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__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__hd(cons(X, Y)) → mark(X)
a__tl(cons(X, Y)) → mark(Y)
mark(nats) → a__nats
mark(adx(X)) → a__adx(mark(X))
mark(zeros) → a__zeros
mark(incr(X)) → a__incr(mark(X))
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
mark(cons(X1, X2)) → cons(X1, X2)
mark(0) → 0
mark(s(X)) → s(X)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__hd(cons(X, Y)) → mark(X)
a__tl(cons(X, Y)) → mark(Y)
mark(nats) → a__nats
mark(adx(X)) → a__adx(mark(X))
mark(zeros) → a__zeros
mark(incr(X)) → a__incr(mark(X))
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
mark(cons(X1, X2)) → cons(X1, X2)
mark(0) → 0
mark(s(X)) → s(X)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(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:

MARK(tl(X)) → A__TL(mark(X))
MARK(hd(X)) → A__HD(mark(X))
MARK(tl(X)) → MARK(X)
A__HD(cons(X, Y)) → MARK(X)
MARK(adx(X)) → A__ADX(mark(X))
A__NATS → A__ADX(a__zeros)
MARK(incr(X)) → A__INCR(mark(X))
MARK(zeros) → A__ZEROS
A__TL(cons(X, Y)) → MARK(Y)
MARK(hd(X)) → MARK(X)
MARK(nats) → A__NATS
A__NATS → A__ZEROS
A__ADX(cons(X, Y)) → A__INCR(cons(X, adx(Y)))
MARK(incr(X)) → MARK(X)
MARK(adx(X)) → MARK(X)

The TRS R consists of the following rules:

a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__hd(cons(X, Y)) → mark(X)
a__tl(cons(X, Y)) → mark(Y)
mark(nats) → a__nats
mark(adx(X)) → a__adx(mark(X))
mark(zeros) → a__zeros
mark(incr(X)) → a__incr(mark(X))
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
mark(cons(X1, X2)) → cons(X1, X2)
mark(0) → 0
mark(s(X)) → s(X)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(X)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

MARK(tl(X)) → A__TL(mark(X))
MARK(hd(X)) → A__HD(mark(X))
MARK(tl(X)) → MARK(X)
A__HD(cons(X, Y)) → MARK(X)
MARK(adx(X)) → A__ADX(mark(X))
A__NATS → A__ADX(a__zeros)
MARK(incr(X)) → A__INCR(mark(X))
MARK(zeros) → A__ZEROS
A__TL(cons(X, Y)) → MARK(Y)
MARK(hd(X)) → MARK(X)
MARK(nats) → A__NATS
A__NATS → A__ZEROS
A__ADX(cons(X, Y)) → A__INCR(cons(X, adx(Y)))
MARK(incr(X)) → MARK(X)
MARK(adx(X)) → MARK(X)

The TRS R consists of the following rules:

a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__hd(cons(X, Y)) → mark(X)
a__tl(cons(X, Y)) → mark(Y)
mark(nats) → a__nats
mark(adx(X)) → a__adx(mark(X))
mark(zeros) → a__zeros
mark(incr(X)) → a__incr(mark(X))
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
mark(cons(X1, X2)) → cons(X1, X2)
mark(0) → 0
mark(s(X)) → s(X)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

MARK(hd(X)) → A__HD(mark(X))
MARK(tl(X)) → A__TL(mark(X))
MARK(tl(X)) → MARK(X)
A__HD(cons(X, Y)) → MARK(X)
MARK(adx(X)) → A__ADX(mark(X))
MARK(incr(X)) → A__INCR(mark(X))
A__NATS → A__ADX(a__zeros)
MARK(zeros) → A__ZEROS
MARK(hd(X)) → MARK(X)
A__TL(cons(X, Y)) → MARK(Y)
MARK(nats) → A__NATS
A__NATS → A__ZEROS
MARK(incr(X)) → MARK(X)
A__ADX(cons(X, Y)) → A__INCR(cons(X, adx(Y)))
MARK(adx(X)) → MARK(X)

The TRS R consists of the following rules:

a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__hd(cons(X, Y)) → mark(X)
a__tl(cons(X, Y)) → mark(Y)
mark(nats) → a__nats
mark(adx(X)) → a__adx(mark(X))
mark(zeros) → a__zeros
mark(incr(X)) → a__incr(mark(X))
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
mark(cons(X1, X2)) → cons(X1, X2)
mark(0) → 0
mark(s(X)) → s(X)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

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

MARK(tl(X)) → A__TL(mark(X))
MARK(hd(X)) → A__HD(mark(X))
MARK(tl(X)) → MARK(X)
A__HD(cons(X, Y)) → MARK(X)
MARK(incr(X)) → MARK(X)
A__TL(cons(X, Y)) → MARK(Y)
MARK(hd(X)) → MARK(X)
MARK(adx(X)) → MARK(X)

The TRS R consists of the following rules:

a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__hd(cons(X, Y)) → mark(X)
a__tl(cons(X, Y)) → mark(Y)
mark(nats) → a__nats
mark(adx(X)) → a__adx(mark(X))
mark(zeros) → a__zeros
mark(incr(X)) → a__incr(mark(X))
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
mark(cons(X1, X2)) → cons(X1, X2)
mark(0) → 0
mark(s(X)) → s(X)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(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.

