Termination w.r.t. Q proof of /tmp/tmpjPeSDx/Ex1_GM99

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

The set Q consists of the following terms:

f(active(x0), x1, x2)
f(x0, x1, mark(x2))
mark(a)
f(x0, active(x1), x2)
active(c)
f(x0, x1, active(x2))
f(x0, mark(x1), x2)
mark(c)
mark(f(x0, x1, x2))
active(f(a, b, x0))
f(mark(x0), x1, x2)
mark(b)

↳ 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(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)

The set Q consists of the following terms:

f(active(x0), x1, x2)
f(x0, x1, mark(x2))
mark(a)
f(x0, active(x1), x2)
active(c)
f(x0, x1, active(x2))
f(x0, mark(x1), x2)
mark(c)
mark(f(x0, x1, x2))
active(f(a, b, x0))
f(mark(x0), x1, x2)
mark(b)

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

The set Q consists of the following terms:

f(active(x0), x1, x2)
f(x0, x1, mark(x2))
mark(a)
f(x0, active(x1), x2)
active(c)
f(x0, x1, active(x2))
f(x0, mark(x1), x2)
mark(c)
mark(f(x0, x1, x2))
active(f(a, b, x0))
f(mark(x0), x1, x2)
mark(b)

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

The set Q consists of the following terms:

f(active(x0), x1, x2)
f(x0, x1, mark(x2))
mark(a)
f(x0, active(x1), x2)
active(c)
f(x0, x1, active(x2))
f(x0, mark(x1), x2)
mark(c)
mark(f(x0, x1, x2))
active(f(a, b, x0))
f(mark(x0), x1, x2)
mark(b)

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(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)

The set Q consists of the following terms:

f(active(x0), x1, x2)
f(x0, x1, mark(x2))
mark(a)
f(x0, active(x1), x2)
active(c)
f(x0, x1, active(x2))
f(x0, mark(x1), x2)
mark(c)
mark(f(x0, x1, x2))
active(f(a, b, x0))
f(mark(x0), x1, x2)
mark(b)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

          ↳ 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)

R is empty.
The set Q consists of the following terms:

f(active(x0), x1, x2)
f(x0, x1, mark(x2))
mark(a)
f(x0, active(x1), x2)
active(c)
f(x0, x1, active(x2))
f(x0, mark(x1), x2)
mark(c)
mark(f(x0, x1, x2))
active(f(a, b, x0))
f(mark(x0), x1, x2)
mark(b)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

f(active(x0), x1, x2)
f(x0, x1, mark(x2))
f(x0, active(x1), x2)
f(x0, x1, active(x2))
f(x0, mark(x1), x2)
f(mark(x0), x1, x2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QReductionProof

                  ↳ 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.
The set Q consists of the following terms:

mark(a)
active(c)
mark(c)
mark(f(x0, x1, x2))
active(f(a, b, x0))
mark(b)

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

            ↳ QDPOrderProof

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

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

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(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)

The set Q consists of the following terms:

f(active(x0), x1, x2)
f(x0, x1, mark(x2))
mark(a)
f(x0, active(x1), x2)
active(c)
f(x0, x1, active(x2))
f(x0, mark(x1), x2)
mark(c)
mark(f(x0, x1, x2))
active(f(a, b, x0))
f(mark(x0), x1, x2)
mark(b)

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(f(X1, X2, X3)) → MARK(X3)

The remaining pairs can at least be oriented weakly.

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

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

POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(a) = 0
POL(active(x₁)) = x₁
POL(b) = 0
POL(c) = 0
POL(f(x₁, x₂, x₃)) = 1 + x₃
POL(mark(x₁)) = x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ Instantiation

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

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

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(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)

The set Q consists of the following terms:

f(active(x0), x1, x2)
f(x0, x1, mark(x2))
mark(a)
f(x0, active(x1), x2)
active(c)
f(x0, x1, active(x2))
f(x0, mark(x1), x2)
mark(c)
mark(f(x0, x1, x2))
active(f(a, b, x0))
f(mark(x0), x1, x2)
mark(b)

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

MARK(f(z0, z0, z0)) → ACTIVE(f(z0, z0, mark(z0)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ Instantiation

                  ↳ QDP

                    ↳ Narrowing

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

MARK(f(z0, z0, z0)) → ACTIVE(f(z0, z0, mark(z0)))
ACTIVE(f(a, b, X)) → MARK(f(X, X, X))

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(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)

The set Q consists of the following terms:

f(active(x0), x1, x2)
f(x0, x1, mark(x2))
mark(a)
f(x0, active(x1), x2)
active(c)
f(x0, x1, active(x2))
f(x0, mark(x1), x2)
mark(c)
mark(f(x0, x1, x2))
active(f(a, b, x0))
f(mark(x0), x1, x2)
mark(b)

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule MARK(f(z0, z0, z0)) → ACTIVE(f(z0, z0, mark(z0))) at position [0] we obtained the following new rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ Instantiation

