Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/TRCSRSLASHExIntrod_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 strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.

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

Used ordering: Combined order from the following AFS and order.
MARK₁(x1) = MARK₁(x1)
tl₁(x1) = tl₁(x1)
A__TL₁(x1) = A__TL₁(x1)
mark₁(x1) = mark₁(x1)
hd₁(x1) = hd₁(x1)
A__HD₁(x1) = A__HD₁(x1)
cons₂(x1, x2) = cons₂(x1, x2)
incr₁(x1) = x1
adx₁(x1) = x1
nats = nats
a__nats = a__nats
a__adx₁(x1) = x1
zeros = zeros
a__zeros = a__zeros
a__incr₁(x1) = x1
a__hd₁(x1) = a__hd₁(x1)
a__tl₁(x1) = a__tl₁(x1)
0 = 0
s₁(x1) = s

Lexicographic Path Order [19].
Precedence:

[tl₁, mark₁, hd₁, a_{_nats}, a_{_hd₁}, a_{_{tl_1]}} > [MARK₁, A_{_TL₁}, A_{_{HD_1]}} > s
[tl₁, mark₁, hd₁, a_{_nats}, a_{_hd₁}, a_{_{tl_1]}} > nats > s
[tl₁, mark₁, hd₁, a_{_nats}, a_{_hd₁}, a_{_{tl_1]}} > a_{_zeros} > cons₂ > s
[tl₁, mark₁, hd₁, a_{_nats}, a_{_hd₁}, a_{_{tl_1]}} > a_{_zeros} > zeros > s
[tl₁, mark₁, hd₁, a_{_nats}, a_{_hd₁}, a_{_{tl_1]}} > a_{_zeros} > 0 > s

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

MARK₁(hd₁(X)) -> A__HD₁(mark₁(X))
MARK₁(tl₁(X)) -> A__TL₁(mark₁(X))
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 2 less nodes.

↳ 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 strictly oriented and are deleted.

MARK₁(adx₁(X)) -> MARK₁(X)

The remaining pairs can at least by weakly be oriented.

MARK₁(incr₁(X)) -> MARK₁(X)

Used ordering: Combined order from the following AFS and order.
MARK₁(x1) = MARK₁(x1)
incr₁(x1) = x1
adx₁(x1) = adx₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

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(incr₁(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 strictly oriented and are deleted.

MARK₁(incr₁(X)) -> MARK₁(X)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MARK₁(x1) = MARK₁(x1)
incr₁(x1) = incr₁(x1)

Lexicographic Path Order [19].
Precedence:

incr₁ > MARK₁

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.