Termination w.r.t. Q proof of TRS/Rubiobn122.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:

plus(plus(X, Y), Z) → plus(X, plus(Y, Z))
times(X, s(Y)) → plus(X, times(Y, X))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

plus(plus(X, Y), Z) → plus(X, plus(Y, Z))
times(X, s(Y)) → plus(X, times(Y, 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:

PLUS(plus(X, Y), Z) → PLUS(X, plus(Y, Z))
PLUS(plus(X, Y), Z) → PLUS(Y, Z)
TIMES(X, s(Y)) → TIMES(Y, X)
TIMES(X, s(Y)) → PLUS(X, times(Y, X))

The TRS R consists of the following rules:

plus(plus(X, Y), Z) → plus(X, plus(Y, Z))
times(X, s(Y)) → plus(X, times(Y, X))

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:

PLUS(plus(X, Y), Z) → PLUS(X, plus(Y, Z))
PLUS(plus(X, Y), Z) → PLUS(Y, Z)
TIMES(X, s(Y)) → TIMES(Y, X)
TIMES(X, s(Y)) → PLUS(X, times(Y, X))

The TRS R consists of the following rules:

plus(plus(X, Y), Z) → plus(X, plus(Y, Z))
times(X, s(Y)) → plus(X, times(Y, X))

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:

PLUS(plus(X, Y), Z) → PLUS(X, plus(Y, Z))
PLUS(plus(X, Y), Z) → PLUS(Y, Z)
TIMES(X, s(Y)) → TIMES(Y, X)
TIMES(X, s(Y)) → PLUS(X, times(Y, X))

The TRS R consists of the following rules:

plus(plus(X, Y), Z) → plus(X, plus(Y, Z))
times(X, s(Y)) → plus(X, times(Y, 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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

PLUS(plus(X, Y), Z) → PLUS(X, plus(Y, Z))
PLUS(plus(X, Y), Z) → PLUS(Y, Z)

The TRS R consists of the following rules:

plus(plus(X, Y), Z) → plus(X, plus(Y, Z))
times(X, s(Y)) → plus(X, times(Y, 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.

PLUS(plus(X, Y), Z) → PLUS(X, plus(Y, Z))
PLUS(plus(X, Y), Z) → PLUS(Y, Z)

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

Recursive path order with status [2].
Quasi-Precedence:

plus₂ > PLUS₁

Status:

plus₂: [1,2]
PLUS₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

plus(plus(X, Y), Z) → plus(X, plus(Y, Z))
times(X, s(Y)) → plus(X, times(Y, 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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

TIMES(X, s(Y)) → TIMES(Y, X)

The TRS R consists of the following rules:

plus(plus(X, Y), Z) → plus(X, plus(Y, Z))
times(X, s(Y)) → plus(X, times(Y, 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.

TIMES(X, s(Y)) → TIMES(Y, X)

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

Recursive path order with status [2].
Quasi-Precedence:

[TIMES₂, s_1]

Status:

s₁: [1]
TIMES₂: multiset

The following usable rules [14] were oriented: none

↳ 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:

plus(plus(X, Y), Z) → plus(X, plus(Y, Z))
times(X, s(Y)) → plus(X, times(Y, 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.