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

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

MARK(f(c, c, c)) → ACTIVE(f(c, c, active(c)))
MARK(f(x0, x0, x0)) → ACTIVE(f(x0, x0, x0))
ACTIVE(f(a, b, X)) → MARK(f(X, X, X))
MARK(f(f(x0, x1, x2), f(x0, x1, x2), f(x0, x1, x2))) → ACTIVE(f(f(x0, x1, x2), f(x0, x1, x2), active(f(x0, x1, mark(x2)))))
MARK(f(b, b, b)) → ACTIVE(f(b, b, active(b)))
MARK(f(a, a, a)) → ACTIVE(f(a, a, active(a)))

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(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)

The set Q consists of the following terms:

f(active(x0), x1, x2)
f(x0, x1, mark(x2))
mark(a)
f(x0, active(x1), x2)
active(c)
f(x0, x1, active(x2))
f(x0, mark(x1), x2)
mark(c)
mark(f(x0, x1, x2))
active(f(a, b, x0))
f(mark(x0), x1, x2)
mark(b)

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

            ↳ QDPOrderProof

              ↳ QDP

                ↳ Instantiation

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

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

MARK(f(c, c, c)) → ACTIVE(f(c, c, active(c)))
ACTIVE(f(a, b, X)) → MARK(f(X, X, X))
MARK(f(f(x0, x1, x2), f(x0, x1, x2), f(x0, x1, x2))) → ACTIVE(f(f(x0, x1, x2), f(x0, x1, x2), active(f(x0, x1, mark(x2)))))
MARK(f(b, b, b)) → ACTIVE(f(b, b, active(b)))
MARK(f(a, a, a)) → ACTIVE(f(a, a, active(a)))

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(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)

The set Q consists of the following terms:

f(active(x0), x1, x2)
f(x0, x1, mark(x2))
mark(a)
f(x0, active(x1), x2)
active(c)
f(x0, x1, active(x2))
f(x0, mark(x1), x2)
mark(c)
mark(f(x0, x1, x2))
active(f(a, b, x0))
f(mark(x0), x1, x2)
mark(b)

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

MARK(f(b, b, b)) → ACTIVE(f(b, b, b))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ Instantiation

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

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

MARK(f(c, c, c)) → ACTIVE(f(c, c, active(c)))
ACTIVE(f(a, b, X)) → MARK(f(X, X, X))
MARK(f(f(x0, x1, x2), f(x0, x1, x2), f(x0, x1, x2))) → ACTIVE(f(f(x0, x1, x2), f(x0, x1, x2), active(f(x0, x1, mark(x2)))))
MARK(f(b, b, b)) → ACTIVE(f(b, b, b))
MARK(f(a, a, a)) → ACTIVE(f(a, a, active(a)))

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(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)

The set Q consists of the following terms:

f(active(x0), x1, x2)
f(x0, x1, mark(x2))
mark(a)
f(x0, active(x1), x2)
active(c)
f(x0, x1, active(x2))
f(x0, mark(x1), x2)
mark(c)
mark(f(x0, x1, x2))
active(f(a, b, x0))
f(mark(x0), x1, x2)
mark(b)

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

            ↳ QDPOrderProof

              ↳ QDP

                ↳ Instantiation

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

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

MARK(f(c, c, c)) → ACTIVE(f(c, c, active(c)))
ACTIVE(f(a, b, X)) → MARK(f(X, X, X))
MARK(f(f(x0, x1, x2), f(x0, x1, x2), f(x0, x1, x2))) → ACTIVE(f(f(x0, x1, x2), f(x0, x1, x2), active(f(x0, x1, mark(x2)))))
MARK(f(a, a, a)) → ACTIVE(f(a, a, active(a)))

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(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)

The set Q consists of the following terms:

f(active(x0), x1, x2)
f(x0, x1, mark(x2))
mark(a)
f(x0, active(x1), x2)
active(c)
f(x0, x1, active(x2))
f(x0, mark(x1), x2)
mark(c)
mark(f(x0, x1, x2))
active(f(a, b, x0))
f(mark(x0), x1, x2)
mark(b)

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

MARK(f(a, a, a)) → ACTIVE(f(a, a, a))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ Instantiation

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

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

MARK(f(c, c, c)) → ACTIVE(f(c, c, active(c)))
ACTIVE(f(a, b, X)) → MARK(f(X, X, X))
MARK(f(f(x0, x1, x2), f(x0, x1, x2), f(x0, x1, x2))) → ACTIVE(f(f(x0, x1, x2), f(x0, x1, x2), active(f(x0, x1, mark(x2)))))
MARK(f(a, a, a)) → ACTIVE(f(a, a, a))

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(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)

The set Q consists of the following terms:

f(active(x0), x1, x2)
f(x0, x1, mark(x2))
mark(a)
f(x0, active(x1), x2)
active(c)
f(x0, x1, active(x2))
f(x0, mark(x1), x2)
mark(c)
mark(f(x0, x1, x2))
active(f(a, b, x0))
f(mark(x0), x1, x2)
mark(b)

