Termination w.r.t. Q proof of /tmp/tmpA6dllE/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)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(a, b, X)) → F(X, X, X)
F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)
ACTIVE(f(X1, X2, X3)) → F(X1, X2, active(X3))
TOP(mark(X)) → PROPER(X)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
TOP(ok(X)) → ACTIVE(X)
PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X2)
ACTIVE(f(X1, X2, X3)) → ACTIVE(X3)
PROPER(f(X1, X2, X3)) → PROPER(X3)
TOP(mark(X)) → TOP(proper(X))
PROPER(f(X1, X2, X3)) → F(proper(X1), proper(X2), proper(X3))
TOP(ok(X)) → TOP(active(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)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(a, b, X)) → F(X, X, X)
F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)
ACTIVE(f(X1, X2, X3)) → F(X1, X2, active(X3))
TOP(mark(X)) → PROPER(X)
F(X1, X2, mark(X3)) → F(X1, X2, X3)
TOP(ok(X)) → ACTIVE(X)
PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X2)
ACTIVE(f(X1, X2, X3)) → ACTIVE(X3)
PROPER(f(X1, X2, X3)) → PROPER(X3)
TOP(mark(X)) → TOP(proper(X))
PROPER(f(X1, X2, X3)) → F(proper(X1), proper(X2), proper(X3))
TOP(ok(X)) → TOP(active(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)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)
F(X1, X2, mark(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)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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

          ↳ QDP

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → PROPER(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)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → PROPER(X3)

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

PROPER(f(X1, X2, X3)) → PROPER(X1)
The graph contains the following edges 1 > 1
PROPER(f(X1, X2, X3)) → PROPER(X2)
The graph contains the following edges 1 > 1
PROPER(f(X1, X2, X3)) → PROPER(X3)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

ACTIVE(f(X1, X2, X3)) → ACTIVE(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)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

ACTIVE(f(X1, X2, X3)) → ACTIVE(X3)

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

ACTIVE(f(X1, X2, X3)) → ACTIVE(X3)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

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

TOP(mark(X)) → TOP(proper(X))
TOP(ok(X)) → TOP(active(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)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(TOP(x₁)) = 2·x₁
POL(a) = 0
POL(active(x₁)) = 2·x₁
POL(b) = 0
POL(c) = 0
POL(f(x₁, x₂, x₃)) = x₁ + x₂ + 2·x₃
POL(mark(x₁)) = x₁
POL(ok(x₁)) = 2·x₁
POL(proper(x₁)) = x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

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

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

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

TOP(mark(b)) → TOP(ok(b))
TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))
TOP(mark(c)) → TOP(ok(c))
TOP(mark(a)) → TOP(ok(a))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

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

TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))
TOP(mark(b)) → TOP(ok(b))
TOP(mark(a)) → TOP(ok(a))
TOP(mark(c)) → TOP(ok(c))
TOP(ok(X)) → TOP(active(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)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)

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

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ Narrowing

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

TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))
TOP(mark(b)) → TOP(ok(b))
TOP(mark(a)) → TOP(ok(a))
TOP(ok(X)) → TOP(active(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)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule TOP(ok(X)) → TOP(active(X)) at position [0] we obtained the following new rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

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

TOP(ok(f(a, b, x0))) → TOP(mark(f(x0, x0, x0)))
TOP(ok(c)) → TOP(mark(a))
TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))
TOP(ok(c)) → TOP(mark(b))
TOP(mark(b)) → TOP(ok(b))
TOP(ok(f(x0, x1, x2))) → TOP(f(x0, x1, active(x2)))
TOP(mark(a)) → TOP(ok(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)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)

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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ SemLabProof

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

TOP(ok(f(a, b, x0))) → TOP(mark(f(x0, x0, x0)))
TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))
TOP(ok(f(x0, x1, x2))) → TOP(f(x0, x1, active(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)
active(f(X1, X2, X3)) → f(X1, X2, active(X3))
f(X1, X2, mark(X3)) → mark(f(X1, X2, X3))
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(a) → ok(a)
proper(b) → ok(b)
proper(c) → ok(c)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We found the following model for the rules of the TRS R. Interpretation over the domain with elements from 0 to 1.c: 0
active: 0
a: 0
f: 0
ok: x₀
mark: 0
b: 1
TOP: 0
proper: x₀
By semantic labelling [33] we obtain the following labelled TRS:Q DP problem:
The TRS P consists of the following rules:

TOP.0(ok.0(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-1-0(x0, x1, active.1(x2)))
TOP.0(ok.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-1-1(x0, x1, x2))) → TOP.0(f.1-1-0(x0, x1, active.1(x2)))
TOP.0(ok.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-0(x0, x1, active.1(x2)))
TOP.0(mark.0(f.1-1-1(x0, x1, x2))) → TOP.0(f.1-1-1(proper.1(x0), proper.1(x1), proper.1(x2)))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, x1, active.0(x2)))
TOP.0(mark.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-1(proper.0(x0), proper.0(x1), proper.1(x2)))
TOP.0(mark.0(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-1-1(proper.0(x0), proper.1(x1), proper.1(x2)))
TOP.0(ok.0(f.0-1-0(a., b., x0))) → TOP.0(mark.0(f.0-0-0(x0, x0, x0)))
TOP.0(mark.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(proper.1(x0), proper.1(x1), proper.0(x2)))
TOP.0(ok.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-0(x0, x1, active.1(x2)))
TOP.0(mark.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(proper.0(x0), proper.1(x1), proper.0(x2)))
TOP.0(mark.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.0(x2)))
TOP.0(mark.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(proper.1(x0), proper.0(x1), proper.0(x2)))
TOP.0(ok.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(x0, x1, active.0(x2)))
TOP.0(mark.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-1(proper.1(x0), proper.0(x1), proper.1(x2)))
TOP.0(ok.0(f.0-1-1(a., b., x0))) → TOP.0(mark.0(f.1-1-1(x0, x0, x0)))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, x1, active.0(x2)))

