+¹(s(x), y) → +¹(x, y)
ODD(s(x)) → ODD(x)

The TRS R consists of the following rules:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

not(true)
not(false)
odd(0)
odd(s(x0))
+(x0, 0)
+(x0, s(x1))
+(s(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

ODD(s(x)) → NOT(odd(x))
+¹(x, s(y)) → +¹(x, y)
+¹(s(x), y) → +¹(x, y)
ODD(s(x)) → ODD(x)

The TRS R consists of the following rules:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

not(true)
not(false)
odd(0)
odd(s(x0))
+(x0, 0)
+(x0, s(x1))
+(s(x0), x1)

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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

+¹(x, s(y)) → +¹(x, y)
+¹(s(x), y) → +¹(x, y)

The TRS R consists of the following rules:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

not(true)
not(false)
odd(0)
odd(s(x0))
+(x0, 0)
+(x0, s(x1))
+(s(x0), x1)

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.

+¹(x, s(y)) → +¹(x, y)

The remaining pairs can at least be oriented weakly.

+¹(s(x), y) → +¹(x, y)

Used ordering: Combined order from the following AFS and order.
+¹(x1, x2) = +¹(x2)
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > +^1₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                  ↳ QDP

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

+¹(s(x), y) → +¹(x, y)

The TRS R consists of the following rules:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

not(true)
not(false)
odd(0)
odd(s(x0))
+(x0, 0)
+(x0, s(x1))
+(s(x0), x1)

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.

+¹(s(x), y) → +¹(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
+¹(x1, x2) = x1
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

                  ↳ QDP

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

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

not(true)
not(false)
odd(0)
odd(s(x0))
+(x0, 0)
+(x0, s(x1))
+(s(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

ODD(s(x)) → ODD(x)

The TRS R consists of the following rules:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

not(true)
not(false)
odd(0)
odd(s(x0))
+(x0, 0)
+(x0, s(x1))
+(s(x0), x1)

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.

ODD(s(x)) → ODD(x)

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

Lexicographic Path Order [19].
Precedence:

[ODD₁, s_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

not(true)
not(false)
odd(0)
odd(s(x0))
+(x0, 0)
+(x0, s(x1))
+(s(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.