Termination w.r.t. Q proof of /tmp/tmpyKK4AB/Ex14_Luc06

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(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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)

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:

F(active(X1), X2) → F(X1, X2)
ACTIVE(h(X)) → G(X, X)
H(active(X)) → H(X)
G(mark(X1), X2) → G(X1, X2)
MARK(g(X1, X2)) → G(mark(X1), X2)
MARK(b) → ACTIVE(b)
ACTIVE(f(X, X)) → H(a)
ACTIVE(g(a, X)) → F(b, X)
ACTIVE(h(X)) → MARK(g(X, X))
MARK(f(X1, X2)) → F(mark(X1), X2)
ACTIVE(a) → MARK(b)
F(X1, active(X2)) → F(X1, X2)
F(mark(X1), X2) → F(X1, X2)
MARK(h(X)) → ACTIVE(h(mark(X)))
F(X1, mark(X2)) → F(X1, X2)
ACTIVE(f(X, X)) → MARK(h(a))
MARK(g(X1, X2)) → MARK(X1)
G(active(X1), X2) → G(X1, X2)
MARK(g(X1, X2)) → ACTIVE(g(mark(X1), X2))
MARK(a) → ACTIVE(a)
MARK(f(X1, X2)) → MARK(X1)
G(X1, mark(X2)) → G(X1, X2)
G(X1, active(X2)) → G(X1, X2)
MARK(h(X)) → MARK(X)
ACTIVE(g(a, X)) → MARK(f(b, X))
H(mark(X)) → H(X)
MARK(h(X)) → H(mark(X))
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))

The TRS R consists of the following rules:

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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)

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:

F(active(X1), X2) → F(X1, X2)
ACTIVE(h(X)) → G(X, X)
H(active(X)) → H(X)
G(mark(X1), X2) → G(X1, X2)
MARK(g(X1, X2)) → G(mark(X1), X2)
MARK(b) → ACTIVE(b)
ACTIVE(f(X, X)) → H(a)
ACTIVE(g(a, X)) → F(b, X)
ACTIVE(h(X)) → MARK(g(X, X))
MARK(f(X1, X2)) → F(mark(X1), X2)
ACTIVE(a) → MARK(b)
F(X1, active(X2)) → F(X1, X2)
F(mark(X1), X2) → F(X1, X2)
MARK(h(X)) → ACTIVE(h(mark(X)))
F(X1, mark(X2)) → F(X1, X2)
ACTIVE(f(X, X)) → MARK(h(a))
MARK(g(X1, X2)) → MARK(X1)
G(active(X1), X2) → G(X1, X2)
MARK(g(X1, X2)) → ACTIVE(g(mark(X1), X2))
MARK(a) → ACTIVE(a)
MARK(f(X1, X2)) → MARK(X1)
G(X1, mark(X2)) → G(X1, X2)
G(X1, active(X2)) → G(X1, X2)
MARK(h(X)) → MARK(X)
ACTIVE(g(a, X)) → MARK(f(b, X))
H(mark(X)) → H(X)
MARK(h(X)) → H(mark(X))
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))

The TRS R consists of the following rules:

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 9 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ 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(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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)

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

          ↳ 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)

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(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

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

H(active(X)) → H(X)
H(mark(X)) → H(X)

The TRS R consists of the following rules:

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

H(active(X)) → H(X)
H(mark(X)) → H(X)

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

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

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

The TRS R consists of the following rules:

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

MARK(f(X1, X2)) → MARK(X1)
MARK(h(X)) → MARK(X)
MARK(g(X1, X2)) → MARK(X1)

Used ordering: POLO with Polynomial interpretation [25]:

POL(ACTIVE(x₁)) = 2·x₁
POL(MARK(x₁)) = 2·x₁
POL(a) = 0
POL(active(x₁)) = x₁
POL(b) = 0
POL(f(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(g(x₁, x₂)) = 2 + x₁ + x₂
POL(h(x₁)) = 2 + 2·x₁
POL(mark(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ Narrowing

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

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

The TRS R consists of the following rules:

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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)