The TRS R consists of the following rules:

f.1-0-0(X1, X2, mark.1(X3)) → mark.0(f.1-0-1(X1, X2, X3))
proper.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(proper.1(X1), proper.0(X2), proper.0(X3))
active.0(f.0-1-0(X1, X2, X3)) → f.0-1-0(X1, X2, active.0(X3))
active.0(f.0-0-1(X1, X2, X3)) → f.0-0-0(X1, X2, active.1(X3))
active.0(f.0-1-0(a., b., X)) → mark.0(f.0-0-0(X, X, X))
active.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(X1, X2, active.0(X3))
proper.0(f.1-1-0(X1, X2, X3)) → f.1-1-0(proper.1(X1), proper.1(X2), proper.0(X3))
f.0-0-1(ok.0(X1), ok.0(X2), ok.1(X3)) → ok.0(f.0-0-1(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.0(X2), ok.0(X3)) → ok.0(f.0-0-0(X1, X2, X3))
f.1-0-0(X1, X2, mark.0(X3)) → mark.0(f.1-0-0(X1, X2, X3))
active.0(f.0-1-1(a., b., X)) → mark.0(f.1-1-1(X, X, X))
proper.0(f.0-1-0(X1, X2, X3)) → f.0-1-0(proper.0(X1), proper.1(X2), proper.0(X3))
proper.0(c.) → ok.0(c.)
f.1-0-0(ok.1(X1), ok.0(X2), ok.0(X3)) → ok.0(f.1-0-0(X1, X2, X3))
f.0-0-0(X1, X2, mark.0(X3)) → mark.0(f.0-0-0(X1, X2, X3))
proper.0(f.0-1-1(X1, X2, X3)) → f.0-1-1(proper.0(X1), proper.1(X2), proper.1(X3))
active.0(f.1-1-0(X1, X2, X3)) → f.1-1-0(X1, X2, active.0(X3))
proper.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.0(X3))
f.1-0-1(ok.1(X1), ok.0(X2), ok.1(X3)) → ok.0(f.1-0-1(X1, X2, X3))
f.1-1-0(X1, X2, mark.0(X3)) → mark.0(f.1-1-0(X1, X2, X3))
proper.1(b.) → ok.1(b.)
f.1-1-0(X1, X2, mark.1(X3)) → mark.0(f.1-1-1(X1, X2, X3))
f.1-1-0(ok.1(X1), ok.1(X2), ok.0(X3)) → ok.0(f.1-1-0(X1, X2, X3))
active.0(c.) → mark.1(b.)
f.0-1-1(ok.0(X1), ok.1(X2), ok.1(X3)) → ok.0(f.0-1-1(X1, X2, X3))
proper.0(f.1-0-1(X1, X2, X3)) → f.1-0-1(proper.1(X1), proper.0(X2), proper.1(X3))
active.0(f.1-0-1(X1, X2, X3)) → f.1-0-0(X1, X2, active.1(X3))
f.0-1-0(ok.0(X1), ok.1(X2), ok.0(X3)) → ok.0(f.0-1-0(X1, X2, X3))
active.0(f.1-1-1(X1, X2, X3)) → f.1-1-0(X1, X2, active.1(X3))
proper.0(a.) → ok.0(a.)
f.0-1-0(X1, X2, mark.0(X3)) → mark.0(f.0-1-0(X1, X2, X3))
proper.0(f.0-0-1(X1, X2, X3)) → f.0-0-1(proper.0(X1), proper.0(X2), proper.1(X3))
f.0-0-0(X1, X2, mark.1(X3)) → mark.0(f.0-0-1(X1, X2, X3))
active.0(f.0-1-1(X1, X2, X3)) → f.0-1-0(X1, X2, active.1(X3))
f.0-1-0(X1, X2, mark.1(X3)) → mark.0(f.0-1-1(X1, X2, X3))
active.0(c.) → mark.0(a.)
proper.0(f.1-1-1(X1, X2, X3)) → f.1-1-1(proper.1(X1), proper.1(X2), proper.1(X3))
f.1-1-1(ok.1(X1), ok.1(X2), ok.1(X3)) → ok.0(f.1-1-1(X1, X2, X3))
active.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(X1, X2, active.0(X3))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ SemLabProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

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

TOP.0(ok.0(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-1-0(x0, x1, active.1(x2)))
TOP.0(ok.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-1-1(x0, x1, x2))) → TOP.0(f.1-1-0(x0, x1, active.1(x2)))
TOP.0(ok.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-0(x0, x1, active.1(x2)))
TOP.0(mark.0(f.1-1-1(x0, x1, x2))) → TOP.0(f.1-1-1(proper.1(x0), proper.1(x1), proper.1(x2)))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, x1, active.0(x2)))
TOP.0(mark.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-1(proper.0(x0), proper.0(x1), proper.1(x2)))
TOP.0(mark.0(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-1-1(proper.0(x0), proper.1(x1), proper.1(x2)))
TOP.0(ok.0(f.0-1-0(a., b., x0))) → TOP.0(mark.0(f.0-0-0(x0, x0, x0)))
TOP.0(mark.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(proper.1(x0), proper.1(x1), proper.0(x2)))
TOP.0(ok.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-0(x0, x1, active.1(x2)))
TOP.0(mark.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(proper.0(x0), proper.1(x1), proper.0(x2)))
TOP.0(mark.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.0(x2)))
TOP.0(mark.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(proper.1(x0), proper.0(x1), proper.0(x2)))
TOP.0(ok.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(x0, x1, active.0(x2)))
TOP.0(mark.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-1(proper.1(x0), proper.0(x1), proper.1(x2)))
TOP.0(ok.0(f.0-1-1(a., b., x0))) → TOP.0(mark.0(f.1-1-1(x0, x0, x0)))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, x1, active.0(x2)))

