Termination w.r.t. Q proof of TRS/TRCSRExIntrod_GM04

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__nats -> a__adx₁(a__zeros)
a__zeros -> cons₂(0, zeros)
a__incr₁(cons₂(X, Y)) -> cons₂(s₁(X), incr₁(Y))
a__adx₁(cons₂(X, Y)) -> a__incr₁(cons₂(X, adx₁(Y)))
a__hd₁(cons₂(X, Y)) -> mark₁(X)
a__tl₁(cons₂(X, Y)) -> mark₁(Y)
mark₁(nats) -> a__nats
mark₁(adx₁(X)) -> a__adx₁(mark₁(X))
mark₁(zeros) -> a__zeros
mark₁(incr₁(X)) -> a__incr₁(mark₁(X))
mark₁(hd₁(X)) -> a__hd₁(mark₁(X))
mark₁(tl₁(X)) -> a__tl₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
a__nats -> nats
a__adx₁(X) -> adx₁(X)
a__zeros -> zeros
a__incr₁(X) -> incr₁(X)
a__hd₁(X) -> hd₁(X)
a__tl₁(X) -> tl₁(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__nats -> a__adx₁(a__zeros)
a__zeros -> cons₂(0, zeros)
a__incr₁(cons₂(X, Y)) -> cons₂(s₁(X), incr₁(Y))
a__adx₁(cons₂(X, Y)) -> a__incr₁(cons₂(X, adx₁(Y)))
a__hd₁(cons₂(X, Y)) -> mark₁(X)
a__tl₁(cons₂(X, Y)) -> mark₁(Y)
mark₁(nats) -> a__nats
mark₁(adx₁(X)) -> a__adx₁(mark₁(X))
mark₁(zeros) -> a__zeros
mark₁(incr₁(X)) -> a__incr₁(mark₁(X))
mark₁(hd₁(X)) -> a__hd₁(mark₁(X))
mark₁(tl₁(X)) -> a__tl₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
a__nats -> nats
a__adx₁(X) -> adx₁(X)
a__zeros -> zeros
a__incr₁(X) -> incr₁(X)
a__hd₁(X) -> hd₁(X)
a__tl₁(X) -> tl₁(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:

MARK₁(tl₁(X)) -> A__TL₁(mark₁(X))
MARK₁(hd₁(X)) -> A__HD₁(mark₁(X))
MARK₁(tl₁(X)) -> MARK₁(X)
A__HD₁(cons₂(X, Y)) -> MARK₁(X)
MARK₁(adx₁(X)) -> A__ADX₁(mark₁(X))
A__NATS -> A__ADX₁(a__zeros)
MARK₁(incr₁(X)) -> A__INCR₁(mark₁(X))
MARK₁(zeros) -> A__ZEROS
A__TL₁(cons₂(X, Y)) -> MARK₁(Y)
MARK₁(hd₁(X)) -> MARK₁(X)
MARK₁(nats) -> A__NATS
A__NATS -> A__ZEROS
A__ADX₁(cons₂(X, Y)) -> A__INCR₁(cons₂(X, adx₁(Y)))
MARK₁(incr₁(X)) -> MARK₁(X)
MARK₁(adx₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__nats -> a__adx₁(a__zeros)
a__zeros -> cons₂(0, zeros)
a__incr₁(cons₂(X, Y)) -> cons₂(s₁(X), incr₁(Y))
a__adx₁(cons₂(X, Y)) -> a__incr₁(cons₂(X, adx₁(Y)))
a__hd₁(cons₂(X, Y)) -> mark₁(X)
a__tl₁(cons₂(X, Y)) -> mark₁(Y)
mark₁(nats) -> a__nats
mark₁(adx₁(X)) -> a__adx₁(mark₁(X))
mark₁(zeros) -> a__zeros
mark₁(incr₁(X)) -> a__incr₁(mark₁(X))
mark₁(hd₁(X)) -> a__hd₁(mark₁(X))
mark₁(tl₁(X)) -> a__tl₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
a__nats -> nats
a__adx₁(X) -> adx₁(X)
a__zeros -> zeros
a__incr₁(X) -> incr₁(X)
a__hd₁(X) -> hd₁(X)
a__tl₁(X) -> tl₁(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:

