Termination w.r.t. Q proof of /tmp/tmpFfybJM/Ex6_Luc98

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__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(0) → 0
mark(nil) → nil
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(0) → 0
mark(nil) → nil
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(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(first(X1, X2)) → MARK(X1)
A__FIRST(s(X), cons(Y, Z)) → MARK(Y)
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
MARK(first(X1, X2)) → A__FIRST(mark(X1), mark(X2))
A__FROM(X) → MARK(X)

The TRS R consists of the following rules:

a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(0) → 0
mark(nil) → nil
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(X)

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:

MARK(first(X1, X2)) → MARK(X1)
A__FIRST(s(X), cons(Y, Z)) → MARK(Y)
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
MARK(first(X1, X2)) → A__FIRST(mark(X1), mark(X2))
A__FROM(X) → MARK(X)

The TRS R consists of the following rules:

a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(0) → 0
mark(nil) → nil
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(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(first(X1, X2)) → MARK(X1)
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(s(X)) → MARK(X)
MARK(first(X1, X2)) → MARK(X2)
MARK(first(X1, X2)) → A__FIRST(mark(X1), mark(X2))

The remaining pairs can at least be oriented weakly.

A__FIRST(s(X), cons(Y, Z)) → MARK(Y)
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(A__FIRST(x₁, x₂)) = x₂
POL(A__FROM(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(a__first(x₁, x₂)) = 1 + x₁ + x₂
POL(a__from(x₁)) = 1 + x₁
POL(cons(x₁, x₂)) = x₁
POL(first(x₁, x₂)) = 1 + x₁ + x₂
POL(from(x₁)) = 1 + x₁
POL(mark(x₁)) = x₁
POL(nil) = 1
POL(s(x₁)) = 1 + x₁

The following usable rules [17] were oriented:

mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(nil) → nil
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(X)
a__first(0, X) → nil
a__from(X) → cons(mark(X), from(s(X)))
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
mark(from(X)) → a__from(mark(X))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

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

A__FIRST(s(X), cons(Y, Z)) → MARK(Y)
MARK(cons(X1, X2)) → MARK(X1)
A__FROM(X) → MARK(X)

The TRS R consists of the following rules:

a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(0) → 0
mark(nil) → nil
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(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

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

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

MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(mark(Y), first(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(0) → 0
mark(nil) → nil
mark(s(X)) → s(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(X)

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

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

                ↳ QDP

                  ↳ QDPSizeChangeProof

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

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