The TRS R consists of the following rules:

f.1-0-0(X1, X2, mark.1(X3)) → mark.0(f.1-0-1(X1, X2, X3))
proper.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(proper.1(X1), proper.0(X2), proper.0(X3))
active.0(f.0-1-0(X1, X2, X3)) → f.0-1-0(X1, X2, active.0(X3))
active.0(f.0-0-1(X1, X2, X3)) → f.0-0-0(X1, X2, active.1(X3))
active.0(f.0-1-0(a., b., X)) → mark.0(f.0-0-0(X, X, X))
active.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(X1, X2, active.0(X3))
proper.0(f.1-1-0(X1, X2, X3)) → f.1-1-0(proper.1(X1), proper.1(X2), proper.0(X3))
f.0-0-1(ok.0(X1), ok.0(X2), ok.1(X3)) → ok.0(f.0-0-1(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.0(X2), ok.0(X3)) → ok.0(f.0-0-0(X1, X2, X3))
f.1-0-0(X1, X2, mark.0(X3)) → mark.0(f.1-0-0(X1, X2, X3))
active.0(f.0-1-1(a., b., X)) → mark.0(f.1-1-1(X, X, X))
proper.0(f.0-1-0(X1, X2, X3)) → f.0-1-0(proper.0(X1), proper.1(X2), proper.0(X3))
proper.0(c.) → ok.0(c.)
f.1-0-0(ok.1(X1), ok.0(X2), ok.0(X3)) → ok.0(f.1-0-0(X1, X2, X3))
f.0-0-0(X1, X2, mark.0(X3)) → mark.0(f.0-0-0(X1, X2, X3))
proper.0(f.0-1-1(X1, X2, X3)) → f.0-1-1(proper.0(X1), proper.1(X2), proper.1(X3))
active.0(f.1-1-0(X1, X2, X3)) → f.1-1-0(X1, X2, active.0(X3))
proper.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.0(X3))
f.1-0-1(ok.1(X1), ok.0(X2), ok.1(X3)) → ok.0(f.1-0-1(X1, X2, X3))
f.1-1-0(X1, X2, mark.0(X3)) → mark.0(f.1-1-0(X1, X2, X3))
proper.1(b.) → ok.1(b.)
f.1-1-0(X1, X2, mark.1(X3)) → mark.0(f.1-1-1(X1, X2, X3))
f.1-1-0(ok.1(X1), ok.1(X2), ok.0(X3)) → ok.0(f.1-1-0(X1, X2, X3))
active.0(c.) → mark.1(b.)
f.0-1-1(ok.0(X1), ok.1(X2), ok.1(X3)) → ok.0(f.0-1-1(X1, X2, X3))
proper.0(f.1-0-1(X1, X2, X3)) → f.1-0-1(proper.1(X1), proper.0(X2), proper.1(X3))
active.0(f.1-0-1(X1, X2, X3)) → f.1-0-0(X1, X2, active.1(X3))
f.0-1-0(ok.0(X1), ok.1(X2), ok.0(X3)) → ok.0(f.0-1-0(X1, X2, X3))
active.0(f.1-1-1(X1, X2, X3)) → f.1-1-0(X1, X2, active.1(X3))
proper.0(a.) → ok.0(a.)
f.0-1-0(X1, X2, mark.0(X3)) → mark.0(f.0-1-0(X1, X2, X3))
proper.0(f.0-0-1(X1, X2, X3)) → f.0-0-1(proper.0(X1), proper.0(X2), proper.1(X3))
f.0-0-0(X1, X2, mark.1(X3)) → mark.0(f.0-0-1(X1, X2, X3))
active.0(f.0-1-1(X1, X2, X3)) → f.0-1-0(X1, X2, active.1(X3))
f.0-1-0(X1, X2, mark.1(X3)) → mark.0(f.0-1-1(X1, X2, X3))
active.0(c.) → mark.0(a.)
proper.0(f.1-1-1(X1, X2, X3)) → f.1-1-1(proper.1(X1), proper.1(X2), proper.1(X3))
f.1-1-1(ok.1(X1), ok.1(X2), ok.1(X3)) → ok.0(f.1-1-1(X1, X2, X3))
active.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(X1, X2, active.0(X3))

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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ SemLabProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ QDPOrderProof

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

TOP.0(ok.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(x0, x1, active.0(x2)))
TOP.0(mark.0(f.1-1-1(x0, x1, x2))) → TOP.0(f.1-1-1(proper.1(x0), proper.1(x1), proper.1(x2)))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, x1, active.0(x2)))
TOP.0(mark.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-1(proper.0(x0), proper.0(x1), proper.1(x2)))
TOP.0(mark.0(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-1-1(proper.0(x0), proper.1(x1), proper.1(x2)))
TOP.0(ok.0(f.0-1-0(a., b., x0))) → TOP.0(mark.0(f.0-0-0(x0, x0, x0)))
TOP.0(mark.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(proper.1(x0), proper.1(x1), proper.0(x2)))
TOP.0(mark.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(proper.0(x0), proper.1(x1), proper.0(x2)))
TOP.0(mark.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.0(x2)))
TOP.0(mark.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(proper.1(x0), proper.0(x1), proper.0(x2)))
TOP.0(ok.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(x0, x1, active.0(x2)))
TOP.0(mark.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-1(proper.1(x0), proper.0(x1), proper.1(x2)))
TOP.0(ok.0(f.0-1-1(a., b., x0))) → TOP.0(mark.0(f.1-1-1(x0, x0, x0)))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, x1, active.0(x2)))

