Termination w.r.t. Q proof of TRS/AG01#3.18.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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(y)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(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:

DOUBLE(s(x)) → DOUBLE(x)
PLUS(s(x), y) → PLUS(x, y)
PLUS(s(x), y) → MINUS(x, y)
PLUS(s(x), y) → DOUBLE(y)
PLUS(s(x), y) → PLUS(x, s(y))
MINUS(s(x), s(y)) → MINUS(x, y)
PLUS(s(x), y) → PLUS(minus(x, y), double(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(y)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

DOUBLE(s(x)) → DOUBLE(x)
PLUS(s(x), y) → PLUS(x, y)
PLUS(s(x), y) → MINUS(x, y)
PLUS(s(x), y) → DOUBLE(y)
PLUS(s(x), y) → PLUS(x, s(y))
MINUS(s(x), s(y)) → MINUS(x, y)
PLUS(s(x), y) → PLUS(minus(x, y), double(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(y)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

DOUBLE(s(x)) → DOUBLE(x)
PLUS(s(x), y) → PLUS(x, y)
PLUS(s(x), y) → MINUS(x, y)
PLUS(s(x), y) → PLUS(x, s(y))
PLUS(s(x), y) → DOUBLE(y)
MINUS(s(x), s(y)) → MINUS(x, y)
PLUS(s(x), y) → PLUS(minus(x, y), double(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

DOUBLE(s(x)) → DOUBLE(x)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(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 oriented strictly and are deleted.

DOUBLE(s(x)) → DOUBLE(x)

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

Lexicographic path order with status [19].
Precedence:

s₁ > DOUBLE₁

Status:

s₁: multiset
DOUBLE₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(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 oriented strictly and are deleted.

MINUS(s(x), s(y)) → MINUS(x, y)

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

Lexicographic path order with status [19].
Precedence:

s₁ > MINUS₁

Status:

MINUS₁: [1]
s₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

PLUS(s(x), y) → PLUS(x, y)
PLUS(s(x), y) → PLUS(x, s(y))
PLUS(s(x), y) → PLUS(minus(x, y), double(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(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 oriented strictly and are deleted.

PLUS(s(x), y) → PLUS(x, y)
PLUS(s(x), y) → PLUS(x, s(y))
PLUS(s(x), y) → PLUS(minus(x, y), double(y))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
PLUS(x1, x2) = x1
s(x1) = s(x1)
minus(x1, x2) = x1
double(x1) = double
0 = 0

Lexicographic path order with status [19].
Precedence:

double > s₁

Status:

s₁: [1]
0: multiset
double: []

The following usable rules [14] were oriented:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
double(0) → 0
double(s(x)) → s(s(double(x)))
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
plus(s(x), y) → plus(x, s(y))
plus(s(x), y) → s(plus(minus(x, y), double(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.