Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/AG01SLASHHASH3.16.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:

times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(0, x) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(0, x) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(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:

PLUS₂(s₁(x), y) -> PLUS₂(x, y)
TIMES₂(x, s₁(y)) -> TIMES₂(x, y)
TIMES₂(x, s₁(y)) -> PLUS₂(times₂(x, y), x)
PLUS₂(x, s₁(y)) -> PLUS₂(x, y)

The TRS R consists of the following rules:

times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(0, x) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(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:

PLUS₂(s₁(x), y) -> PLUS₂(x, y)
TIMES₂(x, s₁(y)) -> TIMES₂(x, y)
TIMES₂(x, s₁(y)) -> PLUS₂(times₂(x, y), x)
PLUS₂(x, s₁(y)) -> PLUS₂(x, y)

The TRS R consists of the following rules:

times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(0, x) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(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:

PLUS₂(s₁(x), y) -> PLUS₂(x, y)
PLUS₂(x, s₁(y)) -> PLUS₂(x, y)

The TRS R consists of the following rules:

times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(0, x) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(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.

PLUS₂(s₁(x), y) -> PLUS₂(x, y)

The remaining pairs can at least by weakly be oriented.

PLUS₂(x, s₁(y)) -> PLUS₂(x, y)

Used ordering: Combined order from the following AFS and order.
PLUS₂(x1, x2) = PLUS₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

[PLUS₁, 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:

PLUS₂(x, s₁(y)) -> PLUS₂(x, y)

The TRS R consists of the following rules:

times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(0, x) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(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.

PLUS₂(x, s₁(y)) -> PLUS₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
PLUS₂(x1, x2) = PLUS₁(x2)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

[PLUS₁, 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:

times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(0, x) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(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:

TIMES₂(x, s₁(y)) -> TIMES₂(x, y)

The TRS R consists of the following rules:

times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(0, x) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(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.

TIMES₂(x, s₁(y)) -> TIMES₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
TIMES₂(x1, x2) = TIMES₁(x2)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

[TIMES₁, s_1]

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:

times₂(x, 0) -> 0
times₂(x, s₁(y)) -> plus₂(times₂(x, y), x)
plus₂(x, 0) -> x
plus₂(0, x) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
plus₂(s₁(x), y) -> s₁(plus₂(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.