Termination w.r.t. Q proof of /tmp/tmpuH7CMB/Ex4_Zan97

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(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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(X)) → G(X)
SEL(mark(X1), X2) → SEL(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
SEL(X1, active(X2)) → SEL(X1, X2)
MARK(s(X)) → MARK(X)
ACTIVE(g(0)) → MARK(s(0))
ACTIVE(f(X)) → CONS(X, f(g(X)))
MARK(cons(X1, X2)) → MARK(X1)
CONS(X1, mark(X2)) → CONS(X1, X2)
ACTIVE(f(X)) → F(g(X))
ACTIVE(f(X)) → MARK(cons(X, f(g(X))))
MARK(g(X)) → MARK(X)
CONS(X1, active(X2)) → CONS(X1, X2)
ACTIVE(sel(s(X), cons(Y, Z))) → MARK(sel(X, Z))
MARK(s(X)) → ACTIVE(s(mark(X)))
SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(active(X1), X2) → SEL(X1, X2)
ACTIVE(sel(0, cons(X, Y))) → MARK(X)
F(active(X)) → F(X)
MARK(f(X)) → MARK(X)
F(mark(X)) → F(X)
S(active(X)) → S(X)
S(mark(X)) → S(X)
MARK(sel(X1, X2)) → MARK(X1)
MARK(s(X)) → S(mark(X))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(g(s(X))) → MARK(s(s(g(X))))
ACTIVE(g(s(X))) → G(X)
MARK(g(X)) → ACTIVE(g(mark(X)))
G(active(X)) → G(X)
MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2)))
G(mark(X)) → G(X)
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(g(X)) → G(mark(X))
ACTIVE(g(0)) → S(0)
MARK(sel(X1, X2)) → SEL(mark(X1), mark(X2))
MARK(0) → ACTIVE(0)
MARK(f(X)) → F(mark(X))
CONS(active(X1), X2) → CONS(X1, X2)
ACTIVE(g(s(X))) → S(g(X))
ACTIVE(sel(s(X), cons(Y, Z))) → SEL(X, Z)
MARK(sel(X1, X2)) → MARK(X2)
ACTIVE(g(s(X))) → S(s(g(X)))

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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(X)) → G(X)
SEL(mark(X1), X2) → SEL(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
SEL(X1, active(X2)) → SEL(X1, X2)
MARK(s(X)) → MARK(X)
ACTIVE(g(0)) → MARK(s(0))
ACTIVE(f(X)) → CONS(X, f(g(X)))
MARK(cons(X1, X2)) → MARK(X1)
CONS(X1, mark(X2)) → CONS(X1, X2)
ACTIVE(f(X)) → F(g(X))
ACTIVE(f(X)) → MARK(cons(X, f(g(X))))
MARK(g(X)) → MARK(X)
CONS(X1, active(X2)) → CONS(X1, X2)
ACTIVE(sel(s(X), cons(Y, Z))) → MARK(sel(X, Z))
MARK(s(X)) → ACTIVE(s(mark(X)))
SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(active(X1), X2) → SEL(X1, X2)
ACTIVE(sel(0, cons(X, Y))) → MARK(X)
F(active(X)) → F(X)
MARK(f(X)) → MARK(X)
F(mark(X)) → F(X)
S(active(X)) → S(X)
S(mark(X)) → S(X)
MARK(sel(X1, X2)) → MARK(X1)
MARK(s(X)) → S(mark(X))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(g(s(X))) → MARK(s(s(g(X))))
ACTIVE(g(s(X))) → G(X)
MARK(g(X)) → ACTIVE(g(mark(X)))
G(active(X)) → G(X)
MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2)))
G(mark(X)) → G(X)
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(g(X)) → G(mark(X))
ACTIVE(g(0)) → S(0)
MARK(sel(X1, X2)) → SEL(mark(X1), mark(X2))
MARK(0) → ACTIVE(0)
MARK(f(X)) → F(mark(X))
CONS(active(X1), X2) → CONS(X1, X2)
ACTIVE(g(s(X))) → S(g(X))
ACTIVE(sel(s(X), cons(Y, Z))) → SEL(X, Z)
MARK(sel(X1, X2)) → MARK(X2)
ACTIVE(g(s(X))) → S(s(g(X)))

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 6 SCCs with 14 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL(mark(X1), X2) → SEL(X1, X2)
SEL(X1, active(X2)) → SEL(X1, X2)
SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(active(X1), X2) → SEL(X1, X2)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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.

SEL(X1, active(X2)) → SEL(X1, X2)
SEL(X1, mark(X2)) → SEL(X1, X2)

The remaining pairs can at least be oriented weakly.

