Termination w.r.t. Q proof of /tmp/tmpXNpK3l/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

            ↳ UsableRulesProof

          ↳ 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 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

          ↳ 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)

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:

SEL(mark(X1), X2) → SEL(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
SEL(X1, active(X2)) → SEL(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
SEL(X1, mark(X2)) → SEL(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
SEL(active(X1), X2) → SEL(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ 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)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

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

S(mark(X)) → S(X)
The graph contains the following edges 1 > 1
S(active(X)) → S(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ 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)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

G(active(X)) → G(X)
G(mark(X)) → G(X)

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

G(active(X)) → G(X)
The graph contains the following edges 1 > 1
G(mark(X)) → G(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ 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)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

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

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

CONS(X1, active(X2)) → CONS(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
CONS(mark(X1), X2) → CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
CONS(active(X1), X2) → CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
CONS(X1, mark(X2)) → CONS(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ 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)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

F(mark(X)) → F(X)
F(active(X)) → F(X)

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

F(mark(X)) → F(X)
The graph contains the following edges 1 > 1
F(active(X)) → F(X)
The graph contains the following edges 1 > 1

↳ 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]:

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

The following usable rules [17] were oriented:

f(mark(X)) → f(X)
f(active(X)) → f(X)
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)
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)
g(active(X)) → g(X)
g(mark(X)) → g(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ Narrowing

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.
By narrowing [15] the rule MARK(g(X)) → ACTIVE(g(mark(X))) at position [0] we obtained the following new rules:

MARK(g(f(x0))) → ACTIVE(g(active(f(mark(x0)))))
MARK(g(sel(x0, x1))) → ACTIVE(g(active(sel(mark(x0), mark(x1)))))
MARK(g(s(x0))) → ACTIVE(g(active(s(mark(x0)))))
MARK(g(cons(x0, x1))) → ACTIVE(g(active(cons(mark(x0), x1))))
MARK(g(x0)) → ACTIVE(g(x0))
MARK(g(g(x0))) → ACTIVE(g(active(g(mark(x0)))))
MARK(g(0)) → ACTIVE(g(active(0)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ Narrowing

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

MARK(g(sel(x0, x1))) → ACTIVE(g(active(sel(mark(x0), mark(x1)))))
MARK(sel(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
ACTIVE(g(s(X))) → MARK(s(s(g(X))))
ACTIVE(g(0)) → MARK(s(0))
MARK(cons(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2)))
MARK(g(X)) → MARK(X)
ACTIVE(f(X)) → MARK(cons(X, f(g(X))))
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(g(f(x0))) → ACTIVE(g(active(f(mark(x0)))))
MARK(g(s(x0))) → ACTIVE(g(active(s(mark(x0)))))
ACTIVE(sel(s(X), cons(Y, Z))) → MARK(sel(X, Z))
MARK(g(cons(x0, x1))) → ACTIVE(g(active(cons(mark(x0), x1))))
MARK(g(x0)) → ACTIVE(g(x0))
MARK(g(g(x0))) → ACTIVE(g(active(g(mark(x0)))))
ACTIVE(sel(0, cons(X, Y))) → MARK(X)
MARK(g(0)) → ACTIVE(g(active(0)))
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.
By narrowing [15] the rule MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2))) at position [0] we obtained the following new rules:

