Termination w.r.t. Q proof of /tmp/tmptY40Ga/JFP

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

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

active(f(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

            ↳ QDPOrderProof

          ↳ 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 use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

F(ok(x), ok(y), ok(z)) → F(x, y, z)
F(x, y, mark(z)) → F(x, y, z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(ok(x₁)) = 1 + (4)x_1
POL(mark(x₁)) = 1 + (4)x_1
POL(F(x₁, x₂, x₃)) = x_2 + x_3

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
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 TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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))

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.

PROPER(f(x, y, z)) → PROPER(x)
PROPER(f(x, y, z)) → PROPER(y)
PROPER(f(x, y, z)) → PROPER(z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(PROPER(x₁)) = (4)x_1
POL(f(x₁, x₂, x₃)) = 1 + (4)x_1 + (4)x_2 + (4)x_3

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
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 TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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))

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.

ACTIVE(f(x, y, z)) → ACTIVE(z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(f(x₁, x₂, x₃)) = 1 + (4)x_3
POL(ACTIVE(x₁)) = (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
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 TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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.