Termination w.r.t. Q proof of /tmp/tmpl9IwtA/JFP

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(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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:

F(ok(x), ok(y), ok(z)) → F(x, y, z)
ACTIVE(f(x, y, z)) → F(x, y, active(z))
TOP(mark(x)) → PROPER(x)
ACTIVE(f(b, c, x)) → F(x, x, x)
F(x, y, mark(z)) → F(x, y, z)
TOP(ok(x)) → ACTIVE(x)
PROPER(f(x, y, z)) → PROPER(x)
PROPER(f(x, y, z)) → PROPER(y)
ACTIVE(f(x, y, z)) → ACTIVE(z)
PROPER(f(x, y, z)) → PROPER(z)
TOP(mark(x)) → TOP(proper(x))
PROPER(f(x, y, z)) → F(proper(x), proper(y), proper(z))
TOP(ok(x)) → TOP(active(x))

The TRS R consists of the following rules:

active(f(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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:

F(ok(x), ok(y), ok(z)) → F(x, y, z)
ACTIVE(f(x, y, z)) → F(x, y, active(z))
TOP(mark(x)) → PROPER(x)
ACTIVE(f(b, c, x)) → F(x, x, x)
F(x, y, mark(z)) → F(x, y, z)
TOP(ok(x)) → ACTIVE(x)
PROPER(f(x, y, z)) → PROPER(x)
PROPER(f(x, y, z)) → PROPER(y)
ACTIVE(f(x, y, z)) → ACTIVE(z)
PROPER(f(x, y, z)) → PROPER(z)
TOP(mark(x)) → TOP(proper(x))
PROPER(f(x, y, z)) → F(proper(x), proper(y), proper(z))
TOP(ok(x)) → TOP(active(x))

The TRS R consists of the following rules:

active(f(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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(x), ok(y), ok(z)) → F(x, y, z)
F(x, y, mark(z)) → F(x, y, z)

The TRS R consists of the following rules:

active(f(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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(x), ok(y), ok(z)) → F(x, y, z)
F(x, y, mark(z)) → F(x, y, z)

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(x), ok(y), ok(z)) → F(x, y, z)
The graph contains the following edges 1 > 1, 2 > 2, 3 > 3
F(x, y, mark(z)) → F(x, y, z)
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(x, y, z)) → PROPER(x)
PROPER(f(x, y, z)) → PROPER(y)
PROPER(f(x, y, z)) → PROPER(z)

The TRS R consists of the following rules:

active(f(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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(x, y, z)) → PROPER(x)
PROPER(f(x, y, z)) → PROPER(y)
PROPER(f(x, y, z)) → PROPER(z)

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

PROPER(f(x, y, z)) → PROPER(x)
The graph contains the following edges 1 > 1
PROPER(f(x, y, z)) → PROPER(y)
The graph contains the following edges 1 > 1
PROPER(f(x, y, z)) → PROPER(z)
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(x, y, z)) → ACTIVE(z)

The TRS R consists of the following rules:

active(f(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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(x, y, z)) → ACTIVE(z)

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

ACTIVE(f(x, y, z)) → ACTIVE(z)
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(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
f(x, y, mark(z)) → mark(f(x, y, z))
active(d) → mark(c)
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
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)) = x₁
POL(active(x₁)) = 2·x₁
POL(b) = 0
POL(c) = 0
POL(d) = 0
POL(f(x₁, x₂, x₃)) = x₁ + x₂ + 2·x₃
POL(m(x₁)) = x₁
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(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
active(d) → mark(c)
f(x, y, mark(z)) → mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))

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

↳ 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(c)) → TOP(ok(c))
TOP(ok(x)) → TOP(active(x))
TOP(mark(d)) → TOP(ok(d))

The TRS R consists of the following rules:

active(f(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
active(d) → mark(c)
f(x, y, mark(z)) → mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))

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 2 less nodes.

