Termination w.r.t. Q proof of /tmp/tmpAzaK_9/Ex4_7_56_Bor03

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__from(X) → cons(mark(X), from(s(X)))
a__after(0, XS) → mark(XS)
a__after(s(N), cons(X, XS)) → a__after(mark(N), mark(XS))
mark(from(X)) → a__from(mark(X))
mark(after(X1, X2)) → a__after(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__after(X1, X2) → after(X1, X2)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__from(X) → cons(mark(X), from(s(X)))
a__after(0, XS) → mark(XS)
a__after(s(N), cons(X, XS)) → a__after(mark(N), mark(XS))
mark(from(X)) → a__from(mark(X))
mark(after(X1, X2)) → a__after(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__after(X1, X2) → after(X1, X2)

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:

A__AFTER(s(N), cons(X, XS)) → MARK(N)
MARK(from(X)) → MARK(X)
MARK(after(X1, X2)) → A__AFTER(mark(X1), mark(X2))
MARK(from(X)) → A__FROM(mark(X))
MARK(s(X)) → MARK(X)
MARK(after(X1, X2)) → MARK(X1)
A__AFTER(0, XS) → MARK(XS)
MARK(cons(X1, X2)) → MARK(X1)
A__AFTER(s(N), cons(X, XS)) → MARK(XS)
A__AFTER(s(N), cons(X, XS)) → A__AFTER(mark(N), mark(XS))
MARK(after(X1, X2)) → MARK(X2)
A__FROM(X) → MARK(X)

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__after(0, XS) → mark(XS)
a__after(s(N), cons(X, XS)) → a__after(mark(N), mark(XS))
mark(from(X)) → a__from(mark(X))
mark(after(X1, X2)) → a__after(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__after(X1, X2) → after(X1, X2)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

A__AFTER(s(N), cons(X, XS)) → MARK(N)
MARK(from(X)) → MARK(X)
MARK(after(X1, X2)) → A__AFTER(mark(X1), mark(X2))
MARK(from(X)) → A__FROM(mark(X))
MARK(s(X)) → MARK(X)
MARK(after(X1, X2)) → MARK(X1)
A__AFTER(0, XS) → MARK(XS)
MARK(cons(X1, X2)) → MARK(X1)
A__AFTER(s(N), cons(X, XS)) → MARK(XS)
A__AFTER(s(N), cons(X, XS)) → A__AFTER(mark(N), mark(XS))
MARK(after(X1, X2)) → MARK(X2)
A__FROM(X) → MARK(X)

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__after(0, XS) → mark(XS)
a__after(s(N), cons(X, XS)) → a__after(mark(N), mark(XS))
mark(from(X)) → a__from(mark(X))
mark(after(X1, X2)) → a__after(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__after(X1, X2) → after(X1, X2)

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__AFTER(0, XS) → MARK(XS)

The remaining pairs can at least be oriented weakly.

A__AFTER(s(N), cons(X, XS)) → MARK(N)
MARK(from(X)) → MARK(X)
MARK(after(X1, X2)) → A__AFTER(mark(X1), mark(X2))
MARK(from(X)) → A__FROM(mark(X))
MARK(s(X)) → MARK(X)
MARK(after(X1, X2)) → MARK(X1)
MARK(cons(X1, X2)) → MARK(X1)
A__AFTER(s(N), cons(X, XS)) → MARK(XS)
A__AFTER(s(N), cons(X, XS)) → A__AFTER(mark(N), mark(XS))
MARK(after(X1, X2)) → MARK(X2)
A__FROM(X) → MARK(X)

Used ordering: Polynomial interpretation with max and min functions [25]:

POL(0) = 1
POL(A__AFTER(x₁, x₂)) = x₁ + x₂
POL(A__FROM(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(a__after(x₁, x₂)) = x₁ + x₂
POL(a__from(x₁)) = x₁
POL(after(x₁, x₂)) = x₁ + x₂
POL(cons(x₁, x₂)) = max(x₁, x₂)
POL(from(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(s(x₁)) = x₁

The following usable rules [17] were oriented:

a__after(s(N), cons(X, XS)) → a__after(mark(N), mark(XS))
mark(after(X1, X2)) → a__after(mark(X1), mark(X2))
a__after(0, XS) → mark(XS)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(from(X)) → a__from(mark(X))
a__from(X) → from(X)
a__after(X1, X2) → after(X1, X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → cons(mark(X), from(s(X)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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