The TRS R consists of the following rules:

f.1-0-0(X1, X2, mark.1(X3)) → mark.0(f.1-0-1(X1, X2, X3))
proper.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(proper.1(X1), proper.0(X2), proper.0(X3))
active.0(f.0-1-0(X1, X2, X3)) → f.0-1-0(X1, X2, active.0(X3))
active.0(f.0-0-1(X1, X2, X3)) → f.0-0-0(X1, X2, active.1(X3))
active.0(f.0-1-0(a., b., X)) → mark.0(f.0-0-0(X, X, X))
active.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(X1, X2, active.0(X3))
proper.0(f.1-1-0(X1, X2, X3)) → f.1-1-0(proper.1(X1), proper.1(X2), proper.0(X3))
f.0-0-1(ok.0(X1), ok.0(X2), ok.1(X3)) → ok.0(f.0-0-1(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.0(X2), ok.0(X3)) → ok.0(f.0-0-0(X1, X2, X3))
f.1-0-0(X1, X2, mark.0(X3)) → mark.0(f.1-0-0(X1, X2, X3))
active.0(f.0-1-1(a., b., X)) → mark.0(f.1-1-1(X, X, X))
proper.0(f.0-1-0(X1, X2, X3)) → f.0-1-0(proper.0(X1), proper.1(X2), proper.0(X3))
proper.0(c.) → ok.0(c.)
f.1-0-0(ok.1(X1), ok.0(X2), ok.0(X3)) → ok.0(f.1-0-0(X1, X2, X3))
f.0-0-0(X1, X2, mark.0(X3)) → mark.0(f.0-0-0(X1, X2, X3))
proper.0(f.0-1-1(X1, X2, X3)) → f.0-1-1(proper.0(X1), proper.1(X2), proper.1(X3))
active.0(f.1-1-0(X1, X2, X3)) → f.1-1-0(X1, X2, active.0(X3))
proper.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.0(X3))
f.1-0-1(ok.1(X1), ok.0(X2), ok.1(X3)) → ok.0(f.1-0-1(X1, X2, X3))
f.1-1-0(X1, X2, mark.0(X3)) → mark.0(f.1-1-0(X1, X2, X3))
proper.1(b.) → ok.1(b.)
f.1-1-0(X1, X2, mark.1(X3)) → mark.0(f.1-1-1(X1, X2, X3))
f.1-1-0(ok.1(X1), ok.1(X2), ok.0(X3)) → ok.0(f.1-1-0(X1, X2, X3))
active.0(c.) → mark.1(b.)
f.0-1-1(ok.0(X1), ok.1(X2), ok.1(X3)) → ok.0(f.0-1-1(X1, X2, X3))
proper.0(f.1-0-1(X1, X2, X3)) → f.1-0-1(proper.1(X1), proper.0(X2), proper.1(X3))
active.0(f.1-0-1(X1, X2, X3)) → f.1-0-0(X1, X2, active.1(X3))
f.0-1-0(ok.0(X1), ok.1(X2), ok.0(X3)) → ok.0(f.0-1-0(X1, X2, X3))
active.0(f.1-1-1(X1, X2, X3)) → f.1-1-0(X1, X2, active.1(X3))
proper.0(a.) → ok.0(a.)
f.0-1-0(X1, X2, mark.0(X3)) → mark.0(f.0-1-0(X1, X2, X3))
proper.0(f.0-0-1(X1, X2, X3)) → f.0-0-1(proper.0(X1), proper.0(X2), proper.1(X3))
f.0-0-0(X1, X2, mark.1(X3)) → mark.0(f.0-0-1(X1, X2, X3))
active.0(f.0-1-1(X1, X2, X3)) → f.0-1-0(X1, X2, active.1(X3))
f.0-1-0(X1, X2, mark.1(X3)) → mark.0(f.0-1-1(X1, X2, X3))
active.0(c.) → mark.0(a.)
proper.0(f.1-1-1(X1, X2, X3)) → f.1-1-1(proper.1(X1), proper.1(X2), proper.1(X3))
f.1-1-1(ok.1(X1), ok.1(X2), ok.1(X3)) → ok.0(f.1-1-1(X1, X2, X3))
active.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(X1, X2, active.0(X3))

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.

TOP.0(mark.0(f.1-1-1(x0, x1, x2))) → TOP.0(f.1-1-1(proper.1(x0), proper.1(x1), proper.1(x2)))
TOP.0(mark.0(f.0-0-1(x0, x1, x2))) → TOP.0(f.0-0-1(proper.0(x0), proper.0(x1), proper.1(x2)))
TOP.0(mark.0(f.0-1-1(x0, x1, x2))) → TOP.0(f.0-1-1(proper.0(x0), proper.1(x1), proper.1(x2)))
TOP.0(mark.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(proper.1(x0), proper.1(x1), proper.0(x2)))
TOP.0(mark.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(proper.0(x0), proper.1(x1), proper.0(x2)))
TOP.0(mark.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(proper.0(x0), proper.0(x1), proper.0(x2)))
TOP.0(mark.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(proper.1(x0), proper.0(x1), proper.0(x2)))
TOP.0(mark.0(f.1-0-1(x0, x1, x2))) → TOP.0(f.1-0-1(proper.1(x0), proper.0(x1), proper.1(x2)))
TOP.0(ok.0(f.0-1-1(a., b., x0))) → TOP.0(mark.0(f.1-1-1(x0, x0, x0)))

The remaining pairs can at least be oriented weakly.