MARK(tl(X)) → MARK(X)
A__HD(cons(X, Y)) → MARK(X)
A__TL(cons(X, Y)) → MARK(Y)
MARK(hd(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.

MARK(tl(X)) → A__TL(mark(X))
MARK(hd(X)) → A__HD(mark(X))
MARK(incr(X)) → MARK(X)
MARK(adx(X)) → MARK(X)

Used ordering: Combined order from the following AFS and order.
MARK(x1) = MARK(x1)
tl(x1) = tl(x1)
A__TL(x1) = A__TL(x1)
mark(x1) = mark(x1)
hd(x1) = hd(x1)
A__HD(x1) = A__HD(x1)
cons(x1, x2) = cons(x1, x2)
incr(x1) = x1
adx(x1) = x1
a__adx(x1) = x1
a__incr(x1) = x1
nats = nats
a__nats = a__nats
a__tl(x1) = a__tl(x1)
0 = 0
a__zeros = a__zeros
a__hd(x1) = a__hd(x1)
s(x1) = s
zeros = zeros

Recursive Path Order [2].
Precedence:

[tl₁, mark₁, hd₁, a_{_tl₁}, a_{_{hd_1]}} > [MARK₁, A_{_TL₁}, A_{_{HD_1]}} > [cons₂, s]
[tl₁, mark₁, hd₁, a_{_tl₁}, a_{_{hd_1]}} > [nats, a_{_nats}, a_{_zeros]} > 0 > [cons₂, s]
[tl₁, mark₁, hd₁, a_{_tl₁}, a_{_{hd_1]}} > [nats, a_{_nats}, a_{_zeros]} > zeros > [cons₂, s]

The following usable rules [14] were oriented:

a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
mark(nats) → a__nats
a__tl(X) → tl(X)
mark(0) → 0
a__nats → a__adx(a__zeros)
a__hd(X) → hd(X)
mark(incr(X)) → a__incr(mark(X))
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
mark(s(X)) → s(X)
mark(hd(X)) → a__hd(mark(X))
a__hd(cons(X, Y)) → mark(X)
mark(tl(X)) → a__tl(mark(X))
a__tl(cons(X, Y)) → mark(Y)
a__incr(X) → incr(X)
a__zeros → cons(0, zeros)
mark(zeros) → a__zeros
a__nats → nats
a__zeros → zeros
mark(adx(X)) → a__adx(mark(X))
mark(cons(X1, X2)) → cons(X1, X2)
a__adx(X) → adx(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

MARK(hd(X)) → A__HD(mark(X))
MARK(tl(X)) → A__TL(mark(X))
MARK(incr(X)) → MARK(X)
MARK(adx(X)) → MARK(X)

The TRS R consists of the following rules:

a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__hd(cons(X, Y)) → mark(X)
a__tl(cons(X, Y)) → mark(Y)
mark(nats) → a__nats
mark(adx(X)) → a__adx(mark(X))
mark(zeros) → a__zeros
mark(incr(X)) → a__incr(mark(X))
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
mark(cons(X1, X2)) → cons(X1, X2)
mark(0) → 0
mark(s(X)) → s(X)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPOrderProof

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

MARK(incr(X)) → MARK(X)
MARK(adx(X)) → MARK(X)

The TRS R consists of the following rules:

a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__hd(cons(X, Y)) → mark(X)
a__tl(cons(X, Y)) → mark(Y)
mark(nats) → a__nats
mark(adx(X)) → a__adx(mark(X))
mark(zeros) → a__zeros
mark(incr(X)) → a__incr(mark(X))
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
mark(cons(X1, X2)) → cons(X1, X2)
mark(0) → 0
mark(s(X)) → s(X)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(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.

MARK(incr(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.

MARK(adx(X)) → MARK(X)

Used ordering: Combined order from the following AFS and order.
MARK(x1) = MARK(x1)
incr(x1) = incr(x1)
adx(x1) = x1

Recursive Path Order [2].
Precedence:

[MARK₁, incr_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ QDPOrderProof

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

MARK(adx(X)) → MARK(X)

The TRS R consists of the following rules:

a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__hd(cons(X, Y)) → mark(X)
a__tl(cons(X, Y)) → mark(Y)
mark(nats) → a__nats
mark(adx(X)) → a__adx(mark(X))
mark(zeros) → a__zeros
mark(incr(X)) → a__incr(mark(X))
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
mark(cons(X1, X2)) → cons(X1, X2)
mark(0) → 0
mark(s(X)) → s(X)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(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.

MARK(adx(X)) → MARK(X)

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

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__hd(cons(X, Y)) → mark(X)
a__tl(cons(X, Y)) → mark(Y)
mark(nats) → a__nats
mark(adx(X)) → a__adx(mark(X))
mark(zeros) → a__zeros
mark(incr(X)) → a__incr(mark(X))
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
mark(cons(X1, X2)) → cons(X1, X2)
mark(0) → 0
mark(s(X)) → s(X)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(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.