Termination w.r.t. Q proof of /tmp/aprove_benchmark_example

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:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(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:

ACTIVE(f(g(X), Y)) → MARK(f(X, f(g(X), Y)))
F(active(X1), X2) → F(X1, X2)
F(X1, mark(X2)) → F(X1, X2)
MARK(g(X)) → ACTIVE(g(mark(X)))
G(active(X)) → G(X)
G(mark(X)) → G(X)
MARK(g(X)) → G(mark(X))
MARK(g(X)) → MARK(X)
ACTIVE(f(g(X), Y)) → F(X, f(g(X), Y))
MARK(f(X1, X2)) → MARK(X1)
MARK(f(X1, X2)) → F(mark(X1), X2)
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
F(mark(X1), X2) → F(X1, X2)
F(X1, active(X2)) → F(X1, X2)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(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:

ACTIVE(f(g(X), Y)) → MARK(f(X, f(g(X), Y)))
F(active(X1), X2) → F(X1, X2)
F(X1, mark(X2)) → F(X1, X2)
MARK(g(X)) → ACTIVE(g(mark(X)))
G(active(X)) → G(X)
G(mark(X)) → G(X)
MARK(g(X)) → G(mark(X))
MARK(g(X)) → MARK(X)
ACTIVE(f(g(X), Y)) → F(X, f(g(X), Y))
MARK(f(X1, X2)) → MARK(X1)
MARK(f(X1, X2)) → F(mark(X1), X2)
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
F(mark(X1), X2) → F(X1, X2)
F(X1, active(X2)) → F(X1, X2)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

G(active(X)) → G(X)
G(mark(X)) → G(X)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

G(active(X)) → G(X)
G(mark(X)) → G(X)

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:

G(active(X)) → G(X)
The graph contains the following edges 1 > 1
G(mark(X)) → G(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

F(active(X1), X2) → F(X1, X2)
F(X1, mark(X2)) → F(X1, X2)
F(X1, active(X2)) → F(X1, X2)
F(mark(X1), X2) → F(X1, X2)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

F(active(X1), X2) → F(X1, X2)
F(X1, mark(X2)) → F(X1, X2)
F(mark(X1), X2) → F(X1, X2)
F(X1, active(X2)) → F(X1, X2)

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

F(active(X1), X2) → F(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
F(X1, mark(X2)) → F(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
F(X1, active(X2)) → F(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
F(mark(X1), X2) → F(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

            ↳ QDPOrderProof

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

ACTIVE(f(g(X), Y)) → MARK(f(X, f(g(X), Y)))
MARK(f(X1, X2)) → MARK(X1)
MARK(g(X)) → ACTIVE(g(mark(X)))
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
MARK(g(X)) → MARK(X)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(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(g(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.

ACTIVE(f(g(X), Y)) → MARK(f(X, f(g(X), Y)))
MARK(f(X1, X2)) → MARK(X1)
MARK(g(X)) → ACTIVE(g(mark(X)))
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))

Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = x1
f(x1, x2) = x1
g(x1) = g(x1)
MARK(x1) = MARK(x1)
mark(x1) = mark(x1)
active(x1) = x1

Recursive path order with status [2].
Quasi-Precedence:

[g₁, MARK₁, mark₁]

Status:

MARK₁: multiset
g₁: multiset
mark₁: multiset

The following usable rules [17] were oriented:

mark(g(X)) → active(g(mark(X)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
f(X1, active(X2)) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(mark(X1), X2) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
g(active(X)) → g(X)
g(mark(X)) → g(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

            ↳ QDPOrderProof

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

ACTIVE(f(g(X), Y)) → MARK(f(X, f(g(X), Y)))
MARK(f(X1, X2)) → MARK(X1)
MARK(g(X)) → ACTIVE(g(mark(X)))
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(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(g(X)) → ACTIVE(g(mark(X)))

The remaining pairs can at least be oriented weakly.

ACTIVE(f(g(X), Y)) → MARK(f(X, f(g(X), Y)))
MARK(f(X1, X2)) → MARK(X1)
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))

Used ordering: Polynomial interpretation [25]:

POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = 1
POL(active(x₁)) = 0
POL(f(x₁, x₂)) = 1
POL(g(x₁)) = 0
POL(mark(x₁)) = 0

The following usable rules [17] were oriented:

f(X1, active(X2)) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(mark(X1), X2) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
g(active(X)) → g(X)
g(mark(X)) → g(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

            ↳ QDPOrderProof

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

ACTIVE(f(g(X), Y)) → MARK(f(X, f(g(X), Y)))
MARK(f(X1, X2)) → MARK(X1)
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(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.

ACTIVE(f(g(X), Y)) → MARK(f(X, f(g(X), Y)))

The remaining pairs can at least be oriented weakly.

MARK(f(X1, X2)) → MARK(X1)
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))

Used ordering: Polynomial interpretation [25]:

POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(active(x₁)) = x₁
POL(f(x₁, x₂)) = x₁
POL(g(x₁)) = 1 + x₁
POL(mark(x₁)) = x₁

The following usable rules [17] were oriented:

mark(g(X)) → active(g(mark(X)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
f(X1, active(X2)) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(mark(X1), X2) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
g(active(X)) → g(X)
g(mark(X)) → g(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

            ↳ QDPOrderProof

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

MARK(f(X1, X2)) → MARK(X1)
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ UsableRulesProof

            ↳ QDPOrderProof

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

MARK(f(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

            ↳ QDPOrderProof

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

MARK(f(X1, X2)) → MARK(X1)

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

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

We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

MARK(g(X)) → ACTIVE(g(mark(X)))

The remaining pairs can at least be oriented weakly.

ACTIVE(f(g(X), Y)) → MARK(f(X, f(g(X), Y)))
MARK(f(X1, X2)) → MARK(X1)
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
MARK(g(X)) → MARK(X)

Used ordering: Polynomial interpretation [25]:

POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = 1
POL(active(x₁)) = 0
POL(f(x₁, x₂)) = 1
POL(g(x₁)) = 0
POL(mark(x₁)) = 0

The following usable rules [17] were oriented:

f(X1, active(X2)) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(mark(X1), X2) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
g(active(X)) → g(X)
g(mark(X)) → g(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

            ↳ QDPOrderProof

              ↳ QDP

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

ACTIVE(f(g(X), Y)) → MARK(f(X, f(g(X), Y)))
MARK(f(X1, X2)) → MARK(X1)
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
MARK(g(X)) → MARK(X)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)

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