Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar18/nonterminSLASHCSRSLASHExIntrod

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

incr₁(nil) -> nil
incr₁(cons₂(X, L)) -> cons₂(s₁(X), incr₁(L))
adx₁(nil) -> nil
adx₁(cons₂(X, L)) -> incr₁(cons₂(X, adx₁(L)))
nats -> adx₁(zeros)
zeros -> cons₂(0, zeros)
head₁(cons₂(X, L)) -> X
tail₁(cons₂(X, L)) -> L

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

incr₁(nil) -> nil
incr₁(cons₂(X, L)) -> cons₂(s₁(X), incr₁(L))
adx₁(nil) -> nil
adx₁(cons₂(X, L)) -> incr₁(cons₂(X, adx₁(L)))
nats -> adx₁(zeros)
zeros -> cons₂(0, zeros)
head₁(cons₂(X, L)) -> X
tail₁(cons₂(X, L)) -> L

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

incr₁(nil) -> nil
incr₁(cons₂(X, L)) -> cons₂(s₁(X), incr₁(L))
adx₁(nil) -> nil
adx₁(cons₂(X, L)) -> incr₁(cons₂(X, adx₁(L)))
nats -> adx₁(zeros)
zeros -> cons₂(0, zeros)
head₁(cons₂(X, L)) -> X
tail₁(cons₂(X, L)) -> L

The set Q consists of the following terms:

incr₁(nil)
incr₁(cons₂(x0, x1))
adx₁(nil)
adx₁(cons₂(x0, x1))
nats
zeros
head₁(cons₂(x0, x1))
tail₁(cons₂(x0, x1))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ADX₁(cons₂(X, L)) -> ADX₁(L)
ADX₁(cons₂(X, L)) -> INCR₁(cons₂(X, adx₁(L)))
NATS -> ADX₁(zeros)
NATS -> ZEROS
ZEROS -> ZEROS
INCR₁(cons₂(X, L)) -> INCR₁(L)

The TRS R consists of the following rules:

incr₁(nil) -> nil
incr₁(cons₂(X, L)) -> cons₂(s₁(X), incr₁(L))
adx₁(nil) -> nil
adx₁(cons₂(X, L)) -> incr₁(cons₂(X, adx₁(L)))
nats -> adx₁(zeros)
zeros -> cons₂(0, zeros)
head₁(cons₂(X, L)) -> X
tail₁(cons₂(X, L)) -> L

The set Q consists of the following terms:

incr₁(nil)
incr₁(cons₂(x0, x1))
adx₁(nil)
adx₁(cons₂(x0, x1))
nats
zeros
head₁(cons₂(x0, x1))
tail₁(cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

ADX₁(cons₂(X, L)) -> ADX₁(L)
ADX₁(cons₂(X, L)) -> INCR₁(cons₂(X, adx₁(L)))
NATS -> ADX₁(zeros)
NATS -> ZEROS
ZEROS -> ZEROS
INCR₁(cons₂(X, L)) -> INCR₁(L)

The TRS R consists of the following rules:

incr₁(nil) -> nil
incr₁(cons₂(X, L)) -> cons₂(s₁(X), incr₁(L))
adx₁(nil) -> nil
adx₁(cons₂(X, L)) -> incr₁(cons₂(X, adx₁(L)))
nats -> adx₁(zeros)
zeros -> cons₂(0, zeros)
head₁(cons₂(X, L)) -> X
tail₁(cons₂(X, L)) -> L

The set Q consists of the following terms:

incr₁(nil)
incr₁(cons₂(x0, x1))
adx₁(nil)
adx₁(cons₂(x0, x1))
nats
zeros
head₁(cons₂(x0, x1))
tail₁(cons₂(x0, x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

