Termination w.r.t. Q proof of TRS/AG01#3.22.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:

times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(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:

TIMES(x, plus(y, s(z))) → TIMES(x, s(z))
TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))
TIMES(x, plus(y, s(z))) → TIMES(s(z), 0)
TIMES(x, s(y)) → TIMES(x, y)
TIMES(x, s(y)) → PLUS(times(x, y), x)
TIMES(x, plus(y, s(z))) → PLUS(y, times(s(z), 0))
PLUS(x, s(y)) → PLUS(x, y)
TIMES(x, plus(y, s(z))) → PLUS(times(x, plus(y, times(s(z), 0))), times(x, s(z)))

The TRS R consists of the following rules:

times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, 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:

TIMES(x, plus(y, s(z))) → TIMES(x, s(z))
TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))
TIMES(x, plus(y, s(z))) → TIMES(s(z), 0)
TIMES(x, s(y)) → TIMES(x, y)
TIMES(x, s(y)) → PLUS(times(x, y), x)
TIMES(x, plus(y, s(z))) → PLUS(y, times(s(z), 0))
PLUS(x, s(y)) → PLUS(x, y)
TIMES(x, plus(y, s(z))) → PLUS(times(x, plus(y, times(s(z), 0))), times(x, s(z)))

The TRS R consists of the following rules:

times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, 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:

TIMES(x, plus(y, s(z))) → TIMES(x, s(z))
TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))
TIMES(x, plus(y, s(z))) → TIMES(s(z), 0)
TIMES(x, s(y)) → TIMES(x, y)
TIMES(x, s(y)) → PLUS(times(x, y), x)
TIMES(x, plus(y, s(z))) → PLUS(y, times(s(z), 0))
PLUS(x, s(y)) → PLUS(x, y)
TIMES(x, plus(y, s(z))) → PLUS(times(x, plus(y, times(s(z), 0))), times(x, s(z)))

The TRS R consists of the following rules:

times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

PLUS(x, s(y)) → PLUS(x, y)

The TRS R consists of the following rules:

times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(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 oriented strictly and are deleted.

PLUS(x, s(y)) → PLUS(x, y)

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

Recursive Path Order [2].
Precedence:

[PLUS₁, s_1]

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:

times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

TIMES(x, plus(y, s(z))) → TIMES(x, s(z))
TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))
TIMES(x, s(y)) → TIMES(x, y)

The TRS R consists of the following rules:

times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(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 oriented strictly and are deleted.

TIMES(x, plus(y, s(z))) → TIMES(x, s(z))
TIMES(x, s(y)) → TIMES(x, y)

The remaining pairs can at least be oriented weakly.

TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))

Used ordering: Combined order from the following AFS and order.
TIMES(x1, x2) = TIMES(x2)
plus(x1, x2) = plus(x1, x2)
s(x1) = s(x1)
times(x1, x2) = x1
0 = 0

Recursive Path Order [2].
Precedence:

plus₂ > s₁ > TIMES₁ > 0

The following usable rules [14] were oriented:

plus(x, s(y)) → s(plus(x, y))
times(x, 0) → 0
plus(x, 0) → x

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

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

TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))

The TRS R consists of the following rules:

times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(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 oriented strictly and are deleted.

TIMES(x, plus(y, s(z))) → TIMES(x, plus(y, times(s(z), 0)))

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

Recursive Path Order [2].
Precedence:

[plus₂, s] > 0

The following usable rules [14] were oriented:

plus(x, s(y)) → s(plus(x, y))
times(x, 0) → 0
plus(x, 0) → x

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

times(x, plus(y, s(z))) → plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) → 0
times(x, s(y)) → plus(times(x, y), x)
plus(x, 0) → x
plus(x, s(y)) → s(plus(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.