A__AFTER(s(N), cons(X, XS)) → MARK(N)
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(after(X1, X2)) → A__AFTER(mark(X1), mark(X2))
MARK(s(X)) → MARK(X)
MARK(after(X1, X2)) → MARK(X1)
MARK(cons(X1, X2)) → MARK(X1)
A__AFTER(s(N), cons(X, XS)) → A__AFTER(mark(N), mark(XS))
A__AFTER(s(N), cons(X, XS)) → MARK(XS)
MARK(after(X1, X2)) → MARK(X2)
A__FROM(X) → MARK(X)

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__after(0, XS) → mark(XS)
a__after(s(N), cons(X, XS)) → a__after(mark(N), mark(XS))
mark(from(X)) → a__from(mark(X))
mark(after(X1, X2)) → a__after(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__after(X1, X2) → after(X1, X2)

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(from(X)) → MARK(X)
A__FROM(X) → MARK(X)

The remaining pairs can at least be oriented weakly.

A__AFTER(s(N), cons(X, XS)) → MARK(N)
MARK(from(X)) → A__FROM(mark(X))
MARK(after(X1, X2)) → A__AFTER(mark(X1), mark(X2))
MARK(s(X)) → MARK(X)
MARK(after(X1, X2)) → MARK(X1)
MARK(cons(X1, X2)) → MARK(X1)
A__AFTER(s(N), cons(X, XS)) → A__AFTER(mark(N), mark(XS))
A__AFTER(s(N), cons(X, XS)) → MARK(XS)
MARK(after(X1, X2)) → MARK(X2)

Used ordering: Polynomial interpretation with max and min functions [25]:

POL(0) = 1
POL(A__AFTER(x₁, x₂)) = x₁ + x₂
POL(A__FROM(x₁)) = 1 + x₁
POL(MARK(x₁)) = x₁
POL(a__after(x₁, x₂)) = x₁ + x₂
POL(a__from(x₁)) = 1 + x₁
POL(after(x₁, x₂)) = x₁ + x₂
POL(cons(x₁, x₂)) = max(x₁, x₂)
POL(from(x₁)) = 1 + x₁
POL(mark(x₁)) = x₁
POL(s(x₁)) = x₁

The following usable rules [17] were oriented:

a__after(s(N), cons(X, XS)) → a__after(mark(N), mark(XS))
mark(after(X1, X2)) → a__after(mark(X1), mark(X2))
a__after(0, XS) → mark(XS)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(from(X)) → a__from(mark(X))
a__from(X) → from(X)
a__after(X1, X2) → after(X1, X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → cons(mark(X), from(s(X)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

A__AFTER(s(N), cons(X, XS)) → MARK(N)
MARK(after(X1, X2)) → A__AFTER(mark(X1), mark(X2))
MARK(from(X)) → A__FROM(mark(X))
MARK(s(X)) → MARK(X)
MARK(after(X1, X2)) → MARK(X1)
MARK(cons(X1, X2)) → MARK(X1)
A__AFTER(s(N), cons(X, XS)) → MARK(XS)
A__AFTER(s(N), cons(X, XS)) → A__AFTER(mark(N), mark(XS))
MARK(after(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__after(0, XS) → mark(XS)
a__after(s(N), cons(X, XS)) → a__after(mark(N), mark(XS))
mark(from(X)) → a__from(mark(X))
mark(after(X1, X2)) → a__after(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__after(X1, X2) → after(X1, X2)

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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

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

A__AFTER(s(N), cons(X, XS)) → MARK(N)
MARK(after(X1, X2)) → A__AFTER(mark(X1), mark(X2))
MARK(s(X)) → MARK(X)
MARK(after(X1, X2)) → MARK(X1)
MARK(cons(X1, X2)) → MARK(X1)
A__AFTER(s(N), cons(X, XS)) → MARK(XS)
A__AFTER(s(N), cons(X, XS)) → A__AFTER(mark(N), mark(XS))
MARK(after(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__after(0, XS) → mark(XS)
a__after(s(N), cons(X, XS)) → a__after(mark(N), mark(XS))
mark(from(X)) → a__from(mark(X))
mark(after(X1, X2)) → a__after(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__after(X1, X2) → after(X1, X2)

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(after(X1, X2)) → A__AFTER(mark(X1), mark(X2))
MARK(after(X1, X2)) → MARK(X1)
MARK(after(X1, X2)) → MARK(X2)

The remaining pairs can at least be oriented weakly.