↳ 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(c)) → TOP(ok(c))
TOP(ok(x)) → TOP(active(x))

The TRS R consists of the following rules:

active(f(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
active(d) → mark(c)
f(x, y, mark(z)) → mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))

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(f(b, c, x0))) → TOP(mark(f(x0, x0, x0)))
TOP(ok(d)) → TOP(mark(c))
TOP(ok(d)) → TOP(m(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(b, c, x0))) → TOP(mark(f(x0, x0, x0)))
TOP(mark(f(x0, x1, x2))) → TOP(f(proper(x0), proper(x1), proper(x2)))
TOP(ok(d)) → TOP(mark(c))
TOP(ok(f(x0, x1, x2))) → TOP(f(x0, x1, active(x2)))
TOP(ok(d)) → TOP(m(b))
TOP(mark(c)) → TOP(ok(c))

The TRS R consists of the following rules:

active(f(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
active(d) → mark(c)
f(x, y, mark(z)) → mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))

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 3 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(b, c, 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(b, c, x)) → mark(f(x, x, x))
active(f(x, y, z)) → f(x, y, active(z))
active(d) → m(b)
active(d) → mark(c)
f(x, y, mark(z)) → mark(f(x, y, z))
f(ok(x), ok(y), ok(z)) → ok(f(x, y, z))
proper(b) → ok(b)
proper(c) → ok(c)
proper(d) → ok(d)
proper(f(x, y, z)) → f(proper(x), proper(y), proper(z))

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: 1
active: 0
m: 0
f: 0
ok: x₀
mark: 0
d: 0
b: 0
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(b., c., 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(ok.0(f.0-1-1(b., c., x0))) → TOP.0(mark.0(f.1-1-1(x0, x0, x0)))
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.0-0-0(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(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
proper.0(f.1-0-0(x, y, z)) → f.1-0-0(proper.1(x), proper.0(y), proper.0(z))
active.0(f.0-1-0(x, y, z)) → f.0-1-0(x, y, active.0(z))
active.0(f.0-0-1(x, y, z)) → f.0-0-0(x, y, active.1(z))
active.0(f.1-0-0(x, y, z)) → f.1-0-0(x, y, active.0(z))
proper.0(f.1-1-0(x, y, z)) → f.1-1-0(proper.1(x), proper.1(y), proper.0(z))
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
proper.0(f.0-1-0(x, y, z)) → f.0-1-0(proper.0(x), proper.1(y), proper.0(z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
proper.0(f.0-1-1(x, y, z)) → f.0-1-1(proper.0(x), proper.1(y), proper.1(z))
active.0(f.1-1-0(x, y, z)) → f.1-1-0(x, y, active.0(z))
active.0(f.0-1-1(b., c., x)) → mark.0(f.1-1-1(x, x, x))
proper.0(f.0-0-0(x, y, z)) → f.0-0-0(proper.0(x), proper.0(y), proper.0(z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
proper.0(f.1-0-1(x, y, z)) → f.1-0-1(proper.1(x), proper.0(y), proper.1(z))
active.0(d.) → m.0(b.)
active.0(f.1-0-1(x, y, z)) → f.1-0-0(x, y, active.1(z))
proper.0(d.) → ok.0(d.)
proper.0(b.) → ok.0(b.)
active.0(f.0-1-0(b., c., x)) → mark.0(f.0-0-0(x, x, x))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
active.0(f.1-1-1(x, y, z)) → f.1-1-0(x, y, active.1(z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
proper.0(f.0-0-1(x, y, z)) → f.0-0-1(proper.0(x), proper.0(y), proper.1(z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
active.0(d.) → mark.1(c.)
active.0(f.0-1-1(x, y, z)) → f.0-1-0(x, y, active.1(z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
proper.1(c.) → ok.1(c.)
proper.0(f.1-1-1(x, y, z)) → f.1-1-1(proper.1(x), proper.1(y), proper.1(z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
active.0(f.0-0-0(x, y, z)) → f.0-0-0(x, y, active.0(z))

