Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/AProVESLASHLPAR

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:

intlist₁(nil) -> nil
int₂(s₁(x), 0) -> nil
int₂(x, x) -> cons₂(x, nil)
intlist₁(cons₂(x, y)) -> cons₂(s₁(x), intlist₁(y))
int₂(s₁(x), s₁(y)) -> intlist₁(int₂(x, y))
int₂(0, s₁(y)) -> cons₂(0, int₂(s₁(0), s₁(y)))
intlist₁(cons₂(x, nil)) -> cons₂(s₁(x), nil)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

intlist₁(nil) -> nil
int₂(s₁(x), 0) -> nil
int₂(x, x) -> cons₂(x, nil)
intlist₁(cons₂(x, y)) -> cons₂(s₁(x), intlist₁(y))
int₂(s₁(x), s₁(y)) -> intlist₁(int₂(x, y))
int₂(0, s₁(y)) -> cons₂(0, int₂(s₁(0), s₁(y)))
intlist₁(cons₂(x, nil)) -> cons₂(s₁(x), nil)

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:

INT₂(s₁(x), s₁(y)) -> INT₂(x, y)
INT₂(s₁(x), s₁(y)) -> INTLIST₁(int₂(x, y))
INTLIST₁(cons₂(x, y)) -> INTLIST₁(y)
INT₂(0, s₁(y)) -> INT₂(s₁(0), s₁(y))

The TRS R consists of the following rules:

intlist₁(nil) -> nil
int₂(s₁(x), 0) -> nil
int₂(x, x) -> cons₂(x, nil)
intlist₁(cons₂(x, y)) -> cons₂(s₁(x), intlist₁(y))
int₂(s₁(x), s₁(y)) -> intlist₁(int₂(x, y))
int₂(0, s₁(y)) -> cons₂(0, int₂(s₁(0), s₁(y)))
intlist₁(cons₂(x, nil)) -> cons₂(s₁(x), nil)

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:

INT₂(s₁(x), s₁(y)) -> INT₂(x, y)
INT₂(s₁(x), s₁(y)) -> INTLIST₁(int₂(x, y))
INTLIST₁(cons₂(x, y)) -> INTLIST₁(y)
INT₂(0, s₁(y)) -> INT₂(s₁(0), s₁(y))

The TRS R consists of the following rules:

intlist₁(nil) -> nil
int₂(s₁(x), 0) -> nil
int₂(x, x) -> cons₂(x, nil)
intlist₁(cons₂(x, y)) -> cons₂(s₁(x), intlist₁(y))
int₂(s₁(x), s₁(y)) -> intlist₁(int₂(x, y))
int₂(0, s₁(y)) -> cons₂(0, int₂(s₁(0), s₁(y)))
intlist₁(cons₂(x, nil)) -> cons₂(s₁(x), nil)

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:

INTLIST₁(cons₂(x, y)) -> INTLIST₁(y)

The TRS R consists of the following rules:

intlist₁(nil) -> nil
int₂(s₁(x), 0) -> nil
int₂(x, x) -> cons₂(x, nil)
intlist₁(cons₂(x, y)) -> cons₂(s₁(x), intlist₁(y))
int₂(s₁(x), s₁(y)) -> intlist₁(int₂(x, y))
int₂(0, s₁(y)) -> cons₂(0, int₂(s₁(0), s₁(y)))
intlist₁(cons₂(x, nil)) -> cons₂(s₁(x), nil)

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 strictly oriented and are deleted.

INTLIST₁(cons₂(x, y)) -> INTLIST₁(y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
INTLIST₁(x1) = INTLIST₁(x1)
cons₂(x1, x2) = cons₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

cons₂ > INTLIST₁

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:

intlist₁(nil) -> nil
int₂(s₁(x), 0) -> nil
int₂(x, x) -> cons₂(x, nil)
intlist₁(cons₂(x, y)) -> cons₂(s₁(x), intlist₁(y))
int₂(s₁(x), s₁(y)) -> intlist₁(int₂(x, y))
int₂(0, s₁(y)) -> cons₂(0, int₂(s₁(0), s₁(y)))
intlist₁(cons₂(x, nil)) -> cons₂(s₁(x), nil)

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:

INT₂(s₁(x), s₁(y)) -> INT₂(x, y)
INT₂(0, s₁(y)) -> INT₂(s₁(0), s₁(y))

The TRS R consists of the following rules:

intlist₁(nil) -> nil
int₂(s₁(x), 0) -> nil
int₂(x, x) -> cons₂(x, nil)
intlist₁(cons₂(x, y)) -> cons₂(s₁(x), intlist₁(y))
int₂(s₁(x), s₁(y)) -> intlist₁(int₂(x, y))
int₂(0, s₁(y)) -> cons₂(0, int₂(s₁(0), s₁(y)))
intlist₁(cons₂(x, nil)) -> cons₂(s₁(x), nil)

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 strictly oriented and are deleted.

INT₂(s₁(x), s₁(y)) -> INT₂(x, y)

The remaining pairs can at least by weakly be oriented.

INT₂(0, s₁(y)) -> INT₂(s₁(0), s₁(y))

Used ordering: Combined order from the following AFS and order.
INT₂(x1, x2) = x2
s₁(x1) = s₁(x1)
0 = 0

Lexicographic Path Order [19].
Precedence:

0 > s₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

INT₂(0, s₁(y)) -> INT₂(s₁(0), s₁(y))

The TRS R consists of the following rules:

intlist₁(nil) -> nil
int₂(s₁(x), 0) -> nil
int₂(x, x) -> cons₂(x, nil)
intlist₁(cons₂(x, y)) -> cons₂(s₁(x), intlist₁(y))
int₂(s₁(x), s₁(y)) -> intlist₁(int₂(x, y))
int₂(0, s₁(y)) -> cons₂(0, int₂(s₁(0), s₁(y)))
intlist₁(cons₂(x, nil)) -> cons₂(s₁(x), nil)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.