Termination w.r.t. Q proof of /tmp/tmpFsyPCf/Ex25_Luc06

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(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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(h(X)) → H(active(X))
H(ok(X)) → H(X)
TOP(mark(X)) → PROPER(X)
ACTIVE(f(f(X))) → C(f(g(f(X))))
ACTIVE(f(X)) → ACTIVE(X)
TOP(ok(X)) → ACTIVE(X)
PROPER(h(X)) → H(proper(X))
ACTIVE(h(X)) → D(X)
PROPER(c(X)) → C(proper(X))
PROPER(d(X)) → PROPER(X)
ACTIVE(h(X)) → ACTIVE(X)
PROPER(g(X)) → G(proper(X))
G(ok(X)) → G(X)
ACTIVE(h(X)) → C(d(X))
PROPER(f(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
PROPER(c(X)) → PROPER(X)
F(mark(X)) → F(X)
ACTIVE(c(X)) → D(X)
PROPER(g(X)) → PROPER(X)
PROPER(d(X)) → D(proper(X))
ACTIVE(f(f(X))) → G(f(X))
ACTIVE(f(X)) → F(active(X))
PROPER(h(X)) → PROPER(X)
D(ok(X)) → D(X)
PROPER(f(X)) → F(proper(X))
TOP(mark(X)) → TOP(proper(X))
F(ok(X)) → F(X)
H(mark(X)) → H(X)
ACTIVE(f(f(X))) → F(g(f(X)))
C(ok(X)) → C(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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(h(X)) → H(active(X))
H(ok(X)) → H(X)
TOP(mark(X)) → PROPER(X)
ACTIVE(f(f(X))) → C(f(g(f(X))))
ACTIVE(f(X)) → ACTIVE(X)
TOP(ok(X)) → ACTIVE(X)
PROPER(h(X)) → H(proper(X))
ACTIVE(h(X)) → D(X)
PROPER(c(X)) → C(proper(X))
PROPER(d(X)) → PROPER(X)
ACTIVE(h(X)) → ACTIVE(X)
PROPER(g(X)) → G(proper(X))
G(ok(X)) → G(X)
ACTIVE(h(X)) → C(d(X))
PROPER(f(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
PROPER(c(X)) → PROPER(X)
F(mark(X)) → F(X)
ACTIVE(c(X)) → D(X)
PROPER(g(X)) → PROPER(X)
PROPER(d(X)) → D(proper(X))
ACTIVE(f(f(X))) → G(f(X))
ACTIVE(f(X)) → F(active(X))
PROPER(h(X)) → PROPER(X)
D(ok(X)) → D(X)
PROPER(f(X)) → F(proper(X))
TOP(mark(X)) → TOP(proper(X))
F(ok(X)) → F(X)
H(mark(X)) → H(X)
ACTIVE(f(f(X))) → F(g(f(X)))
C(ok(X)) → C(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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 8 SCCs with 15 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

D(ok(X)) → D(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

D(ok(X)) → D(X)

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

POL(D(x₁)) = (4)x_1
POL(ok(x₁)) = 1 + (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

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

G(ok(X)) → G(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

G(ok(X)) → G(X)

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

POL(ok(x₁)) = 1 + (4)x_1
POL(G(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

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

C(ok(X)) → C(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

C(ok(X)) → C(X)

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

POL(C(x₁)) = (4)x_1
POL(ok(x₁)) = 1 + (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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

H(ok(X)) → H(X)
H(mark(X)) → H(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

H(ok(X)) → H(X)
H(mark(X)) → H(X)

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

POL(H(x₁)) = (4)x_1
POL(ok(x₁)) = 4 + (4)x_1
POL(mark(x₁)) = 4 + x_1

The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

F(mark(X)) → F(X)
F(ok(X)) → F(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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(mark(X)) → F(X)
F(ok(X)) → F(X)

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

POL(mark(x₁)) = 4 + x_1
POL(ok(x₁)) = 4 + (4)x_1
POL(F(x₁)) = (4)x_1

The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

PROPER(g(X)) → PROPER(X)
PROPER(d(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)
PROPER(h(X)) → PROPER(X)
PROPER(c(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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(g(X)) → PROPER(X)
PROPER(d(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)
PROPER(h(X)) → PROPER(X)
PROPER(c(X)) → PROPER(X)

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)) = 4 + (4)x_1
POL(c(x₁)) = 4 + (4)x_1
POL(g(x₁)) = 4 + (2)x_1
POL(h(x₁)) = 4 + (4)x_1
POL(d(x₁)) = 4 + (4)x_1

The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

ACTIVE(h(X)) → ACTIVE(X)
ACTIVE(f(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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(h(X)) → ACTIVE(X)
ACTIVE(f(X)) → ACTIVE(X)

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

POL(f(x₁)) = 4 + x_1
POL(h(x₁)) = 4 + (4)x_1
POL(ACTIVE(x₁)) = (4)x_1

The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.

TOP(mark(X)) → TOP(proper(X))
TOP(ok(X)) → TOP(active(X))

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

POL(active(x₁)) = 1 + x_1
POL(c(x₁)) = (4)x_1
POL(f(x₁)) = x_1
POL(g(x₁)) = (4)x_1
POL(h(x₁)) = x_1
POL(mark(x₁)) = 1
POL(ok(x₁)) = 4 + (2)x_1
POL(TOP(x₁)) = (4)x_1
POL(proper(x₁)) = 0
POL(d(x₁)) = x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented:

active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
h(ok(X)) → ok(h(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

active(f(f(X))) → mark(c(f(g(f(X)))))
active(c(X)) → mark(d(X))
active(h(X)) → mark(c(d(X)))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(c(X)) → c(proper(X))
proper(g(X)) → g(proper(X))
proper(d(X)) → d(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
c(ok(X)) → ok(c(X))
g(ok(X)) → ok(g(X))
d(ok(X)) → ok(d(X))
h(ok(X)) → ok(h(X))
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.