Termination w.r.t. Q proof of TRS/nontermin/CSRExIntrod

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:

nats -> adx₁(zeros)
zeros -> cons₂(0, zeros)
incr₁(cons₂(X, Y)) -> cons₂(s₁(X), incr₁(Y))
adx₁(cons₂(X, Y)) -> incr₁(cons₂(X, adx₁(Y)))
hd₁(cons₂(X, Y)) -> X
tl₁(cons₂(X, Y)) -> Y

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

nats -> adx₁(zeros)
zeros -> cons₂(0, zeros)
incr₁(cons₂(X, Y)) -> cons₂(s₁(X), incr₁(Y))
adx₁(cons₂(X, Y)) -> incr₁(cons₂(X, adx₁(Y)))
hd₁(cons₂(X, Y)) -> X
tl₁(cons₂(X, Y)) -> Y

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:

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

The TRS R consists of the following rules:

nats -> adx₁(zeros)
zeros -> cons₂(0, zeros)
incr₁(cons₂(X, Y)) -> cons₂(s₁(X), incr₁(Y))
adx₁(cons₂(X, Y)) -> incr₁(cons₂(X, adx₁(Y)))
hd₁(cons₂(X, Y)) -> X
tl₁(cons₂(X, Y)) -> Y

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:

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

The TRS R consists of the following rules:

nats -> adx₁(zeros)
zeros -> cons₂(0, zeros)
incr₁(cons₂(X, Y)) -> cons₂(s₁(X), incr₁(Y))
adx₁(cons₂(X, Y)) -> incr₁(cons₂(X, adx₁(Y)))
hd₁(cons₂(X, Y)) -> X
tl₁(cons₂(X, Y)) -> Y

Q is empty.
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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

INCR₁(cons₂(X, Y)) -> INCR₁(Y)

The TRS R consists of the following rules:

nats -> adx₁(zeros)
zeros -> cons₂(0, zeros)
incr₁(cons₂(X, Y)) -> cons₂(s₁(X), incr₁(Y))
adx₁(cons₂(X, Y)) -> incr₁(cons₂(X, adx₁(Y)))
hd₁(cons₂(X, Y)) -> X
tl₁(cons₂(X, Y)) -> Y

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.

INCR₁(cons₂(X, Y)) -> INCR₁(Y)

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

POL(INCR₁(x₁)) = 3·x₁
POL(cons₂(x₁, x₂)) = 1 + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

nats -> adx₁(zeros)
zeros -> cons₂(0, zeros)
incr₁(cons₂(X, Y)) -> cons₂(s₁(X), incr₁(Y))
adx₁(cons₂(X, Y)) -> incr₁(cons₂(X, adx₁(Y)))
hd₁(cons₂(X, Y)) -> X
tl₁(cons₂(X, Y)) -> Y

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

          ↳ QDP

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

ADX₁(cons₂(X, Y)) -> ADX₁(Y)

The TRS R consists of the following rules:

nats -> adx₁(zeros)
zeros -> cons₂(0, zeros)
incr₁(cons₂(X, Y)) -> cons₂(s₁(X), incr₁(Y))
adx₁(cons₂(X, Y)) -> incr₁(cons₂(X, adx₁(Y)))
hd₁(cons₂(X, Y)) -> X
tl₁(cons₂(X, Y)) -> Y

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.

ADX₁(cons₂(X, Y)) -> ADX₁(Y)

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

POL(ADX₁(x₁)) = 3·x₁
POL(cons₂(x₁, x₂)) = 1 + x₂

The following usable rules [14] were oriented: none

↳ 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:

nats -> adx₁(zeros)
zeros -> cons₂(0, zeros)
incr₁(cons₂(X, Y)) -> cons₂(s₁(X), incr₁(Y))
adx₁(cons₂(X, Y)) -> incr₁(cons₂(X, adx₁(Y)))
hd₁(cons₂(X, Y)) -> X
tl₁(cons₂(X, Y)) -> Y

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

          ↳ QDP

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

ZEROS -> ZEROS

The TRS R consists of the following rules:

nats -> adx₁(zeros)
zeros -> cons₂(0, zeros)
incr₁(cons₂(X, Y)) -> cons₂(s₁(X), incr₁(Y))
adx₁(cons₂(X, Y)) -> incr₁(cons₂(X, adx₁(Y)))
hd₁(cons₂(X, Y)) -> X
tl₁(cons₂(X, Y)) -> Y

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