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

            ↳ QDPOrderProof

              ↳ QDP

                ↳ Instantiation

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ ForwardInstantiation

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

MARK(f(c, c, c)) → ACTIVE(f(c, c, active(c)))
ACTIVE(f(a, b, X)) → MARK(f(X, X, X))
MARK(f(f(x0, x1, x2), f(x0, x1, x2), f(x0, x1, x2))) → ACTIVE(f(f(x0, x1, x2), f(x0, x1, x2), active(f(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(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)

The set Q consists of the following terms:

f(active(x0), x1, x2)
f(x0, x1, mark(x2))
mark(a)
f(x0, active(x1), x2)
active(c)
f(x0, x1, active(x2))
f(x0, mark(x1), x2)
mark(c)
mark(f(x0, x1, x2))
active(f(a, b, x0))
f(mark(x0), x1, x2)
mark(b)

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule ACTIVE(f(a, b, X)) → MARK(f(X, X, X)) we obtained the following new rules:

ACTIVE(f(a, b, c)) → MARK(f(c, c, c))
ACTIVE(f(a, b, f(y_0, y_1, y_2))) → MARK(f(f(y_0, y_1, y_2), f(y_0, y_1, y_2), f(y_0, y_1, y_2)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ Instantiation

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ ForwardInstantiation

                                              ↳ QDP

                                                ↳ QDPOrderProof

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

MARK(f(c, c, c)) → ACTIVE(f(c, c, active(c)))
MARK(f(f(x0, x1, x2), f(x0, x1, x2), f(x0, x1, x2))) → ACTIVE(f(f(x0, x1, x2), f(x0, x1, x2), active(f(x0, x1, mark(x2)))))
ACTIVE(f(a, b, c)) → MARK(f(c, c, c))
ACTIVE(f(a, b, f(y_0, y_1, y_2))) → MARK(f(f(y_0, y_1, y_2), f(y_0, y_1, y_2), f(y_0, y_1, y_2)))

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(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)

The set Q consists of the following terms:

f(active(x0), x1, x2)
f(x0, x1, mark(x2))
mark(a)
f(x0, active(x1), x2)
active(c)
f(x0, x1, active(x2))
f(x0, mark(x1), x2)
mark(c)
mark(f(x0, x1, x2))
active(f(a, b, x0))
f(mark(x0), x1, x2)
mark(b)

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(a, b, c)) → MARK(f(c, c, c))
ACTIVE(f(a, b, f(y_0, y_1, y_2))) → MARK(f(f(y_0, y_1, y_2), f(y_0, y_1, y_2), f(y_0, y_1, y_2)))

The remaining pairs can at least be oriented weakly.

MARK(f(c, c, c)) → ACTIVE(f(c, c, active(c)))
MARK(f(f(x0, x1, x2), f(x0, x1, x2), f(x0, x1, x2))) → ACTIVE(f(f(x0, x1, x2), f(x0, x1, x2), active(f(x0, x1, mark(x2)))))

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

M( active(x₁) ) =

/	0	\
\	0	/

/	0	0	\
\	0	1	/

x₁

M( c ) =

/	0	\
\	0	/

M( a ) =

/	0	\
\	1	/

M( f(x₁, ..., x₃) ) =

/	0	\
\	0	/

/	0	1	\
\	0	0	/

x₁

/	0	0	\
\	0	0	/

x₂

/	0	0	\
\	0	0	/

x₃

M( mark(x₁) ) =

/	1	\
\	1	/

/	0	0	\
\	0	1	/

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:

f(X1, X2, active(X3)) → f(X1, X2, X3)
f(mark(X1), X2, 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, mark(X3)) → f(X1, X2, X3)
f(X1, mark(X2), X3) → f(X1, X2, X3)
mark(c) → active(c)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ Instantiation

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ Narrowing

                              ↳ QDP

                                ↳ DependencyGraphProof

                                  ↳ QDP

                                    ↳ Narrowing

                                      ↳ QDP

                                        ↳ DependencyGraphProof

                                          ↳ QDP

                                            ↳ ForwardInstantiation

                                              ↳ QDP

                                                ↳ QDPOrderProof

                                                  ↳ QDP

                                                    ↳ DependencyGraphProof

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

MARK(f(c, c, c)) → ACTIVE(f(c, c, active(c)))
MARK(f(f(x0, x1, x2), f(x0, x1, x2), f(x0, x1, x2))) → ACTIVE(f(f(x0, x1, x2), f(x0, x1, x2), active(f(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(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)

The set Q consists of the following terms:

f(active(x0), x1, x2)
f(x0, x1, mark(x2))
mark(a)
f(x0, active(x1), x2)
active(c)
f(x0, x1, active(x2))
f(x0, mark(x1), x2)
mark(c)
mark(f(x0, x1, x2))
active(f(a, b, x0))
f(mark(x0), x1, x2)
mark(b)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.