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(b., c., 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(ok.0(f.0-1-1(b., c., x0))) → TOP.0(mark.0(f.1-1-1(x0, x0, x0)))
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.0-0-0(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(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
proper.0(f.1-0-0(x, y, z)) → f.1-0-0(proper.1(x), proper.0(y), proper.0(z))
active.0(f.0-1-0(x, y, z)) → f.0-1-0(x, y, active.0(z))
active.0(f.0-0-1(x, y, z)) → f.0-0-0(x, y, active.1(z))
active.0(f.1-0-0(x, y, z)) → f.1-0-0(x, y, active.0(z))
proper.0(f.1-1-0(x, y, z)) → f.1-1-0(proper.1(x), proper.1(y), proper.0(z))
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
proper.0(f.0-1-0(x, y, z)) → f.0-1-0(proper.0(x), proper.1(y), proper.0(z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
proper.0(f.0-1-1(x, y, z)) → f.0-1-1(proper.0(x), proper.1(y), proper.1(z))
active.0(f.1-1-0(x, y, z)) → f.1-1-0(x, y, active.0(z))
active.0(f.0-1-1(b., c., x)) → mark.0(f.1-1-1(x, x, x))
proper.0(f.0-0-0(x, y, z)) → f.0-0-0(proper.0(x), proper.0(y), proper.0(z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
proper.0(f.1-0-1(x, y, z)) → f.1-0-1(proper.1(x), proper.0(y), proper.1(z))
active.0(d.) → m.0(b.)
active.0(f.1-0-1(x, y, z)) → f.1-0-0(x, y, active.1(z))
proper.0(d.) → ok.0(d.)
proper.0(b.) → ok.0(b.)
active.0(f.0-1-0(b., c., x)) → mark.0(f.0-0-0(x, x, x))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
active.0(f.1-1-1(x, y, z)) → f.1-1-0(x, y, active.1(z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
proper.0(f.0-0-1(x, y, z)) → f.0-0-1(proper.0(x), proper.0(y), proper.1(z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
active.0(d.) → mark.1(c.)
active.0(f.0-1-1(x, y, z)) → f.0-1-0(x, y, active.1(z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
proper.1(c.) → ok.1(c.)
proper.0(f.1-1-1(x, y, z)) → f.1-1-1(proper.1(x), proper.1(y), proper.1(z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
active.0(f.0-0-0(x, y, z)) → f.0-0-0(x, y, active.0(z))

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(b., c., 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(ok.0(f.0-1-1(b., c., x0))) → TOP.0(mark.0(f.1-1-1(x0, x0, x0)))
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.0-0-0(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(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
proper.0(f.1-0-0(x, y, z)) → f.1-0-0(proper.1(x), proper.0(y), proper.0(z))
active.0(f.0-1-0(x, y, z)) → f.0-1-0(x, y, active.0(z))
active.0(f.0-0-1(x, y, z)) → f.0-0-0(x, y, active.1(z))
active.0(f.1-0-0(x, y, z)) → f.1-0-0(x, y, active.0(z))
proper.0(f.1-1-0(x, y, z)) → f.1-1-0(proper.1(x), proper.1(y), proper.0(z))
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
proper.0(f.0-1-0(x, y, z)) → f.0-1-0(proper.0(x), proper.1(y), proper.0(z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
proper.0(f.0-1-1(x, y, z)) → f.0-1-1(proper.0(x), proper.1(y), proper.1(z))
active.0(f.1-1-0(x, y, z)) → f.1-1-0(x, y, active.0(z))
active.0(f.0-1-1(b., c., x)) → mark.0(f.1-1-1(x, x, x))
proper.0(f.0-0-0(x, y, z)) → f.0-0-0(proper.0(x), proper.0(y), proper.0(z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
proper.0(f.1-0-1(x, y, z)) → f.1-0-1(proper.1(x), proper.0(y), proper.1(z))
active.0(d.) → m.0(b.)
active.0(f.1-0-1(x, y, z)) → f.1-0-0(x, y, active.1(z))
proper.0(d.) → ok.0(d.)
proper.0(b.) → ok.0(b.)
active.0(f.0-1-0(b., c., x)) → mark.0(f.0-0-0(x, x, x))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
active.0(f.1-1-1(x, y, z)) → f.1-1-0(x, y, active.1(z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
proper.0(f.0-0-1(x, y, z)) → f.0-0-1(proper.0(x), proper.0(y), proper.1(z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
active.0(d.) → mark.1(c.)
active.0(f.0-1-1(x, y, z)) → f.0-1-0(x, y, active.1(z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
proper.1(c.) → ok.1(c.)
proper.0(f.1-1-1(x, y, z)) → f.1-1-1(proper.1(x), proper.1(y), proper.1(z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
active.0(f.0-0-0(x, y, z)) → f.0-0-0(x, y, active.0(z))