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

MARK(g(g(x0, x1), y1)) → ACTIVE(g(active(g(mark(x0), x1)), y1))
MARK(g(b, y1)) → ACTIVE(g(active(b), y1))
MARK(g(f(x0, x1), y1)) → ACTIVE(g(active(f(mark(x0), x1)), y1))
MARK(g(h(x0), y1)) → ACTIVE(g(active(h(mark(x0))), y1))
MARK(g(y0, mark(x1))) → ACTIVE(g(mark(y0), x1))
MARK(g(x0, x1)) → ACTIVE(g(x0, x1))
MARK(g(a, y1)) → ACTIVE(g(active(a), y1))
MARK(g(y0, active(x1))) → ACTIVE(g(mark(y0), x1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ QDPOrderProof

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

MARK(h(X)) → ACTIVE(h(mark(X)))
MARK(g(f(x0, x1), y1)) → ACTIVE(g(active(f(mark(x0), x1)), y1))
MARK(g(h(x0), y1)) → ACTIVE(g(active(h(mark(x0))), y1))
ACTIVE(f(X, X)) → MARK(h(a))
MARK(g(a, y1)) → ACTIVE(g(active(a), y1))
MARK(g(y0, active(x1))) → ACTIVE(g(mark(y0), x1))
MARK(g(g(x0, x1), y1)) → ACTIVE(g(active(g(mark(x0), x1)), y1))
MARK(g(b, y1)) → ACTIVE(g(active(b), y1))
MARK(g(y0, mark(x1))) → ACTIVE(g(mark(y0), x1))
ACTIVE(h(X)) → MARK(g(X, X))
ACTIVE(g(a, X)) → MARK(f(b, X))
MARK(g(x0, x1)) → ACTIVE(g(x0, x1))
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))

The TRS R consists of the following rules:

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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)

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(f(x0, x1), y1)) → ACTIVE(g(active(f(mark(x0), x1)), y1))
MARK(g(h(x0), y1)) → ACTIVE(g(active(h(mark(x0))), y1))
MARK(g(g(x0, x1), y1)) → ACTIVE(g(active(g(mark(x0), x1)), y1))

The remaining pairs can at least be oriented weakly.

MARK(h(X)) → ACTIVE(h(mark(X)))
ACTIVE(f(X, X)) → MARK(h(a))
MARK(g(a, y1)) → ACTIVE(g(active(a), y1))
MARK(g(y0, active(x1))) → ACTIVE(g(mark(y0), x1))
MARK(g(b, y1)) → ACTIVE(g(active(b), y1))
MARK(g(y0, mark(x1))) → ACTIVE(g(mark(y0), x1))
ACTIVE(h(X)) → MARK(g(X, X))
ACTIVE(g(a, X)) → MARK(f(b, X))
MARK(g(x0, x1)) → ACTIVE(g(x0, x1))
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))

Used ordering: Matrix interpretation [3]:
Non-tuple symbols:

M( active(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	0	/

x₁

M( a ) =

/	0	\
\	0	/

M( g(x₁, x₂) ) =

/	1	\
\	0	/

/	1	0	\
\	0	0	/

x₁

/	0	0	\
\	0	0	/

x₂

M( f(x₁, x₂) ) =

/	1	\
\	0	/

/	0	0	\
\	0	0	/

x₁

/	0	0	\
\	0	0	/

x₂

M( h(x₁) ) =

/	1	\
\	0	/

/	1	0	\
\	1	0	/

x₁

M( mark(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	0	/

x₁

M( b ) =

/	0	\
\	0	/

Tuple symbols:

M( MARK(x₁) ) =

[

]

x₁

M( ACTIVE(x₁) ) =

[

]

x₁

Matrix type:
We used a basic matrix type which is not further parametrizeable.

As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

active(a) → mark(b)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(h(X)) → active(h(mark(X)))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(h(X)) → mark(g(X, X))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(b) → active(b)
mark(a) → active(a)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
g(mark(X1), X2) → g(X1, X2)
h(active(X)) → h(X)
h(mark(X)) → h(X)
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)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ RuleRemovalProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

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

MARK(h(X)) → ACTIVE(h(mark(X)))
MARK(g(b, y1)) → ACTIVE(g(active(b), y1))
MARK(g(y0, mark(x1))) → ACTIVE(g(mark(y0), x1))
ACTIVE(f(X, X)) → MARK(h(a))
ACTIVE(h(X)) → MARK(g(X, X))
ACTIVE(g(a, X)) → MARK(f(b, X))
MARK(g(x0, x1)) → ACTIVE(g(x0, x1))
MARK(g(a, y1)) → ACTIVE(g(active(a), y1))
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
MARK(g(y0, active(x1))) → ACTIVE(g(mark(y0), x1))

The TRS R consists of the following rules:

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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)

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