Termination w.r.t. Q proof of TRS/AG01#3.21.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, 1)) → plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) → x
plus(x, 0) → x
times(x, 0) → 0

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

times(x, plus(y, 1)) → plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) → x
plus(x, 0) → x
times(x, 0) → 0

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:

times(x, plus(y, 1)) → plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) → x
plus(x, 0) → x
times(x, 0) → 0

The set Q consists of the following terms:

times(x0, plus(x1, 1))
times(x0, 1)
plus(x0, 0)
times(x0, 0)

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, 1)) → TIMES(x, plus(y, times(1, 0)))
TIMES(x, plus(y, 1)) → TIMES(1, 0)
TIMES(x, plus(y, 1)) → PLUS(times(x, plus(y, times(1, 0))), x)
TIMES(x, plus(y, 1)) → PLUS(y, times(1, 0))

The TRS R consists of the following rules:

times(x, plus(y, 1)) → plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) → x
plus(x, 0) → x
times(x, 0) → 0

The set Q consists of the following terms:

times(x0, plus(x1, 1))
times(x0, 1)
plus(x0, 0)
times(x0, 0)

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:

TIMES(x, plus(y, 1)) → TIMES(x, plus(y, times(1, 0)))
TIMES(x, plus(y, 1)) → TIMES(1, 0)
TIMES(x, plus(y, 1)) → PLUS(times(x, plus(y, times(1, 0))), x)
TIMES(x, plus(y, 1)) → PLUS(y, times(1, 0))

The TRS R consists of the following rules:

times(x, plus(y, 1)) → plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) → x
plus(x, 0) → x
times(x, 0) → 0

The set Q consists of the following terms:

times(x0, plus(x1, 1))
times(x0, 1)
plus(x0, 0)
times(x0, 0)

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:

TIMES(x, plus(y, 1)) → TIMES(x, plus(y, times(1, 0)))
TIMES(x, plus(y, 1)) → TIMES(1, 0)
TIMES(x, plus(y, 1)) → PLUS(times(x, plus(y, times(1, 0))), x)
TIMES(x, plus(y, 1)) → PLUS(y, times(1, 0))

The TRS R consists of the following rules:

times(x, plus(y, 1)) → plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) → x
plus(x, 0) → x
times(x, 0) → 0

The set Q consists of the following terms:

times(x0, plus(x1, 1))
times(x0, 1)
plus(x0, 0)
times(x0, 0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

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

TIMES(x, plus(y, 1)) → TIMES(x, plus(y, times(1, 0)))

The TRS R consists of the following rules:

times(x, plus(y, 1)) → plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) → x
plus(x, 0) → x
times(x, 0) → 0

The set Q consists of the following terms:

times(x0, plus(x1, 1))
times(x0, 1)
plus(x0, 0)
times(x0, 0)

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, 1)) → TIMES(x, plus(y, times(1, 0)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
TIMES(x1, x2) = TIMES(x2)
plus(x1, x2) = plus(x1, x2)
1 = 1
times(x1, x2) = x2
0 = 0

Lexicographic Path Order [19].
Precedence:

1 > TIMES₁ > plus₂
1 > TIMES₁ > 0

The following usable rules [14] were oriented:

plus(x, 0) → x
times(x, 0) → 0

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ PisEmptyProof

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

times(x, plus(y, 1)) → plus(times(x, plus(y, times(1, 0))), x)
times(x, 1) → x
plus(x, 0) → x
times(x, 0) → 0

The set Q consists of the following terms:

times(x0, plus(x1, 1))
times(x0, 1)
plus(x0, 0)
times(x0, 0)

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