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

The following usable rules [17] were oriented:

proper.0(b.) → ok.0(b.)
active.0(f.0-1-0(b., c., x)) → mark.0(f.0-0-0(x, x, x))
active.0(f.1-0-1(x, y, z)) → f.1-0-0(x, y, active.1(z))
proper.0(d.) → ok.0(d.)
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
proper.0(f.0-0-1(x, y, z)) → f.0-0-1(proper.0(x), proper.0(y), proper.1(z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
active.0(f.1-1-1(x, y, z)) → f.1-1-0(x, y, active.1(z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
proper.0(f.0-0-0(x, y, z)) → f.0-0-0(proper.0(x), proper.0(y), proper.0(z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))
proper.0(f.1-0-1(x, y, z)) → f.1-0-1(proper.1(x), proper.0(y), proper.1(z))
active.0(d.) → m.0(b.)
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
active.0(f.0-1-1(x, y, z)) → f.0-1-0(x, y, active.1(z))
active.0(d.) → mark.1(c.)
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
active.0(f.0-0-0(x, y, z)) → f.0-0-0(x, y, active.0(z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
proper.0(f.1-1-1(x, y, z)) → f.1-1-1(proper.1(x), proper.1(y), proper.1(z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
proper.0(f.0-1-0(x, y, z)) → f.0-1-0(proper.0(x), proper.1(y), proper.0(z))
proper.0(f.0-1-1(x, y, z)) → f.0-1-1(proper.0(x), proper.1(y), proper.1(z))
f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
active.0(f.0-1-1(b., c., x)) → mark.0(f.1-1-1(x, x, x))
active.0(f.1-1-0(x, y, z)) → f.1-1-0(x, y, active.0(z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
active.0(f.0-1-0(x, y, z)) → f.0-1-0(x, y, active.0(z))
proper.0(f.1-0-0(x, y, z)) → f.1-0-0(proper.1(x), proper.0(y), proper.0(z))
active.0(f.1-0-0(x, y, z)) → f.1-0-0(x, y, active.0(z))
active.0(f.0-0-1(x, y, z)) → f.0-0-0(x, y, active.1(z))
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
proper.0(f.1-1-0(x, y, z)) → f.1-1-0(proper.1(x), proper.1(y), proper.0(z))