TOP.0(ok.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.0-1-0(a., b., x0))) → TOP.0(mark.0(f.0-0-0(x0, x0, x0)))
TOP.0(ok.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, x1, active.0(x2)))

Used ordering: Polynomial interpretation [25]:

POL(TOP.0(x₁)) = x₁
POL(a.) = 0
POL(active.0(x₁)) = x₁
POL(active.1(x₁)) = 0
POL(b.) = 1
POL(c.) = 1
POL(f.0-0-0(x₁, x₂, x₃)) = 1 + x₃
POL(f.0-0-1(x₁, x₂, x₃)) = 1
POL(f.0-1-0(x₁, x₂, x₃)) = 1 + x₂ + x₃
POL(f.0-1-1(x₁, x₂, x₃)) = 1 + x₂
POL(f.1-0-0(x₁, x₂, x₃)) = x₂ + x₃
POL(f.1-0-1(x₁, x₂, x₃)) = x₂
POL(f.1-1-0(x₁, x₂, x₃)) = x₃
POL(f.1-1-1(x₁, x₂, x₃)) = 0
POL(mark.0(x₁)) = 1 + x₁
POL(mark.1(x₁)) = 1
POL(ok.0(x₁)) = x₁
POL(ok.1(x₁)) = x₁
POL(proper.0(x₁)) = x₁
POL(proper.1(x₁)) = x₁

The following usable rules [17] were oriented:

f.1-1-1(ok.1(X1), ok.1(X2), ok.1(X3)) → ok.0(f.1-1-1(X1, X2, X3))
f.1-0-0(ok.1(X1), ok.0(X2), ok.0(X3)) → ok.0(f.1-0-0(X1, X2, X3))
active.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(X1, X2, active.0(X3))
f.0-0-0(X1, X2, mark.0(X3)) → mark.0(f.0-0-0(X1, X2, X3))
proper.0(f.0-1-1(X1, X2, X3)) → f.0-1-1(proper.0(X1), proper.1(X2), proper.1(X3))
active.0(f.1-1-0(X1, X2, X3)) → f.1-1-0(X1, X2, active.0(X3))
f.1-0-0(X1, X2, mark.0(X3)) → mark.0(f.1-0-0(X1, X2, X3))
active.0(f.0-1-1(X1, X2, X3)) → f.0-1-0(X1, X2, active.1(X3))
f.0-1-0(X1, X2, mark.1(X3)) → mark.0(f.0-1-1(X1, X2, X3))
active.0(f.0-1-1(a., b., X)) → mark.0(f.1-1-1(X, X, X))
active.0(c.) → mark.0(a.)
proper.0(f.0-1-0(X1, X2, X3)) → f.0-1-0(proper.0(X1), proper.1(X2), proper.0(X3))
proper.0(c.) → ok.0(c.)
proper.0(f.1-1-1(X1, X2, X3)) → f.1-1-1(proper.1(X1), proper.1(X2), proper.1(X3))
f.1-1-0(X1, X2, mark.1(X3)) → mark.0(f.1-1-1(X1, X2, X3))
proper.0(a.) → ok.0(a.)
f.0-1-0(X1, X2, mark.0(X3)) → mark.0(f.0-1-0(X1, X2, X3))
f.1-1-0(ok.1(X1), ok.1(X2), ok.0(X3)) → ok.0(f.1-1-0(X1, X2, X3))
proper.0(f.0-0-1(X1, X2, X3)) → f.0-0-1(proper.0(X1), proper.0(X2), proper.1(X3))
active.0(c.) → mark.1(b.)
f.0-1-1(ok.0(X1), ok.1(X2), ok.1(X3)) → ok.0(f.0-1-1(X1, X2, X3))
f.0-0-0(X1, X2, mark.1(X3)) → mark.0(f.0-0-1(X1, X2, X3))
proper.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.0(X3))
proper.0(f.1-0-1(X1, X2, X3)) → f.1-0-1(proper.1(X1), proper.0(X2), proper.1(X3))
f.1-0-1(ok.1(X1), ok.0(X2), ok.1(X3)) → ok.0(f.1-0-1(X1, X2, X3))
active.0(f.1-0-1(X1, X2, X3)) → f.1-0-0(X1, X2, active.1(X3))
f.0-1-0(ok.0(X1), ok.1(X2), ok.0(X3)) → ok.0(f.0-1-0(X1, X2, X3))
f.1-1-0(X1, X2, mark.0(X3)) → mark.0(f.1-1-0(X1, X2, X3))
proper.1(b.) → ok.1(b.)
active.0(f.1-1-1(X1, X2, X3)) → f.1-1-0(X1, X2, active.1(X3))
f.1-0-0(X1, X2, mark.1(X3)) → mark.0(f.1-0-1(X1, X2, X3))
proper.0(f.1-1-0(X1, X2, X3)) → f.1-1-0(proper.1(X1), proper.1(X2), proper.0(X3))
active.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(X1, X2, active.0(X3))
f.0-0-0(ok.0(X1), ok.0(X2), ok.0(X3)) → ok.0(f.0-0-0(X1, X2, X3))
f.0-0-1(ok.0(X1), ok.0(X2), ok.1(X3)) → ok.0(f.0-0-1(X1, X2, X3))
active.0(f.0-1-0(X1, X2, X3)) → f.0-1-0(X1, X2, active.0(X3))
proper.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(proper.1(X1), proper.0(X2), proper.0(X3))
active.0(f.0-1-0(a., b., X)) → mark.0(f.0-0-0(X, X, X))
active.0(f.0-0-1(X1, X2, X3)) → f.0-0-0(X1, X2, active.1(X3))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ SemLabProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ QDPOrderProof