ZEROS -> ZEROS

The TRS R consists of the following rules:

incr₁(nil) -> nil
incr₁(cons₂(X, L)) -> cons₂(s₁(X), incr₁(L))
adx₁(nil) -> nil
adx₁(cons₂(X, L)) -> incr₁(cons₂(X, adx₁(L)))
nats -> adx₁(zeros)
zeros -> cons₂(0, zeros)
head₁(cons₂(X, L)) -> X
tail₁(cons₂(X, L)) -> L

The set Q consists of the following terms:

incr₁(nil)
incr₁(cons₂(x0, x1))
adx₁(nil)
adx₁(cons₂(x0, x1))
nats
zeros
head₁(cons₂(x0, x1))
tail₁(cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

INCR₁(cons₂(X, L)) -> INCR₁(L)

The TRS R consists of the following rules:

incr₁(nil) -> nil
incr₁(cons₂(X, L)) -> cons₂(s₁(X), incr₁(L))
adx₁(nil) -> nil
adx₁(cons₂(X, L)) -> incr₁(cons₂(X, adx₁(L)))
nats -> adx₁(zeros)
zeros -> cons₂(0, zeros)
head₁(cons₂(X, L)) -> X
tail₁(cons₂(X, L)) -> L

The set Q consists of the following terms:

incr₁(nil)
incr₁(cons₂(x0, x1))
adx₁(nil)
adx₁(cons₂(x0, x1))
nats
zeros
head₁(cons₂(x0, x1))
tail₁(cons₂(x0, x1))

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.

INCR₁(cons₂(X, L)) -> INCR₁(L)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
INCR₁(x1) = INCR₁(x1)
cons₂(x1, x2) = cons₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

cons₂ > INCR₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

incr₁(nil) -> nil
incr₁(cons₂(X, L)) -> cons₂(s₁(X), incr₁(L))
adx₁(nil) -> nil
adx₁(cons₂(X, L)) -> incr₁(cons₂(X, adx₁(L)))
nats -> adx₁(zeros)
zeros -> cons₂(0, zeros)
head₁(cons₂(X, L)) -> X
tail₁(cons₂(X, L)) -> L

The set Q consists of the following terms:

incr₁(nil)
incr₁(cons₂(x0, x1))
adx₁(nil)
adx₁(cons₂(x0, x1))
nats
zeros
head₁(cons₂(x0, x1))
tail₁(cons₂(x0, x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

ADX₁(cons₂(X, L)) -> ADX₁(L)

The TRS R consists of the following rules:

incr₁(nil) -> nil
incr₁(cons₂(X, L)) -> cons₂(s₁(X), incr₁(L))
adx₁(nil) -> nil
adx₁(cons₂(X, L)) -> incr₁(cons₂(X, adx₁(L)))
nats -> adx₁(zeros)
zeros -> cons₂(0, zeros)
head₁(cons₂(X, L)) -> X
tail₁(cons₂(X, L)) -> L

The set Q consists of the following terms:

incr₁(nil)
incr₁(cons₂(x0, x1))
adx₁(nil)
adx₁(cons₂(x0, x1))
nats
zeros
head₁(cons₂(x0, x1))
tail₁(cons₂(x0, x1))

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.

ADX₁(cons₂(X, L)) -> ADX₁(L)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ADX₁(x1) = ADX₁(x1)
cons₂(x1, x2) = cons₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

cons₂ > ADX₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

incr₁(nil) -> nil
incr₁(cons₂(X, L)) -> cons₂(s₁(X), incr₁(L))
adx₁(nil) -> nil
adx₁(cons₂(X, L)) -> incr₁(cons₂(X, adx₁(L)))
nats -> adx₁(zeros)
zeros -> cons₂(0, zeros)
head₁(cons₂(X, L)) -> X
tail₁(cons₂(X, L)) -> L

The set Q consists of the following terms:

incr₁(nil)
incr₁(cons₂(x0, x1))
adx₁(nil)
adx₁(cons₂(x0, x1))
nats
zeros
head₁(cons₂(x0, x1))
tail₁(cons₂(x0, x1))

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