↳ 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

                                            ↳ 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(b., c., 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.0-0-0(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(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
proper.0(f.1-0-0(x, y, z)) → f.1-0-0(proper.1(x), proper.0(y), proper.0(z))
active.0(f.0-1-0(x, y, z)) → f.0-1-0(x, y, active.0(z))
active.0(f.0-0-1(x, y, z)) → f.0-0-0(x, y, active.1(z))
active.0(f.1-0-0(x, y, z)) → f.1-0-0(x, y, active.0(z))
proper.0(f.1-1-0(x, y, z)) → f.1-1-0(proper.1(x), proper.1(y), proper.0(z))
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
proper.0(f.0-1-0(x, y, z)) → f.0-1-0(proper.0(x), proper.1(y), proper.0(z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
proper.0(f.0-1-1(x, y, z)) → f.0-1-1(proper.0(x), proper.1(y), proper.1(z))
active.0(f.1-1-0(x, y, z)) → f.1-1-0(x, y, active.0(z))
active.0(f.0-1-1(b., c., x)) → mark.0(f.1-1-1(x, x, x))
proper.0(f.0-0-0(x, y, z)) → f.0-0-0(proper.0(x), proper.0(y), proper.0(z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
proper.0(f.1-0-1(x, y, z)) → f.1-0-1(proper.1(x), proper.0(y), proper.1(z))
active.0(d.) → m.0(b.)
active.0(f.1-0-1(x, y, z)) → f.1-0-0(x, y, active.1(z))
proper.0(d.) → ok.0(d.)
proper.0(b.) → ok.0(b.)
active.0(f.0-1-0(b., c., x)) → mark.0(f.0-0-0(x, x, x))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
active.0(f.1-1-1(x, y, z)) → f.1-1-0(x, y, active.1(z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
proper.0(f.0-0-1(x, y, z)) → f.0-0-1(proper.0(x), proper.0(y), proper.1(z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
active.0(d.) → mark.1(c.)
active.0(f.0-1-1(x, y, z)) → f.0-1-0(x, y, active.1(z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
proper.1(c.) → ok.1(c.)
proper.0(f.1-1-1(x, y, z)) → f.1-1-1(proper.1(x), proper.1(y), proper.1(z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
active.0(f.0-0-0(x, y, z)) → f.0-0-0(x, y, active.0(z))

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

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(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(active.0(x₁)) = x₁
POL(active.1(x₁)) = 0
POL(b.) = 1
POL(c.) = 0
POL(d.) = 1
POL(f.0-0-0(x₁, x₂, x₃)) = x₃
POL(f.0-0-1(x₁, x₂, x₃)) = 0
POL(f.0-1-0(x₁, x₂, x₃)) = 1 + x₁ + x₃
POL(f.0-1-1(x₁, x₂, x₃)) = 1 + x₁ + x₃
POL(f.1-0-0(x₁, x₂, x₃)) = x₃
POL(f.1-0-1(x₁, x₂, x₃)) = 0
POL(f.1-1-0(x₁, x₂, x₃)) = 1 + x₁ + x₃
POL(f.1-1-1(x₁, x₂, x₃)) = 1 + x₁
POL(m.0(x₁)) = x₁
POL(mark.0(x₁)) = 1 + x₁
POL(mark.1(x₁)) = 1 + x₁
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:

proper.0(b.) → ok.0(b.)
active.0(f.0-1-0(b., c., x)) → mark.0(f.0-0-0(x, x, x))
active.0(f.1-0-1(x, y, z)) → f.1-0-0(x, y, active.1(z))
proper.0(d.) → ok.0(d.)
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
proper.0(f.0-0-1(x, y, z)) → f.0-0-1(proper.0(x), proper.0(y), proper.1(z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
active.0(f.1-1-1(x, y, z)) → f.1-1-0(x, y, active.1(z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
proper.0(f.0-0-0(x, y, z)) → f.0-0-0(proper.0(x), proper.0(y), proper.0(z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))
proper.0(f.1-0-1(x, y, z)) → f.1-0-1(proper.1(x), proper.0(y), proper.1(z))
active.0(d.) → m.0(b.)
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
active.0(f.0-1-1(x, y, z)) → f.0-1-0(x, y, active.1(z))
active.0(d.) → mark.1(c.)
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
active.0(f.0-0-0(x, y, z)) → f.0-0-0(x, y, active.0(z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
proper.0(f.1-1-1(x, y, z)) → f.1-1-1(proper.1(x), proper.1(y), proper.1(z))
proper.1(c.) → ok.1(c.)
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
proper.0(f.0-1-0(x, y, z)) → f.0-1-0(proper.0(x), proper.1(y), proper.0(z))
proper.0(f.0-1-1(x, y, z)) → f.0-1-1(proper.0(x), proper.1(y), proper.1(z))
f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
active.0(f.0-1-1(b., c., x)) → mark.0(f.1-1-1(x, x, x))
active.0(f.1-1-0(x, y, z)) → f.1-1-0(x, y, active.0(z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
active.0(f.0-1-0(x, y, z)) → f.0-1-0(x, y, active.0(z))
proper.0(f.1-0-0(x, y, z)) → f.1-0-0(proper.1(x), proper.0(y), proper.0(z))
active.0(f.1-0-0(x, y, z)) → f.1-0-0(x, y, active.0(z))
active.0(f.0-0-1(x, y, z)) → f.0-0-0(x, y, active.1(z))
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
proper.0(f.1-1-0(x, y, z)) → f.1-1-0(proper.1(x), proper.1(y), proper.0(z))