A__AFTER(s(N), cons(X, XS)) → MARK(N)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
A__AFTER(s(N), cons(X, XS)) → MARK(XS)
A__AFTER(s(N), cons(X, XS)) → A__AFTER(mark(N), mark(XS))

Used ordering: Polynomial interpretation with max and min functions [25]:

POL(0) = 0
POL(A__AFTER(x₁, x₂)) = x₁ + x₂
POL(MARK(x₁)) = x₁
POL(a__after(x₁, x₂)) = 1 + x₁ + x₂
POL(a__from(x₁)) = x₁
POL(after(x₁, x₂)) = 1 + x₁ + x₂
POL(cons(x₁, x₂)) = max(x₁, x₂)
POL(from(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(s(x₁)) = x₁

The following usable rules [17] were oriented:

a__after(s(N), cons(X, XS)) → a__after(mark(N), mark(XS))
mark(after(X1, X2)) → a__after(mark(X1), mark(X2))
a__after(0, XS) → mark(XS)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(from(X)) → a__from(mark(X))
a__from(X) → from(X)
a__after(X1, X2) → after(X1, X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → cons(mark(X), from(s(X)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

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

A__AFTER(s(N), cons(X, XS)) → MARK(N)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
A__AFTER(s(N), cons(X, XS)) → A__AFTER(mark(N), mark(XS))
A__AFTER(s(N), cons(X, XS)) → MARK(XS)

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__after(0, XS) → mark(XS)
a__after(s(N), cons(X, XS)) → a__after(mark(N), mark(XS))
mark(from(X)) → a__from(mark(X))
mark(after(X1, X2)) → a__after(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__after(X1, X2) → after(X1, X2)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                            ↳ UsableRulesProof

                          ↳ QDP

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

MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__after(0, XS) → mark(XS)
a__after(s(N), cons(X, XS)) → a__after(mark(N), mark(XS))
mark(from(X)) → a__from(mark(X))
mark(after(X1, X2)) → a__after(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__after(X1, X2) → after(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                          ↳ QDP

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

MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

MARK(s(X)) → MARK(X)
The graph contains the following edges 1 > 1
MARK(cons(X1, X2)) → MARK(X1)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                            ↳ QDPOrderProof

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

A__AFTER(s(N), cons(X, XS)) → A__AFTER(mark(N), mark(XS))

The TRS R consists of the following rules:

a__from(X) → cons(mark(X), from(s(X)))
a__after(0, XS) → mark(XS)
a__after(s(N), cons(X, XS)) → a__after(mark(N), mark(XS))
mark(from(X)) → a__from(mark(X))
mark(after(X1, X2)) → a__after(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__after(X1, X2) → after(X1, X2)

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__AFTER(s(N), cons(X, XS)) → A__AFTER(mark(N), mark(XS))

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

POL(0) = 0
POL(A__AFTER(x₁, x₂)) = x₁
POL(a__after(x₁, x₂)) = x₂
POL(a__from(x₁)) = 0
POL(after(x₁, x₂)) = x₂
POL(cons(x₁, x₂)) = x₂
POL(from(x₁)) = 0
POL(mark(x₁)) = x₁
POL(s(x₁)) = 1 + x₁

The following usable rules [17] were oriented:

a__after(s(N), cons(X, XS)) → a__after(mark(N), mark(XS))
mark(after(X1, X2)) → a__after(mark(X1), mark(X2))
a__after(0, XS) → mark(XS)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(from(X)) → a__from(mark(X))
a__from(X) → from(X)
a__after(X1, X2) → after(X1, X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → cons(mark(X), from(s(X)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ PisEmptyProof

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

a__from(X) → cons(mark(X), from(s(X)))
a__after(0, XS) → mark(XS)
a__after(s(N), cons(X, XS)) → a__after(mark(N), mark(XS))
mark(from(X)) → a__from(mark(X))
mark(after(X1, X2)) → a__after(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
a__from(X) → from(X)
a__after(X1, X2) → after(X1, X2)

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.