Termination w.r.t. Q proof of TRS/TRCSREx4_7_15_Bor03

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:

a__f₁(0) -> cons₂(0, f₁(s₁(0)))
a__f₁(s₁(0)) -> a__f₁(a__p₁(s₁(0)))
a__p₁(s₁(0)) -> 0
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(0) -> 0
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__p₁(X) -> p₁(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__f₁(0) -> cons₂(0, f₁(s₁(0)))
a__f₁(s₁(0)) -> a__f₁(a__p₁(s₁(0)))
a__p₁(s₁(0)) -> 0
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(0) -> 0
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__p₁(X) -> p₁(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:

A__F₁(s₁(0)) -> A__P₁(s₁(0))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(f₁(X)) -> A__F₁(mark₁(X))
A__F₁(s₁(0)) -> A__F₁(a__p₁(s₁(0)))
MARK₁(p₁(X)) -> A__P₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(p₁(X)) -> MARK₁(X)
MARK₁(f₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__f₁(0) -> cons₂(0, f₁(s₁(0)))
a__f₁(s₁(0)) -> a__f₁(a__p₁(s₁(0)))
a__p₁(s₁(0)) -> 0
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(0) -> 0
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__p₁(X) -> p₁(X)

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:

A__F₁(s₁(0)) -> A__P₁(s₁(0))
MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(f₁(X)) -> A__F₁(mark₁(X))
A__F₁(s₁(0)) -> A__F₁(a__p₁(s₁(0)))
MARK₁(p₁(X)) -> A__P₁(mark₁(X))
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(p₁(X)) -> MARK₁(X)
MARK₁(f₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__f₁(0) -> cons₂(0, f₁(s₁(0)))
a__f₁(s₁(0)) -> a__f₁(a__p₁(s₁(0)))
a__p₁(s₁(0)) -> 0
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(0) -> 0
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__p₁(X) -> p₁(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 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

A__F₁(s₁(0)) -> A__F₁(a__p₁(s₁(0)))

The TRS R consists of the following rules:

a__f₁(0) -> cons₂(0, f₁(s₁(0)))
a__f₁(s₁(0)) -> a__f₁(a__p₁(s₁(0)))
a__p₁(s₁(0)) -> 0
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(0) -> 0
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__p₁(X) -> p₁(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.

A__F₁(s₁(0)) -> A__F₁(a__p₁(s₁(0)))

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

POL(0) = 0
POL(A__F₁(x₁)) = 2·x₁²
POL(a__p₁(x₁)) = 0
POL(p₁(x₁)) = 0
POL(s₁(x₁)) = 2

The following usable rules [14] were oriented:

a__p₁(X) -> p₁(X)
a__p₁(s₁(0)) -> 0

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

a__f₁(0) -> cons₂(0, f₁(s₁(0)))
a__f₁(s₁(0)) -> a__f₁(a__p₁(s₁(0)))
a__p₁(s₁(0)) -> 0
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(0) -> 0
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__p₁(X) -> p₁(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(cons₂(X1, X2)) -> MARK₁(X1)
MARK₁(f₁(X)) -> MARK₁(X)
MARK₁(p₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__f₁(0) -> cons₂(0, f₁(s₁(0)))
a__f₁(s₁(0)) -> a__f₁(a__p₁(s₁(0)))
a__p₁(s₁(0)) -> 0
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(0) -> 0
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__p₁(X) -> p₁(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.

MARK₁(s₁(X)) -> MARK₁(X)
MARK₁(f₁(X)) -> MARK₁(X)
MARK₁(p₁(X)) -> MARK₁(X)

The remaining pairs can at least be oriented weakly.

MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

Used ordering: Polynomial interpretation [21]:

POL(MARK₁(x₁)) = 2·x₁²
POL(cons₂(x₁, x₂)) = 2·x₁
POL(f₁(x₁)) = 2 + 2·x₁²
POL(p₁(x₁)) = 2 + 2·x₁²
POL(s₁(x₁)) = 2 + 2·x₁²

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:

MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The TRS R consists of the following rules:

a__f₁(0) -> cons₂(0, f₁(s₁(0)))
a__f₁(s₁(0)) -> a__f₁(a__p₁(s₁(0)))
a__p₁(s₁(0)) -> 0
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(0) -> 0
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__p₁(X) -> p₁(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.

MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

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

POL(MARK₁(x₁)) = 2·x₁²
POL(cons₂(x₁, x₂)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ 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:

a__f₁(0) -> cons₂(0, f₁(s₁(0)))
a__f₁(s₁(0)) -> a__f₁(a__p₁(s₁(0)))
a__p₁(s₁(0)) -> 0
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(p₁(X)) -> a__p₁(mark₁(X))
mark₁(0) -> 0
mark₁(cons₂(X1, X2)) -> cons₂(mark₁(X1), X2)
mark₁(s₁(X)) -> s₁(mark₁(X))
a__f₁(X) -> f₁(X)
a__p₁(X) -> p₁(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.