                                          ↳ QDP

                                            ↳ DependencyGraphProof

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

TOP.0(ok.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.0-1-0(a., b., x0))) → TOP.0(mark.0(f.0-0-0(x0, x0, x0)))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, x1, active.0(x2)))

The TRS R consists of the following rules:

f.1-0-0(X1, X2, mark.1(X3)) → mark.0(f.1-0-1(X1, X2, X3))
proper.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(proper.1(X1), proper.0(X2), proper.0(X3))
active.0(f.0-1-0(X1, X2, X3)) → f.0-1-0(X1, X2, active.0(X3))
active.0(f.0-0-1(X1, X2, X3)) → f.0-0-0(X1, X2, active.1(X3))
active.0(f.0-1-0(a., b., X)) → mark.0(f.0-0-0(X, X, X))
active.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(X1, X2, active.0(X3))
proper.0(f.1-1-0(X1, X2, X3)) → f.1-1-0(proper.1(X1), proper.1(X2), proper.0(X3))
f.0-0-1(ok.0(X1), ok.0(X2), ok.1(X3)) → ok.0(f.0-0-1(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.0(X2), ok.0(X3)) → ok.0(f.0-0-0(X1, X2, X3))
f.1-0-0(X1, X2, mark.0(X3)) → mark.0(f.1-0-0(X1, X2, X3))
active.0(f.0-1-1(a., b., X)) → mark.0(f.1-1-1(X, X, X))
proper.0(f.0-1-0(X1, X2, X3)) → f.0-1-0(proper.0(X1), proper.1(X2), proper.0(X3))
proper.0(c.) → ok.0(c.)
f.1-0-0(ok.1(X1), ok.0(X2), ok.0(X3)) → ok.0(f.1-0-0(X1, X2, X3))
f.0-0-0(X1, X2, mark.0(X3)) → mark.0(f.0-0-0(X1, X2, X3))
proper.0(f.0-1-1(X1, X2, X3)) → f.0-1-1(proper.0(X1), proper.1(X2), proper.1(X3))
active.0(f.1-1-0(X1, X2, X3)) → f.1-1-0(X1, X2, active.0(X3))
proper.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.0(X3))
f.1-0-1(ok.1(X1), ok.0(X2), ok.1(X3)) → ok.0(f.1-0-1(X1, X2, X3))
f.1-1-0(X1, X2, mark.0(X3)) → mark.0(f.1-1-0(X1, X2, X3))
proper.1(b.) → ok.1(b.)
f.1-1-0(X1, X2, mark.1(X3)) → mark.0(f.1-1-1(X1, X2, X3))
f.1-1-0(ok.1(X1), ok.1(X2), ok.0(X3)) → ok.0(f.1-1-0(X1, X2, X3))
active.0(c.) → mark.1(b.)
f.0-1-1(ok.0(X1), ok.1(X2), ok.1(X3)) → ok.0(f.0-1-1(X1, X2, X3))
proper.0(f.1-0-1(X1, X2, X3)) → f.1-0-1(proper.1(X1), proper.0(X2), proper.1(X3))
active.0(f.1-0-1(X1, X2, X3)) → f.1-0-0(X1, X2, active.1(X3))
f.0-1-0(ok.0(X1), ok.1(X2), ok.0(X3)) → ok.0(f.0-1-0(X1, X2, X3))
active.0(f.1-1-1(X1, X2, X3)) → f.1-1-0(X1, X2, active.1(X3))
proper.0(a.) → ok.0(a.)
f.0-1-0(X1, X2, mark.0(X3)) → mark.0(f.0-1-0(X1, X2, X3))
proper.0(f.0-0-1(X1, X2, X3)) → f.0-0-1(proper.0(X1), proper.0(X2), proper.1(X3))
f.0-0-0(X1, X2, mark.1(X3)) → mark.0(f.0-0-1(X1, X2, X3))
active.0(f.0-1-1(X1, X2, X3)) → f.0-1-0(X1, X2, active.1(X3))
f.0-1-0(X1, X2, mark.1(X3)) → mark.0(f.0-1-1(X1, X2, X3))
active.0(c.) → mark.0(a.)
proper.0(f.1-1-1(X1, X2, X3)) → f.1-1-1(proper.1(X1), proper.1(X2), proper.1(X3))
f.1-1-1(ok.1(X1), ok.1(X2), ok.1(X3)) → ok.0(f.1-1-1(X1, X2, X3))
active.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(X1, X2, active.0(X3))

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

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ SemLabProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ QDPOrderProof

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ QDPOrderProof

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

TOP.0(ok.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, x1, active.0(x2)))