↳ 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

                                            ↳ QDPOrderProof

                                              ↳ 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(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
proper.0(f.1-0-0(x, y, z)) → f.1-0-0(proper.1(x), proper.0(y), proper.0(z))
active.0(f.0-1-0(x, y, z)) → f.0-1-0(x, y, active.0(z))
active.0(f.0-0-1(x, y, z)) → f.0-0-0(x, y, active.1(z))
active.0(f.1-0-0(x, y, z)) → f.1-0-0(x, y, active.0(z))
proper.0(f.1-1-0(x, y, z)) → f.1-1-0(proper.1(x), proper.1(y), proper.0(z))
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
proper.0(f.0-1-0(x, y, z)) → f.0-1-0(proper.0(x), proper.1(y), proper.0(z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
proper.0(f.0-1-1(x, y, z)) → f.0-1-1(proper.0(x), proper.1(y), proper.1(z))
active.0(f.1-1-0(x, y, z)) → f.1-1-0(x, y, active.0(z))
active.0(f.0-1-1(b., c., x)) → mark.0(f.1-1-1(x, x, x))
proper.0(f.0-0-0(x, y, z)) → f.0-0-0(proper.0(x), proper.0(y), proper.0(z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
proper.0(f.1-0-1(x, y, z)) → f.1-0-1(proper.1(x), proper.0(y), proper.1(z))
active.0(d.) → m.0(b.)
active.0(f.1-0-1(x, y, z)) → f.1-0-0(x, y, active.1(z))
proper.0(d.) → ok.0(d.)
proper.0(b.) → ok.0(b.)
active.0(f.0-1-0(b., c., x)) → mark.0(f.0-0-0(x, x, x))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
active.0(f.1-1-1(x, y, z)) → f.1-1-0(x, y, active.1(z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
proper.0(f.0-0-1(x, y, z)) → f.0-0-1(proper.0(x), proper.0(y), proper.1(z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
active.0(d.) → mark.1(c.)
active.0(f.0-1-1(x, y, z)) → f.0-1-0(x, y, active.1(z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
proper.1(c.) → ok.1(c.)
proper.0(f.1-1-1(x, y, z)) → f.1-1-1(proper.1(x), proper.1(y), proper.1(z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
active.0(f.0-0-0(x, y, z)) → f.0-0-0(x, y, active.0(z))

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(active.0(x₁)) = 0
POL(active.1(x₁)) = 0
POL(b.) = 0
POL(c.) = 0
POL(d.) = 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 + x₂
POL(f.0-1-1(x₁, x₂, x₃)) = 0
POL(f.1-0-0(x₁, x₂, x₃)) = 1 + x₁ + x₂
POL(f.1-0-1(x₁, x₂, x₃)) = 0
POL(f.1-1-0(x₁, x₂, x₃)) = x₂
POL(f.1-1-1(x₁, x₂, x₃)) = 0
POL(m.0(x₁)) = 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.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))

