Termination w.r.t. Q proof of /tmp/tmpLAg5Xx/Ex9_Luc04

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
mark(f(X1, X2, X3)) → active(f(mark(X1), X2, mark(X3)))
mark(a) → active(a)
mark(b) → active(b)
mark(c) → active(c)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
mark(f(X1, X2, X3)) → active(f(mark(X1), X2, mark(X3)))
mark(a) → active(a)
mark(b) → active(b)
mark(c) → active(c)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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(c) → MARK(a)
MARK(f(X1, X2, X3)) → MARK(X3)
ACTIVE(f(a, b, X)) → F(X, X, X)
ACTIVE(f(a, b, X)) → MARK(f(X, X, X))
F(X1, mark(X2), X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
MARK(f(X1, X2, X3)) → MARK(X1)
MARK(b) → ACTIVE(b)
F(active(X1), X2, X3) → F(X1, X2, X3)
MARK(a) → ACTIVE(a)
F(X1, active(X2), X3) → F(X1, X2, X3)
MARK(c) → ACTIVE(c)
MARK(f(X1, X2, X3)) → F(mark(X1), X2, mark(X3))
ACTIVE(c) → MARK(b)
F(X1, X2, active(X3)) → F(X1, X2, X3)
F(mark(X1), X2, X3) → F(X1, X2, X3)
MARK(f(X1, X2, X3)) → ACTIVE(f(mark(X1), X2, mark(X3)))

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
mark(f(X1, X2, X3)) → active(f(mark(X1), X2, mark(X3)))
mark(a) → active(a)
mark(b) → active(b)
mark(c) → active(c)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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(c) → MARK(a)
MARK(f(X1, X2, X3)) → MARK(X3)
ACTIVE(f(a, b, X)) → F(X, X, X)
ACTIVE(f(a, b, X)) → MARK(f(X, X, X))
F(X1, mark(X2), X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
MARK(f(X1, X2, X3)) → MARK(X1)
MARK(b) → ACTIVE(b)
F(active(X1), X2, X3) → F(X1, X2, X3)
MARK(a) → ACTIVE(a)
F(X1, active(X2), X3) → F(X1, X2, X3)
MARK(c) → ACTIVE(c)
MARK(f(X1, X2, X3)) → F(mark(X1), X2, mark(X3))
ACTIVE(c) → MARK(b)
F(X1, X2, active(X3)) → F(X1, X2, X3)
F(mark(X1), X2, X3) → F(X1, X2, X3)
MARK(f(X1, X2, X3)) → ACTIVE(f(mark(X1), X2, mark(X3)))

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
mark(f(X1, X2, X3)) → active(f(mark(X1), X2, mark(X3)))
mark(a) → active(a)
mark(b) → active(b)
mark(c) → active(c)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

F(X1, active(X2), X3) → F(X1, X2, X3)
F(X1, X2, active(X3)) → F(X1, X2, X3)
F(mark(X1), X2, X3) → F(X1, X2, X3)
F(X1, mark(X2), X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
F(active(X1), X2, X3) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
mark(f(X1, X2, X3)) → active(f(mark(X1), X2, mark(X3)))
mark(a) → active(a)
mark(b) → active(b)
mark(c) → active(c)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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

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

F(X1, active(X2), X3) → F(X1, X2, X3)
F(mark(X1), X2, X3) → F(X1, X2, X3)
F(X1, X2, active(X3)) → F(X1, X2, X3)
F(X1, mark(X2), X3) → F(X1, X2, X3)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
F(active(X1), X2, X3) → F(X1, X2, X3)

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:

F(X1, active(X2), X3) → F(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3
F(X1, X2, active(X3)) → F(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3
F(mark(X1), X2, X3) → F(X1, X2, X3)
The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3
F(X1, mark(X2), X3) → F(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 > 2, 3 >= 3
F(X1, X2, mark(X3)) → F(X1, X2, X3)
The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3
F(active(X1), X2, X3) → F(X1, X2, X3)
The graph contains the following edges 1 > 1, 2 >= 2, 3 >= 3