The TRS R consists of the following rules:

f.1-0-0(X1, X2, mark.1(X3)) → mark.0(f.1-0-1(X1, X2, X3))
proper.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(proper.1(X1), proper.0(X2), proper.0(X3))
active.0(f.0-1-0(X1, X2, X3)) → f.0-1-0(X1, X2, active.0(X3))
active.0(f.0-0-1(X1, X2, X3)) → f.0-0-0(X1, X2, active.1(X3))
active.0(f.0-1-0(a., b., X)) → mark.0(f.0-0-0(X, X, X))
active.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(X1, X2, active.0(X3))
proper.0(f.1-1-0(X1, X2, X3)) → f.1-1-0(proper.1(X1), proper.1(X2), proper.0(X3))
f.0-0-1(ok.0(X1), ok.0(X2), ok.1(X3)) → ok.0(f.0-0-1(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.0(X2), ok.0(X3)) → ok.0(f.0-0-0(X1, X2, X3))
f.1-0-0(X1, X2, mark.0(X3)) → mark.0(f.1-0-0(X1, X2, X3))
active.0(f.0-1-1(a., b., X)) → mark.0(f.1-1-1(X, X, X))
proper.0(f.0-1-0(X1, X2, X3)) → f.0-1-0(proper.0(X1), proper.1(X2), proper.0(X3))
proper.0(c.) → ok.0(c.)
f.1-0-0(ok.1(X1), ok.0(X2), ok.0(X3)) → ok.0(f.1-0-0(X1, X2, X3))
f.0-0-0(X1, X2, mark.0(X3)) → mark.0(f.0-0-0(X1, X2, X3))
proper.0(f.0-1-1(X1, X2, X3)) → f.0-1-1(proper.0(X1), proper.1(X2), proper.1(X3))
active.0(f.1-1-0(X1, X2, X3)) → f.1-1-0(X1, X2, active.0(X3))
proper.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.0(X3))
f.1-0-1(ok.1(X1), ok.0(X2), ok.1(X3)) → ok.0(f.1-0-1(X1, X2, X3))
f.1-1-0(X1, X2, mark.0(X3)) → mark.0(f.1-1-0(X1, X2, X3))
proper.1(b.) → ok.1(b.)
f.1-1-0(X1, X2, mark.1(X3)) → mark.0(f.1-1-1(X1, X2, X3))
f.1-1-0(ok.1(X1), ok.1(X2), ok.0(X3)) → ok.0(f.1-1-0(X1, X2, X3))
active.0(c.) → mark.1(b.)
f.0-1-1(ok.0(X1), ok.1(X2), ok.1(X3)) → ok.0(f.0-1-1(X1, X2, X3))
proper.0(f.1-0-1(X1, X2, X3)) → f.1-0-1(proper.1(X1), proper.0(X2), proper.1(X3))
active.0(f.1-0-1(X1, X2, X3)) → f.1-0-0(X1, X2, active.1(X3))
f.0-1-0(ok.0(X1), ok.1(X2), ok.0(X3)) → ok.0(f.0-1-0(X1, X2, X3))
active.0(f.1-1-1(X1, X2, X3)) → f.1-1-0(X1, X2, active.1(X3))
proper.0(a.) → ok.0(a.)
f.0-1-0(X1, X2, mark.0(X3)) → mark.0(f.0-1-0(X1, X2, X3))
proper.0(f.0-0-1(X1, X2, X3)) → f.0-0-1(proper.0(X1), proper.0(X2), proper.1(X3))
f.0-0-0(X1, X2, mark.1(X3)) → mark.0(f.0-0-1(X1, X2, X3))
active.0(f.0-1-1(X1, X2, X3)) → f.0-1-0(X1, X2, active.1(X3))
f.0-1-0(X1, X2, mark.1(X3)) → mark.0(f.0-1-1(X1, X2, X3))
active.0(c.) → mark.0(a.)
proper.0(f.1-1-1(X1, X2, X3)) → f.1-1-1(proper.1(X1), proper.1(X2), proper.1(X3))
f.1-1-1(ok.1(X1), ok.1(X2), ok.1(X3)) → ok.0(f.1-1-1(X1, X2, X3))
active.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(X1, X2, active.0(X3))

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.

TOP.0(ok.0(f.1-1-0(x0, x1, x2))) → TOP.0(f.1-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.1-0-0(x0, x1, x2))) → TOP.0(f.1-0-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.0-1-0(x0, x1, x2))) → TOP.0(f.0-1-0(x0, x1, active.0(x2)))
TOP.0(ok.0(f.0-0-0(x0, x1, x2))) → TOP.0(f.0-0-0(x0, x1, active.0(x2)))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25]:

POL(TOP.0(x₁)) = x₁
POL(a.) = 0
POL(active.0(x₁)) = 0
POL(active.1(x₁)) = 0
POL(b.) = 0
POL(c.) = 0
POL(f.0-0-0(x₁, x₂, x₃)) = 1 + x₁ + x₂
POL(f.0-0-1(x₁, x₂, x₃)) = 0
POL(f.0-1-0(x₁, x₂, x₃)) = x₁
POL(f.0-1-1(x₁, x₂, x₃)) = 0
POL(f.1-0-0(x₁, x₂, x₃)) = 1 + x₁
POL(f.1-0-1(x₁, x₂, x₃)) = 0
POL(f.1-1-0(x₁, x₂, x₃)) = 1 + x₂
POL(f.1-1-1(x₁, x₂, x₃)) = 0
POL(mark.0(x₁)) = 0
POL(mark.1(x₁)) = 0
POL(ok.0(x₁)) = 1 + x₁
POL(ok.1(x₁)) = 1 + x₁