↳ 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

                                            ↳ QDPOrderProof

                                              ↳ QDP

                                                ↳ QDPOrderProof

                                                  ↳ QDP

                                                    ↳ PisEmptyProof

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

f.1-0-0(x, y, mark.1(z)) → mark.0(f.1-0-1(x, y, z))
proper.0(f.1-0-0(x, y, z)) → f.1-0-0(proper.1(x), proper.0(y), proper.0(z))
active.0(f.0-1-0(x, y, z)) → f.0-1-0(x, y, active.0(z))
active.0(f.0-0-1(x, y, z)) → f.0-0-0(x, y, active.1(z))
active.0(f.1-0-0(x, y, z)) → f.1-0-0(x, y, active.0(z))
proper.0(f.1-1-0(x, y, z)) → f.1-1-0(proper.1(x), proper.1(y), proper.0(z))
f.0-0-1(ok.0(x), ok.0(y), ok.1(z)) → ok.0(f.0-0-1(x, y, z))
f.0-0-0(ok.0(x), ok.0(y), ok.0(z)) → ok.0(f.0-0-0(x, y, z))
f.1-0-0(x, y, mark.0(z)) → mark.0(f.1-0-0(x, y, z))
proper.0(f.0-1-0(x, y, z)) → f.0-1-0(proper.0(x), proper.1(y), proper.0(z))
f.1-0-0(ok.1(x), ok.0(y), ok.0(z)) → ok.0(f.1-0-0(x, y, z))
f.0-0-0(x, y, mark.0(z)) → mark.0(f.0-0-0(x, y, z))
proper.0(f.0-1-1(x, y, z)) → f.0-1-1(proper.0(x), proper.1(y), proper.1(z))
active.0(f.1-1-0(x, y, z)) → f.1-1-0(x, y, active.0(z))
active.0(f.0-1-1(b., c., x)) → mark.0(f.1-1-1(x, x, x))
proper.0(f.0-0-0(x, y, z)) → f.0-0-0(proper.0(x), proper.0(y), proper.0(z))
f.1-0-1(ok.1(x), ok.0(y), ok.1(z)) → ok.0(f.1-0-1(x, y, z))
f.1-1-0(x, y, mark.0(z)) → mark.0(f.1-1-0(x, y, z))
f.1-1-0(x, y, mark.1(z)) → mark.0(f.1-1-1(x, y, z))
f.1-1-0(ok.1(x), ok.1(y), ok.0(z)) → ok.0(f.1-1-0(x, y, z))
f.0-1-1(ok.0(x), ok.1(y), ok.1(z)) → ok.0(f.0-1-1(x, y, z))
proper.0(f.1-0-1(x, y, z)) → f.1-0-1(proper.1(x), proper.0(y), proper.1(z))
active.0(d.) → m.0(b.)
active.0(f.1-0-1(x, y, z)) → f.1-0-0(x, y, active.1(z))
proper.0(d.) → ok.0(d.)
proper.0(b.) → ok.0(b.)
active.0(f.0-1-0(b., c., x)) → mark.0(f.0-0-0(x, x, x))
f.0-1-0(ok.0(x), ok.1(y), ok.0(z)) → ok.0(f.0-1-0(x, y, z))
active.0(f.1-1-1(x, y, z)) → f.1-1-0(x, y, active.1(z))
f.0-1-0(x, y, mark.0(z)) → mark.0(f.0-1-0(x, y, z))
proper.0(f.0-0-1(x, y, z)) → f.0-0-1(proper.0(x), proper.0(y), proper.1(z))
f.0-0-0(x, y, mark.1(z)) → mark.0(f.0-0-1(x, y, z))
active.0(d.) → mark.1(c.)
active.0(f.0-1-1(x, y, z)) → f.0-1-0(x, y, active.1(z))
f.0-1-0(x, y, mark.1(z)) → mark.0(f.0-1-1(x, y, z))
proper.1(c.) → ok.1(c.)
proper.0(f.1-1-1(x, y, z)) → f.1-1-1(proper.1(x), proper.1(y), proper.1(z))
f.1-1-1(ok.1(x), ok.1(y), ok.1(z)) → ok.0(f.1-1-1(x, y, z))
active.0(f.0-0-0(x, y, z)) → f.0-0-0(x, y, active.0(z))

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.