Termination w.r.t. Q proof of TRS/nontermin/AG01#4.33.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:

sum₂(cons₂(s₁(n), x), cons₂(m, y)) -> sum₂(cons₂(n, x), cons₂(s₁(m), y))
sum₂(cons₂(0, x), y) -> sum₂(x, y)
sum₂(nil, y) -> y
weight₁(cons₂(n, cons₂(m, x))) -> weight₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
weight₁(cons₂(n, nil)) -> n

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

sum₂(cons₂(s₁(n), x), cons₂(m, y)) -> sum₂(cons₂(n, x), cons₂(s₁(m), y))
sum₂(cons₂(0, x), y) -> sum₂(x, y)
sum₂(nil, y) -> y
weight₁(cons₂(n, cons₂(m, x))) -> weight₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
weight₁(cons₂(n, nil)) -> n

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:

WEIGHT₁(cons₂(n, cons₂(m, x))) -> SUM₂(cons₂(n, cons₂(m, x)), cons₂(0, x))
WEIGHT₁(cons₂(n, cons₂(m, x))) -> WEIGHT₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
SUM₂(cons₂(0, x), y) -> SUM₂(x, y)
SUM₂(cons₂(s₁(n), x), cons₂(m, y)) -> SUM₂(cons₂(n, x), cons₂(s₁(m), y))

The TRS R consists of the following rules:

sum₂(cons₂(s₁(n), x), cons₂(m, y)) -> sum₂(cons₂(n, x), cons₂(s₁(m), y))
sum₂(cons₂(0, x), y) -> sum₂(x, y)
sum₂(nil, y) -> y
weight₁(cons₂(n, cons₂(m, x))) -> weight₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
weight₁(cons₂(n, nil)) -> n

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:

WEIGHT₁(cons₂(n, cons₂(m, x))) -> SUM₂(cons₂(n, cons₂(m, x)), cons₂(0, x))
WEIGHT₁(cons₂(n, cons₂(m, x))) -> WEIGHT₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
SUM₂(cons₂(0, x), y) -> SUM₂(x, y)
SUM₂(cons₂(s₁(n), x), cons₂(m, y)) -> SUM₂(cons₂(n, x), cons₂(s₁(m), y))

The TRS R consists of the following rules:

sum₂(cons₂(s₁(n), x), cons₂(m, y)) -> sum₂(cons₂(n, x), cons₂(s₁(m), y))
sum₂(cons₂(0, x), y) -> sum₂(x, y)
sum₂(nil, y) -> y
weight₁(cons₂(n, cons₂(m, x))) -> weight₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
weight₁(cons₂(n, nil)) -> n

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:

SUM₂(cons₂(s₁(n), x), cons₂(m, y)) -> SUM₂(cons₂(n, x), cons₂(s₁(m), y))
SUM₂(cons₂(0, x), y) -> SUM₂(x, y)

The TRS R consists of the following rules:

sum₂(cons₂(s₁(n), x), cons₂(m, y)) -> sum₂(cons₂(n, x), cons₂(s₁(m), y))
sum₂(cons₂(0, x), y) -> sum₂(x, y)
sum₂(nil, y) -> y
weight₁(cons₂(n, cons₂(m, x))) -> weight₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
weight₁(cons₂(n, nil)) -> n

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.

SUM₂(cons₂(0, x), y) -> SUM₂(x, y)

The remaining pairs can at least be oriented weakly.

SUM₂(cons₂(s₁(n), x), cons₂(m, y)) -> SUM₂(cons₂(n, x), cons₂(s₁(m), y))

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(SUM₂(x₁, x₂)) = x₁
POL(cons₂(x₁, x₂)) = 1 + x₂
POL(s₁(x₁)) = 0

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:

SUM₂(cons₂(s₁(n), x), cons₂(m, y)) -> SUM₂(cons₂(n, x), cons₂(s₁(m), y))

The TRS R consists of the following rules:

sum₂(cons₂(s₁(n), x), cons₂(m, y)) -> sum₂(cons₂(n, x), cons₂(s₁(m), y))
sum₂(cons₂(0, x), y) -> sum₂(x, y)
sum₂(nil, y) -> y
weight₁(cons₂(n, cons₂(m, x))) -> weight₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
weight₁(cons₂(n, nil)) -> n

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.

SUM₂(cons₂(s₁(n), x), cons₂(m, y)) -> SUM₂(cons₂(n, x), cons₂(s₁(m), y))

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

POL(SUM₂(x₁, x₂)) = x₁
POL(cons₂(x₁, x₂)) = x₁
POL(s₁(x₁)) = 1 + x₁

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:

sum₂(cons₂(s₁(n), x), cons₂(m, y)) -> sum₂(cons₂(n, x), cons₂(s₁(m), y))
sum₂(cons₂(0, x), y) -> sum₂(x, y)
sum₂(nil, y) -> y
weight₁(cons₂(n, cons₂(m, x))) -> weight₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
weight₁(cons₂(n, nil)) -> n

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:

WEIGHT₁(cons₂(n, cons₂(m, x))) -> WEIGHT₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))

The TRS R consists of the following rules:

sum₂(cons₂(s₁(n), x), cons₂(m, y)) -> sum₂(cons₂(n, x), cons₂(s₁(m), y))
sum₂(cons₂(0, x), y) -> sum₂(x, y)
sum₂(nil, y) -> y
weight₁(cons₂(n, cons₂(m, x))) -> weight₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
weight₁(cons₂(n, nil)) -> n

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.

WEIGHT₁(cons₂(n, cons₂(m, x))) -> WEIGHT₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))

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

POL(0) = 0
POL(WEIGHT₁(x₁)) = x₁
POL(cons₂(x₁, x₂)) = 1 + x₂
POL(nil) = 0
POL(s₁(x₁)) = 0
POL(sum₂(x₁, x₂)) = x₂

The following usable rules [14] were oriented:

sum₂(nil, y) -> y
sum₂(cons₂(0, x), y) -> sum₂(x, y)
sum₂(cons₂(s₁(n), x), cons₂(m, y)) -> sum₂(cons₂(n, x), cons₂(s₁(m), y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

sum₂(cons₂(s₁(n), x), cons₂(m, y)) -> sum₂(cons₂(n, x), cons₂(s₁(m), y))
sum₂(cons₂(0, x), y) -> sum₂(x, y)
sum₂(nil, y) -> y
weight₁(cons₂(n, cons₂(m, x))) -> weight₁(sum₂(cons₂(n, cons₂(m, x)), cons₂(0, x)))
weight₁(cons₂(n, nil)) -> n

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.