MARK₁(tl₁(X)) -> A__TL₁(mark₁(X))
MARK₁(hd₁(X)) -> A__HD₁(mark₁(X))
MARK₁(tl₁(X)) -> MARK₁(X)
A__HD₁(cons₂(X, Y)) -> MARK₁(X)
MARK₁(adx₁(X)) -> A__ADX₁(mark₁(X))
A__NATS -> A__ADX₁(a__zeros)
MARK₁(incr₁(X)) -> A__INCR₁(mark₁(X))
MARK₁(zeros) -> A__ZEROS
A__TL₁(cons₂(X, Y)) -> MARK₁(Y)
MARK₁(hd₁(X)) -> MARK₁(X)
MARK₁(nats) -> A__NATS
A__NATS -> A__ZEROS
A__ADX₁(cons₂(X, Y)) -> A__INCR₁(cons₂(X, adx₁(Y)))
MARK₁(incr₁(X)) -> MARK₁(X)
MARK₁(adx₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__nats -> a__adx₁(a__zeros)
a__zeros -> cons₂(0, zeros)
a__incr₁(cons₂(X, Y)) -> cons₂(s₁(X), incr₁(Y))
a__adx₁(cons₂(X, Y)) -> a__incr₁(cons₂(X, adx₁(Y)))
a__hd₁(cons₂(X, Y)) -> mark₁(X)
a__tl₁(cons₂(X, Y)) -> mark₁(Y)
mark₁(nats) -> a__nats
mark₁(adx₁(X)) -> a__adx₁(mark₁(X))
mark₁(zeros) -> a__zeros
mark₁(incr₁(X)) -> a__incr₁(mark₁(X))
mark₁(hd₁(X)) -> a__hd₁(mark₁(X))
mark₁(tl₁(X)) -> a__tl₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
a__nats -> nats
a__adx₁(X) -> adx₁(X)
a__zeros -> zeros
a__incr₁(X) -> incr₁(X)
a__hd₁(X) -> hd₁(X)
a__tl₁(X) -> tl₁(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

MARK₁(tl₁(X)) -> A__TL₁(mark₁(X))
MARK₁(hd₁(X)) -> A__HD₁(mark₁(X))
MARK₁(tl₁(X)) -> MARK₁(X)
A__HD₁(cons₂(X, Y)) -> MARK₁(X)
MARK₁(incr₁(X)) -> MARK₁(X)
A__TL₁(cons₂(X, Y)) -> MARK₁(Y)
MARK₁(hd₁(X)) -> MARK₁(X)
MARK₁(adx₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__nats -> a__adx₁(a__zeros)
a__zeros -> cons₂(0, zeros)
a__incr₁(cons₂(X, Y)) -> cons₂(s₁(X), incr₁(Y))
a__adx₁(cons₂(X, Y)) -> a__incr₁(cons₂(X, adx₁(Y)))
a__hd₁(cons₂(X, Y)) -> mark₁(X)
a__tl₁(cons₂(X, Y)) -> mark₁(Y)
mark₁(nats) -> a__nats
mark₁(adx₁(X)) -> a__adx₁(mark₁(X))
mark₁(zeros) -> a__zeros
mark₁(incr₁(X)) -> a__incr₁(mark₁(X))
mark₁(hd₁(X)) -> a__hd₁(mark₁(X))
mark₁(tl₁(X)) -> a__tl₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
a__nats -> nats
a__adx₁(X) -> adx₁(X)
a__zeros -> zeros
a__incr₁(X) -> incr₁(X)
a__hd₁(X) -> hd₁(X)
a__tl₁(X) -> tl₁(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(tl₁(X)) -> A__TL₁(mark₁(X))
MARK₁(hd₁(X)) -> A__HD₁(mark₁(X))
MARK₁(tl₁(X)) -> MARK₁(X)
A__HD₁(cons₂(X, Y)) -> MARK₁(X)
MARK₁(hd₁(X)) -> MARK₁(X)

The remaining pairs can at least be oriented weakly.