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ Narrowing

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

MARK(f(X1, X2, X3)) → MARK(X3)
ACTIVE(f(a, b, X)) → MARK(f(X, X, X))
MARK(f(X1, X2, X3)) → ACTIVE(f(mark(X1), X2, mark(X3)))
MARK(f(X1, X2, X3)) → MARK(X1)

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
mark(f(X1, X2, X3)) → active(f(mark(X1), X2, mark(X3)))
mark(a) → active(a)
mark(b) → active(b)
mark(c) → active(c)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule MARK(f(X1, X2, X3)) → ACTIVE(f(mark(X1), X2, mark(X3))) at position [0] we obtained the following new rules:

MARK(f(f(x0, x1, x2), y1, y2)) → ACTIVE(f(active(f(mark(x0), x1, mark(x2))), y1, mark(y2)))
MARK(f(y0, y1, c)) → ACTIVE(f(mark(y0), y1, active(c)))
MARK(f(b, y1, y2)) → ACTIVE(f(active(b), y1, mark(y2)))
MARK(f(x0, x1, y2)) → ACTIVE(f(x0, x1, mark(y2)))
MARK(f(y0, active(x1), y2)) → ACTIVE(f(mark(y0), x1, mark(y2)))
MARK(f(y0, y1, b)) → ACTIVE(f(mark(y0), y1, active(b)))
MARK(f(y0, x1, x2)) → ACTIVE(f(mark(y0), x1, x2))
MARK(f(c, y1, y2)) → ACTIVE(f(active(c), y1, mark(y2)))
MARK(f(a, y1, y2)) → ACTIVE(f(active(a), y1, mark(y2)))
MARK(f(y0, y1, a)) → ACTIVE(f(mark(y0), y1, active(a)))
MARK(f(y0, y1, f(x0, x1, x2))) → ACTIVE(f(mark(y0), y1, active(f(mark(x0), x1, mark(x2)))))
MARK(f(y0, mark(x1), y2)) → ACTIVE(f(mark(y0), x1, mark(y2)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ Narrowing

              ↳ QDP

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

MARK(f(X1, X2, X3)) → MARK(X3)
MARK(f(y0, y1, c)) → ACTIVE(f(mark(y0), y1, active(c)))
MARK(f(b, y1, y2)) → ACTIVE(f(active(b), y1, mark(y2)))
MARK(f(y0, active(x1), y2)) → ACTIVE(f(mark(y0), x1, mark(y2)))
ACTIVE(f(a, b, X)) → MARK(f(X, X, X))
MARK(f(y0, y1, b)) → ACTIVE(f(mark(y0), y1, active(b)))
MARK(f(y0, x1, x2)) → ACTIVE(f(mark(y0), x1, x2))
MARK(f(a, y1, y2)) → ACTIVE(f(active(a), y1, mark(y2)))
MARK(f(X1, X2, X3)) → MARK(X1)
MARK(f(y0, mark(x1), y2)) → ACTIVE(f(mark(y0), x1, mark(y2)))
MARK(f(f(x0, x1, x2), y1, y2)) → ACTIVE(f(active(f(mark(x0), x1, mark(x2))), y1, mark(y2)))
MARK(f(x0, x1, y2)) → ACTIVE(f(x0, x1, mark(y2)))
MARK(f(c, y1, y2)) → ACTIVE(f(active(c), y1, mark(y2)))
MARK(f(y0, y1, a)) → ACTIVE(f(mark(y0), y1, active(a)))
MARK(f(y0, y1, f(x0, x1, x2))) → ACTIVE(f(mark(y0), y1, active(f(mark(x0), x1, mark(x2)))))

The TRS R consists of the following rules:

active(f(a, b, X)) → mark(f(X, X, X))
active(c) → mark(a)
active(c) → mark(b)
mark(f(X1, X2, X3)) → active(f(mark(X1), X2, mark(X3)))
mark(a) → active(a)
mark(b) → active(b)
mark(c) → active(c)
f(mark(X1), X2, X3) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
f(X1, X2, mark(X3)) → f(X1, X2, X3)
f(active(X1), X2, X3) → f(X1, X2, X3)
f(X1, active(X2), X3) → f(X1, X2, X3)
f(X1, X2, active(X3)) → f(X1, X2, X3)

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