Termination w.r.t. Q proof of TRS/nontermin/AG01#4.23.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:

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, plus₂(z, s₁(0)), s₁(z)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, plus₂(z, s₁(0)), s₁(z)))

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)
QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)
QUOT₃(x, 0, s₁(z)) -> PLUS₂(z, s₁(0))
QUOT₃(x, 0, s₁(z)) -> QUOT₃(x, plus₂(z, s₁(0)), s₁(z))

The TRS R consists of the following rules:

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, plus₂(z, s₁(0)), s₁(z)))

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)
QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)
QUOT₃(x, 0, s₁(z)) -> PLUS₂(z, s₁(0))
QUOT₃(x, 0, s₁(z)) -> QUOT₃(x, plus₂(z, s₁(0)), s₁(z))

The TRS R consists of the following rules:

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, plus₂(z, s₁(0)), s₁(z)))

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)

The TRS R consists of the following rules:

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, plus₂(z, s₁(0)), s₁(z)))

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.

PLUS₂(s₁(x), y) -> PLUS₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PLUS₂(x₁, x₂)) = x₁
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, plus₂(z, s₁(0)), s₁(z)))

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:

QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)
QUOT₃(x, 0, s₁(z)) -> QUOT₃(x, plus₂(z, s₁(0)), s₁(z))

The TRS R consists of the following rules:

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, plus₂(z, s₁(0)), s₁(z)))

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.

QUOT₃(s₁(x), s₁(y), z) -> QUOT₃(x, y, z)

The remaining pairs can at least be oriented weakly.

QUOT₃(x, 0, s₁(z)) -> QUOT₃(x, plus₂(z, s₁(0)), s₁(z))

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(QUOT₃(x₁, x₂, x₃)) = 3·x₁
POL(plus₂(x₁, x₂)) = 0
POL(s₁(x₁)) = 3 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

QUOT₃(x, 0, s₁(z)) -> QUOT₃(x, plus₂(z, s₁(0)), s₁(z))

The TRS R consists of the following rules:

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, plus₂(z, s₁(0)), s₁(z)))

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.

QUOT₃(x, 0, s₁(z)) -> QUOT₃(x, plus₂(z, s₁(0)), s₁(z))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 3
POL(QUOT₃(x₁, x₂, x₃)) = 3·x₂
POL(plus₂(x₁, x₂)) = 1 + 2·x₂
POL(s₁(x₁)) = 0

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

quot₃(0, s₁(y), s₁(z)) -> 0
quot₃(s₁(x), s₁(y), z) -> quot₃(x, y, z)
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
quot₃(x, 0, s₁(z)) -> s₁(quot₃(x, plus₂(z, s₁(0)), s₁(z)))

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.