MARK₁(incr₁(X)) -> MARK₁(X)
A__TL₁(cons₂(X, Y)) -> MARK₁(Y)
MARK₁(adx₁(X)) -> MARK₁(X)

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(A__HD₁(x₁)) = 2 + 2·x₁ + 2·x₁²
POL(A__TL₁(x₁)) = 1 + 2·x₁ + 3·x₁²
POL(MARK₁(x₁)) = 1 + x₁ + 2·x₁²
POL(a__adx₁(x₁)) = 2·x₁
POL(a__hd₁(x₁)) = 2 + x₁
POL(a__incr₁(x₁)) = x₁
POL(a__nats) = 2
POL(a__tl₁(x₁)) = 3 + 2·x₁
POL(a__zeros) = 0
POL(adx₁(x₁)) = 2·x₁
POL(cons₂(x₁, x₂)) = x₁ + x₂
POL(hd₁(x₁)) = 2 + x₁
POL(incr₁(x₁)) = x₁
POL(mark₁(x₁)) = x₁
POL(nats) = 2
POL(s₁(x₁)) = x₁
POL(tl₁(x₁)) = 3 + 2·x₁
POL(zeros) = 0

The following usable rules [14] were oriented:

mark₁(0) -> 0
a__adx₁(cons₂(X, Y)) -> a__incr₁(cons₂(X, adx₁(Y)))
mark₁(incr₁(X)) -> a__incr₁(mark₁(X))
a__adx₁(X) -> adx₁(X)
mark₁(s₁(X)) -> s₁(X)
a__zeros -> zeros
a__tl₁(X) -> tl₁(X)
mark₁(adx₁(X)) -> a__adx₁(mark₁(X))
a__nats -> a__adx₁(a__zeros)
mark₁(hd₁(X)) -> a__hd₁(mark₁(X))
a__hd₁(cons₂(X, Y)) -> mark₁(X)
mark₁(tl₁(X)) -> a__tl₁(mark₁(X))
a__tl₁(cons₂(X, Y)) -> mark₁(Y)
a__nats -> nats
a__hd₁(X) -> hd₁(X)
a__zeros -> cons₂(0, zeros)
mark₁(nats) -> a__nats
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
a__incr₁(X) -> incr₁(X)
a__incr₁(cons₂(X, Y)) -> cons₂(s₁(X), incr₁(Y))
mark₁(zeros) -> a__zeros

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

MARK₁(incr₁(X)) -> MARK₁(X)
MARK₁(adx₁(X)) -> MARK₁(X)
A__TL₁(cons₂(X, Y)) -> MARK₁(Y)

The TRS R consists of the following rules:

a__nats -> a__adx₁(a__zeros)
a__zeros -> cons₂(0, zeros)
a__incr₁(cons₂(X, Y)) -> cons₂(s₁(X), incr₁(Y))
a__adx₁(cons₂(X, Y)) -> a__incr₁(cons₂(X, adx₁(Y)))
a__hd₁(cons₂(X, Y)) -> mark₁(X)
a__tl₁(cons₂(X, Y)) -> mark₁(Y)
mark₁(nats) -> a__nats
mark₁(adx₁(X)) -> a__adx₁(mark₁(X))
mark₁(zeros) -> a__zeros
mark₁(incr₁(X)) -> a__incr₁(mark₁(X))
mark₁(hd₁(X)) -> a__hd₁(mark₁(X))
mark₁(tl₁(X)) -> a__tl₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
a__nats -> nats
a__adx₁(X) -> adx₁(X)
a__zeros -> zeros
a__incr₁(X) -> incr₁(X)
a__hd₁(X) -> hd₁(X)
a__tl₁(X) -> tl₁(X)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

