Termination w.r.t. Q proof of /tmp/tmplZuhhR/ExIntrod_GM01

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__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MARK(tail(X)) → A__TAIL(mark(X))
MARK(incr(X)) → A__INCR(mark(X))
MARK(zeros) → A__ZEROS
MARK(s(X)) → MARK(X)
A__NATS → A__ADX(a__zeros)
MARK(tail(X)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
A__NATS → A__ZEROS
A__ADX(cons(X, L)) → MARK(X)
A__HEAD(cons(X, L)) → MARK(X)
MARK(adx(X)) → A__ADX(mark(X))
MARK(adx(X)) → MARK(X)
A__TAIL(cons(X, L)) → MARK(L)
MARK(head(X)) → A__HEAD(mark(X))
A__ADX(cons(X, L)) → A__INCR(cons(mark(X), adx(L)))
MARK(nats) → A__NATS
A__INCR(cons(X, L)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(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:

MARK(tail(X)) → A__TAIL(mark(X))
MARK(incr(X)) → A__INCR(mark(X))
MARK(zeros) → A__ZEROS
MARK(s(X)) → MARK(X)
A__NATS → A__ADX(a__zeros)
MARK(tail(X)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
A__NATS → A__ZEROS
A__ADX(cons(X, L)) → MARK(X)
A__HEAD(cons(X, L)) → MARK(X)
MARK(adx(X)) → A__ADX(mark(X))
MARK(adx(X)) → MARK(X)
A__TAIL(cons(X, L)) → MARK(L)
MARK(head(X)) → A__HEAD(mark(X))
A__ADX(cons(X, L)) → A__INCR(cons(mark(X), adx(L)))
MARK(nats) → A__NATS
A__INCR(cons(X, L)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

MARK(incr(X)) → A__INCR(mark(X))
MARK(tail(X)) → A__TAIL(mark(X))
MARK(s(X)) → MARK(X)
A__NATS → A__ADX(a__zeros)
MARK(tail(X)) → MARK(X)
MARK(head(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
A__ADX(cons(X, L)) → MARK(X)
A__HEAD(cons(X, L)) → MARK(X)
MARK(adx(X)) → A__ADX(mark(X))
MARK(adx(X)) → MARK(X)
A__TAIL(cons(X, L)) → MARK(L)
MARK(head(X)) → A__HEAD(mark(X))
A__ADX(cons(X, L)) → A__INCR(cons(mark(X), adx(L)))
MARK(nats) → A__NATS
A__INCR(cons(X, L)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

A__NATS → A__ADX(a__zeros)
MARK(head(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
A__ADX(cons(X, L)) → MARK(X)
A__HEAD(cons(X, L)) → MARK(X)
MARK(adx(X)) → MARK(X)
A__TAIL(cons(X, L)) → MARK(L)
MARK(head(X)) → A__HEAD(mark(X))
A__ADX(cons(X, L)) → A__INCR(cons(mark(X), adx(L)))
A__INCR(cons(X, L)) → MARK(X)

The remaining pairs can at least be oriented weakly.

MARK(incr(X)) → A__INCR(mark(X))
MARK(tail(X)) → A__TAIL(mark(X))
MARK(s(X)) → MARK(X)
MARK(tail(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(adx(X)) → A__ADX(mark(X))
MARK(nats) → A__NATS

Used ordering: Polynomial interpretation [25,35]:

POL(A__NATS) = 4
POL(a__head(x₁)) = 2 + (4)x_1
POL(A__INCR(x₁)) = (1/4)x_1
POL(tail(x₁)) = (4)x_1
POL(a__adx(x₁)) = 1 + (4)x_1
POL(a__tail(x₁)) = (4)x_1
POL(A__TAIL(x₁)) = (4)x_1
POL(a__zeros) = 1/2
POL(head(x₁)) = 2 + (4)x_1
POL(mark(x₁)) = x_1
POL(0) = 0
POL(cons(x₁, x₂)) = 1/4 + (4)x_1 + (1/2)x_2
POL(MARK(x₁)) = x_1
POL(A__HEAD(x₁)) = 1/2 + (1/4)x_1
POL(a__nats) = 4
POL(adx(x₁)) = 1 + (4)x_1
POL(incr(x₁)) = x_1
POL(a__incr(x₁)) = x_1
POL(zeros) = 1/2
POL(A__ADX(x₁)) = 1 + (4)x_1
POL(s(x₁)) = x_1
POL(nats) = 4
POL(nil) = 0

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__tail(cons(X, L)) → mark(L)
mark(head(X)) → a__head(mark(X))
a__head(cons(X, L)) → mark(X)
mark(tail(X)) → a__tail(mark(X))
a__incr(X) → incr(X)
a__adx(X) → adx(X)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__head(X) → head(X)
a__tail(X) → tail(X)
a__nats → nats
a__zeros → zeros
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__incr(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__adx(nil) → nil
a__zeros → cons(0, zeros)
a__nats → a__adx(a__zeros)
mark(adx(X)) → a__adx(mark(X))
mark(incr(X)) → a__incr(mark(X))
mark(zeros) → a__zeros
mark(nats) → a__nats

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

MARK(adx(X)) → A__ADX(mark(X))
MARK(tail(X)) → A__TAIL(mark(X))
MARK(incr(X)) → A__INCR(mark(X))
MARK(s(X)) → MARK(X)
MARK(tail(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(nats) → A__NATS

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

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

MARK(s(X)) → MARK(X)
MARK(tail(X)) → MARK(X)
MARK(incr(X)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

MARK(tail(X)) → MARK(X)
MARK(incr(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.

MARK(s(X)) → MARK(X)

Used ordering: Polynomial interpretation [25,35]:

POL(MARK(x₁)) = (2)x_1
POL(tail(x₁)) = 4 + (4)x_1
POL(incr(x₁)) = 4 + (4)x_1
POL(s(x₁)) = x_1

The value of delta used in the strict ordering is 8.
The following usable rules [17] 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:

MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

MARK(s(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(MARK(x₁)) = (1/4)x_1
POL(s(x₁)) = 1/4 + (2)x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] 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__incr(nil) → nil
a__incr(cons(X, L)) → cons(s(mark(X)), incr(L))
a__adx(nil) → nil
a__adx(cons(X, L)) → a__incr(cons(mark(X), adx(L)))
a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__head(cons(X, L)) → mark(X)
a__tail(cons(X, L)) → mark(L)
mark(incr(X)) → a__incr(mark(X))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(zeros) → a__zeros
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__incr(X) → incr(X)
a__adx(X) → adx(X)
a__nats → nats
a__zeros → zeros
a__head(X) → head(X)
a__tail(X) → tail(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.