MARK(sel(y0, f(x0))) → ACTIVE(sel(mark(y0), active(f(mark(x0)))))
MARK(sel(sel(x0, x1), y1)) → ACTIVE(sel(active(sel(mark(x0), mark(x1))), mark(y1)))
MARK(sel(cons(x0, x1), y1)) → ACTIVE(sel(active(cons(mark(x0), x1)), mark(y1)))
MARK(sel(g(x0), y1)) → ACTIVE(sel(active(g(mark(x0))), mark(y1)))
MARK(sel(f(x0), y1)) → ACTIVE(sel(active(f(mark(x0))), mark(y1)))
MARK(sel(y0, x1)) → ACTIVE(sel(mark(y0), x1))
MARK(sel(x0, y1)) → ACTIVE(sel(x0, mark(y1)))
MARK(sel(y0, cons(x0, x1))) → ACTIVE(sel(mark(y0), active(cons(mark(x0), x1))))
MARK(sel(y0, sel(x0, x1))) → ACTIVE(sel(mark(y0), active(sel(mark(x0), mark(x1)))))
MARK(sel(0, y1)) → ACTIVE(sel(active(0), mark(y1)))
MARK(sel(y0, g(x0))) → ACTIVE(sel(mark(y0), active(g(mark(x0)))))
MARK(sel(y0, 0)) → ACTIVE(sel(mark(y0), active(0)))
MARK(sel(y0, s(x0))) → ACTIVE(sel(mark(y0), active(s(mark(x0)))))
MARK(sel(s(x0), y1)) → ACTIVE(sel(active(s(mark(x0))), mark(y1)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

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

MARK(sel(y0, f(x0))) → ACTIVE(sel(mark(y0), active(f(mark(x0)))))
MARK(s(X)) → MARK(X)
ACTIVE(g(0)) → MARK(s(0))
MARK(cons(X1, X2)) → MARK(X1)
MARK(sel(cons(x0, x1), y1)) → ACTIVE(sel(active(cons(mark(x0), x1)), mark(y1)))
MARK(g(X)) → MARK(X)
ACTIVE(f(X)) → MARK(cons(X, f(g(X))))
MARK(sel(y0, x1)) → ACTIVE(sel(mark(y0), x1))
MARK(g(f(x0))) → ACTIVE(g(active(f(mark(x0)))))
ACTIVE(sel(s(X), cons(Y, Z))) → MARK(sel(X, Z))
MARK(sel(y0, cons(x0, x1))) → ACTIVE(sel(mark(y0), active(cons(mark(x0), x1))))
MARK(g(cons(x0, x1))) → ACTIVE(g(active(cons(mark(x0), x1))))
MARK(sel(y0, 0)) → ACTIVE(sel(mark(y0), active(0)))
MARK(sel(0, y1)) → ACTIVE(sel(active(0), mark(y1)))
MARK(g(g(x0))) → ACTIVE(g(active(g(mark(x0)))))
ACTIVE(sel(0, cons(X, Y))) → MARK(X)
MARK(sel(y0, s(x0))) → ACTIVE(sel(mark(y0), active(s(mark(x0)))))
MARK(f(X)) → MARK(X)
MARK(g(sel(x0, x1))) → ACTIVE(g(active(sel(mark(x0), mark(x1)))))
MARK(sel(X1, X2)) → MARK(X1)
MARK(sel(sel(x0, x1), y1)) → ACTIVE(sel(active(sel(mark(x0), mark(x1))), mark(y1)))
ACTIVE(g(s(X))) → MARK(s(s(g(X))))
MARK(sel(g(x0), y1)) → ACTIVE(sel(active(g(mark(x0))), mark(y1)))
MARK(sel(f(x0), y1)) → ACTIVE(sel(active(f(mark(x0))), mark(y1)))
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(sel(x0, y1)) → ACTIVE(sel(x0, mark(y1)))
MARK(g(s(x0))) → ACTIVE(g(active(s(mark(x0)))))
MARK(sel(y0, sel(x0, x1))) → ACTIVE(sel(mark(y0), active(sel(mark(x0), mark(x1)))))
MARK(sel(y0, g(x0))) → ACTIVE(sel(mark(y0), active(g(mark(x0)))))
MARK(g(x0)) → ACTIVE(g(x0))
MARK(g(0)) → ACTIVE(g(active(0)))
MARK(sel(X1, X2)) → MARK(X2)
MARK(sel(s(x0), y1)) → ACTIVE(sel(active(s(mark(x0))), mark(y1)))

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.