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

MARK₁(incr₁(X)) -> MARK₁(X)
MARK₁(adx₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__nats -> a__adx₁(a__zeros)
a__zeros -> cons₂(0, zeros)
a__incr₁(cons₂(X, Y)) -> cons₂(s₁(X), incr₁(Y))
a__adx₁(cons₂(X, Y)) -> a__incr₁(cons₂(X, adx₁(Y)))
a__hd₁(cons₂(X, Y)) -> mark₁(X)
a__tl₁(cons₂(X, Y)) -> mark₁(Y)
mark₁(nats) -> a__nats
mark₁(adx₁(X)) -> a__adx₁(mark₁(X))
mark₁(zeros) -> a__zeros
mark₁(incr₁(X)) -> a__incr₁(mark₁(X))
mark₁(hd₁(X)) -> a__hd₁(mark₁(X))
mark₁(tl₁(X)) -> a__tl₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
a__nats -> nats
a__adx₁(X) -> adx₁(X)
a__zeros -> zeros
a__incr₁(X) -> incr₁(X)
a__hd₁(X) -> hd₁(X)
a__tl₁(X) -> tl₁(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(incr₁(X)) -> MARK₁(X)

The remaining pairs can at least be oriented weakly.

MARK₁(adx₁(X)) -> MARK₁(X)

Used ordering: Polynomial interpretation [21]:

POL(MARK₁(x₁)) = 2·x₁
POL(adx₁(x₁)) = x₁
POL(incr₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPOrderProof

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

MARK₁(adx₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__nats -> a__adx₁(a__zeros)
a__zeros -> cons₂(0, zeros)
a__incr₁(cons₂(X, Y)) -> cons₂(s₁(X), incr₁(Y))
a__adx₁(cons₂(X, Y)) -> a__incr₁(cons₂(X, adx₁(Y)))
a__hd₁(cons₂(X, Y)) -> mark₁(X)
a__tl₁(cons₂(X, Y)) -> mark₁(Y)
mark₁(nats) -> a__nats
mark₁(adx₁(X)) -> a__adx₁(mark₁(X))
mark₁(zeros) -> a__zeros
mark₁(incr₁(X)) -> a__incr₁(mark₁(X))
mark₁(hd₁(X)) -> a__hd₁(mark₁(X))
mark₁(tl₁(X)) -> a__tl₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
a__nats -> nats
a__adx₁(X) -> adx₁(X)
a__zeros -> zeros
a__incr₁(X) -> incr₁(X)
a__hd₁(X) -> hd₁(X)
a__tl₁(X) -> tl₁(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(adx₁(X)) -> MARK₁(X)

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

POL(MARK₁(x₁)) = 2·x₁
POL(adx₁(x₁)) = 2 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ PisEmptyProof

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

a__nats -> a__adx₁(a__zeros)
a__zeros -> cons₂(0, zeros)
a__incr₁(cons₂(X, Y)) -> cons₂(s₁(X), incr₁(Y))
a__adx₁(cons₂(X, Y)) -> a__incr₁(cons₂(X, adx₁(Y)))
a__hd₁(cons₂(X, Y)) -> mark₁(X)
a__tl₁(cons₂(X, Y)) -> mark₁(Y)
mark₁(nats) -> a__nats
mark₁(adx₁(X)) -> a__adx₁(mark₁(X))
mark₁(zeros) -> a__zeros
mark₁(incr₁(X)) -> a__incr₁(mark₁(X))
mark₁(hd₁(X)) -> a__hd₁(mark₁(X))
mark₁(tl₁(X)) -> a__tl₁(mark₁(X))
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
a__nats -> nats
a__adx₁(X) -> adx₁(X)
a__zeros -> zeros
a__incr₁(X) -> incr₁(X)
a__hd₁(X) -> hd₁(X)
a__tl₁(X) -> tl₁(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.