SEL(mark(X1), X2) → SEL(X1, X2)
SEL(active(X1), X2) → SEL(X1, X2)

Used ordering: Polynomial interpretation [25,35]:

POL(active(x₁)) = 4 + (2)x_1
POL(mark(x₁)) = 1/4 + (2)x_1
POL(SEL(x₁, x₂)) = (1/4)x_2

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL(mark(X1), X2) → SEL(X1, X2)
SEL(active(X1), X2) → SEL(X1, X2)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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.

SEL(mark(X1), X2) → SEL(X1, X2)
SEL(active(X1), X2) → SEL(X1, X2)

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

POL(active(x₁)) = 4 + (4)x_1
POL(mark(x₁)) = 1 + x_1
POL(SEL(x₁, x₂)) = (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

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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

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

S(mark(X)) → S(X)
S(active(X)) → S(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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.

S(mark(X)) → S(X)
S(active(X)) → S(X)

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

POL(active(x₁)) = 4 + (4)x_1
POL(mark(x₁)) = 1 + x_1
POL(S(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

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

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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

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

G(active(X)) → G(X)
G(mark(X)) → G(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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(active(X)) → G(X)
G(mark(X)) → G(X)

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

POL(active(x₁)) = 1 + x_1
POL(mark(x₁)) = 4 + (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

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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

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

CONS(X1, active(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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.

CONS(X1, active(X2)) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.

CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)

Used ordering: Polynomial interpretation [25,35]:

POL(active(x₁)) = 1/4 + (2)x_1
POL(CONS(x₁, x₂)) = (1/4)x_2
POL(mark(x₁)) = 4 + (2)x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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.

CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)

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

POL(active(x₁)) = 1 + x_1
POL(CONS(x₁, x₂)) = (4)x_1
POL(mark(x₁)) = 4 + (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

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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

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

F(mark(X)) → F(X)
F(active(X)) → F(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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(active(X)) → F(X)

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

POL(active(x₁)) = 4 + (4)x_1
POL(mark(x₁)) = 1 + x_1
POL(F(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

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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

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

MARK(sel(X1, X2)) → MARK(X1)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(s(X)) → MARK(X)
ACTIVE(g(0)) → MARK(s(0))
ACTIVE(g(s(X))) → MARK(s(s(g(X))))
MARK(cons(X1, X2)) → MARK(X1)
MARK(g(X)) → ACTIVE(g(mark(X)))
MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2)))
MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(X)) → MARK(cons(X, f(g(X))))
MARK(g(X)) → MARK(X)
ACTIVE(sel(s(X), cons(Y, Z))) → MARK(sel(X, Z))
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(sel(0, cons(X, Y))) → MARK(X)
MARK(sel(X1, X2)) → MARK(X2)
MARK(f(X)) → MARK(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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.

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(s(X)) → ACTIVE(s(mark(X)))

The remaining pairs can at least be oriented weakly.

MARK(sel(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
ACTIVE(g(0)) → MARK(s(0))
ACTIVE(g(s(X))) → MARK(s(s(g(X))))
MARK(cons(X1, X2)) → MARK(X1)
MARK(g(X)) → ACTIVE(g(mark(X)))
MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2)))
MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(X)) → MARK(cons(X, f(g(X))))
MARK(g(X)) → MARK(X)
ACTIVE(sel(s(X), cons(Y, Z))) → MARK(sel(X, Z))
ACTIVE(sel(0, cons(X, Y))) → MARK(X)
MARK(sel(X1, X2)) → MARK(X2)
MARK(f(X)) → MARK(X)

Used ordering: Polynomial interpretation [25,35]:

POL(sel(x₁, x₂)) = 1
POL(cons(x₁, x₂)) = 0
POL(active(x₁)) = 0
POL(MARK(x₁)) = 1/4
POL(f(x₁)) = 1
POL(g(x₁)) = 1
POL(mark(x₁)) = 0
POL(s(x₁)) = 0
POL(0) = 0
POL(ACTIVE(x₁)) = (1/4)x_1

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
sel(X1, mark(X2)) → sel(X1, X2)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
f(mark(X)) → f(X)
f(active(X)) → f(X)
g(active(X)) → g(X)
g(mark(X)) → g(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

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

MARK(sel(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
ACTIVE(g(0)) → MARK(s(0))
ACTIVE(g(s(X))) → MARK(s(s(g(X))))
MARK(cons(X1, X2)) → MARK(X1)
MARK(g(X)) → ACTIVE(g(mark(X)))
MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2)))
MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(X)) → MARK(cons(X, f(g(X))))
MARK(g(X)) → MARK(X)
ACTIVE(sel(s(X), cons(Y, Z))) → MARK(sel(X, Z))
ACTIVE(sel(0, cons(X, Y))) → MARK(X)
MARK(sel(X1, X2)) → MARK(X2)
MARK(f(X)) → MARK(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(X)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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