The following usable rules [17] were oriented:

f.1-0-0(ok.1(X1), ok.0(X2), ok.0(X3)) → ok.0(f.1-0-0(X1, X2, X3))
f.0-0-0(X1, X2, mark.0(X3)) → mark.0(f.0-0-0(X1, X2, X3))
f.1-0-0(X1, X2, mark.0(X3)) → mark.0(f.1-0-0(X1, X2, X3))
f.0-1-0(X1, X2, mark.1(X3)) → mark.0(f.0-1-1(X1, X2, X3))
f.1-1-0(X1, X2, mark.1(X3)) → mark.0(f.1-1-1(X1, X2, X3))
f.1-1-0(ok.1(X1), ok.1(X2), ok.0(X3)) → ok.0(f.1-1-0(X1, X2, X3))
f.0-1-0(X1, X2, mark.0(X3)) → mark.0(f.0-1-0(X1, X2, X3))
f.0-0-0(X1, X2, mark.1(X3)) → mark.0(f.0-0-1(X1, X2, X3))
f.1-1-0(X1, X2, mark.0(X3)) → mark.0(f.1-1-0(X1, X2, X3))
f.0-1-0(ok.0(X1), ok.1(X2), ok.0(X3)) → ok.0(f.0-1-0(X1, X2, X3))
f.1-0-0(X1, X2, mark.1(X3)) → mark.0(f.1-0-1(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.0(X2), ok.0(X3)) → ok.0(f.0-0-0(X1, X2, X3))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ SemLabProof

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ QDPOrderProof

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ QDPOrderProof

                                                  ↳ QDP

                                                    ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f.1-0-0(X1, X2, mark.1(X3)) → mark.0(f.1-0-1(X1, X2, X3))
proper.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(proper.1(X1), proper.0(X2), proper.0(X3))
active.0(f.0-1-0(X1, X2, X3)) → f.0-1-0(X1, X2, active.0(X3))
active.0(f.0-0-1(X1, X2, X3)) → f.0-0-0(X1, X2, active.1(X3))
active.0(f.0-1-0(a., b., X)) → mark.0(f.0-0-0(X, X, X))
active.0(f.1-0-0(X1, X2, X3)) → f.1-0-0(X1, X2, active.0(X3))
proper.0(f.1-1-0(X1, X2, X3)) → f.1-1-0(proper.1(X1), proper.1(X2), proper.0(X3))
f.0-0-1(ok.0(X1), ok.0(X2), ok.1(X3)) → ok.0(f.0-0-1(X1, X2, X3))
f.0-0-0(ok.0(X1), ok.0(X2), ok.0(X3)) → ok.0(f.0-0-0(X1, X2, X3))
f.1-0-0(X1, X2, mark.0(X3)) → mark.0(f.1-0-0(X1, X2, X3))
active.0(f.0-1-1(a., b., X)) → mark.0(f.1-1-1(X, X, X))
proper.0(f.0-1-0(X1, X2, X3)) → f.0-1-0(proper.0(X1), proper.1(X2), proper.0(X3))
proper.0(c.) → ok.0(c.)
f.1-0-0(ok.1(X1), ok.0(X2), ok.0(X3)) → ok.0(f.1-0-0(X1, X2, X3))
f.0-0-0(X1, X2, mark.0(X3)) → mark.0(f.0-0-0(X1, X2, X3))
proper.0(f.0-1-1(X1, X2, X3)) → f.0-1-1(proper.0(X1), proper.1(X2), proper.1(X3))
active.0(f.1-1-0(X1, X2, X3)) → f.1-1-0(X1, X2, active.0(X3))
proper.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(proper.0(X1), proper.0(X2), proper.0(X3))
f.1-0-1(ok.1(X1), ok.0(X2), ok.1(X3)) → ok.0(f.1-0-1(X1, X2, X3))
f.1-1-0(X1, X2, mark.0(X3)) → mark.0(f.1-1-0(X1, X2, X3))
proper.1(b.) → ok.1(b.)
f.1-1-0(X1, X2, mark.1(X3)) → mark.0(f.1-1-1(X1, X2, X3))
f.1-1-0(ok.1(X1), ok.1(X2), ok.0(X3)) → ok.0(f.1-1-0(X1, X2, X3))
active.0(c.) → mark.1(b.)
f.0-1-1(ok.0(X1), ok.1(X2), ok.1(X3)) → ok.0(f.0-1-1(X1, X2, X3))
proper.0(f.1-0-1(X1, X2, X3)) → f.1-0-1(proper.1(X1), proper.0(X2), proper.1(X3))
active.0(f.1-0-1(X1, X2, X3)) → f.1-0-0(X1, X2, active.1(X3))
f.0-1-0(ok.0(X1), ok.1(X2), ok.0(X3)) → ok.0(f.0-1-0(X1, X2, X3))
active.0(f.1-1-1(X1, X2, X3)) → f.1-1-0(X1, X2, active.1(X3))
proper.0(a.) → ok.0(a.)
f.0-1-0(X1, X2, mark.0(X3)) → mark.0(f.0-1-0(X1, X2, X3))
proper.0(f.0-0-1(X1, X2, X3)) → f.0-0-1(proper.0(X1), proper.0(X2), proper.1(X3))
f.0-0-0(X1, X2, mark.1(X3)) → mark.0(f.0-0-1(X1, X2, X3))
active.0(f.0-1-1(X1, X2, X3)) → f.0-1-0(X1, X2, active.1(X3))
f.0-1-0(X1, X2, mark.1(X3)) → mark.0(f.0-1-1(X1, X2, X3))
active.0(c.) → mark.0(a.)
proper.0(f.1-1-1(X1, X2, X3)) → f.1-1-1(proper.1(X1), proper.1(X2), proper.1(X3))
f.1-1-1(ok.1(X1), ok.1(X2), ok.1(X3)) → ok.0(f.1-1-1(X1, X2, X3))
active.0(f.0-0-0(X1, X2, X3)) → f.0-0-0(X1, X2, active.0(X3))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.