Termination w.r.t. Q proof of /tmp/tmpgY02BH/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))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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(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))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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:

S(ok(X)) → S(X)
ACTIVE(f(X)) → G(X)
ACTIVE(sel(X1, X2)) → SEL(active(X1), X2)
CONS(mark(X1), X2) → CONS(X1, X2)
SEL(mark(X1), X2) → SEL(X1, X2)
TOP(mark(X)) → PROPER(X)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(f(X)) → CONS(X, f(g(X)))
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(f(X)) → F(g(X))
TOP(ok(X)) → ACTIVE(X)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
ACTIVE(g(X)) → ACTIVE(X)
SEL(X1, mark(X2)) → SEL(X1, X2)
PROPER(cons(X1, X2)) → PROPER(X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
G(ok(X)) → G(X)
PROPER(g(X)) → G(proper(X))
PROPER(f(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
F(mark(X)) → F(X)
PROPER(g(X)) → PROPER(X)
ACTIVE(sel(X1, X2)) → SEL(X1, active(X2))
S(mark(X)) → S(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
ACTIVE(g(X)) → G(active(X))
PROPER(sel(X1, X2)) → SEL(proper(X1), proper(X2))
PROPER(sel(X1, X2)) → PROPER(X2)
ACTIVE(g(s(X))) → G(X)
PROPER(s(X)) → S(proper(X))
ACTIVE(f(X)) → F(active(X))
G(mark(X)) → G(X)
PROPER(f(X)) → F(proper(X))
ACTIVE(g(0)) → S(0)
ACTIVE(s(X)) → ACTIVE(X)
PROPER(sel(X1, X2)) → PROPER(X1)
TOP(mark(X)) → TOP(proper(X))
F(ok(X)) → F(X)
ACTIVE(g(s(X))) → S(g(X))
ACTIVE(sel(s(X), cons(Y, Z))) → SEL(X, Z)
ACTIVE(g(s(X))) → S(s(g(X)))
ACTIVE(s(X)) → S(active(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))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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:

S(ok(X)) → S(X)
ACTIVE(f(X)) → G(X)
ACTIVE(sel(X1, X2)) → SEL(active(X1), X2)
CONS(mark(X1), X2) → CONS(X1, X2)
SEL(mark(X1), X2) → SEL(X1, X2)
TOP(mark(X)) → PROPER(X)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(f(X)) → CONS(X, f(g(X)))
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(f(X)) → F(g(X))
TOP(ok(X)) → ACTIVE(X)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
ACTIVE(g(X)) → ACTIVE(X)
SEL(X1, mark(X2)) → SEL(X1, X2)
PROPER(cons(X1, X2)) → PROPER(X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
G(ok(X)) → G(X)
PROPER(g(X)) → G(proper(X))
PROPER(f(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
F(mark(X)) → F(X)
PROPER(g(X)) → PROPER(X)
ACTIVE(sel(X1, X2)) → SEL(X1, active(X2))
S(mark(X)) → S(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
ACTIVE(g(X)) → G(active(X))
PROPER(sel(X1, X2)) → SEL(proper(X1), proper(X2))
PROPER(sel(X1, X2)) → PROPER(X2)
ACTIVE(g(s(X))) → G(X)
PROPER(s(X)) → S(proper(X))
ACTIVE(f(X)) → F(active(X))
G(mark(X)) → G(X)
PROPER(f(X)) → F(proper(X))
ACTIVE(g(0)) → S(0)
ACTIVE(s(X)) → ACTIVE(X)
PROPER(sel(X1, X2)) → PROPER(X1)
TOP(mark(X)) → TOP(proper(X))
F(ok(X)) → F(X)
ACTIVE(g(s(X))) → S(g(X))
ACTIVE(sel(s(X), cons(Y, Z))) → SEL(X, Z)
ACTIVE(g(s(X))) → S(s(g(X)))
ACTIVE(s(X)) → S(active(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))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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 21 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL(mark(X1), X2) → SEL(X1, X2)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
SEL(X1, mark(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))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S(ok(X)) → S(X)
S(mark(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))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S(ok(X)) → S(X)
S(mark(X)) → S(X)

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

G(ok(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))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

G(ok(X)) → G(X)
G(mark(X)) → G(X)

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

G(ok(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

          ↳ QDP

          ↳ QDP

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

CONS(mark(X1), X2) → CONS(X1, X2)
CONS(ok(X1), ok(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))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

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

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

          ↳ 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(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))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

PROPER(g(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X2)
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(f(X)) → PROPER(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))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

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

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

PROPER(g(X)) → PROPER(X)
The graph contains the following edges 1 > 1
PROPER(s(X)) → PROPER(X)
The graph contains the following edges 1 > 1
PROPER(cons(X1, X2)) → PROPER(X1)
The graph contains the following edges 1 > 1
PROPER(sel(X1, X2)) → PROPER(X2)
The graph contains the following edges 1 > 1
PROPER(sel(X1, X2)) → PROPER(X1)
The graph contains the following edges 1 > 1
PROPER(cons(X1, X2)) → PROPER(X2)
The graph contains the following edges 1 > 1
PROPER(f(X)) → PROPER(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

ACTIVE(g(X)) → ACTIVE(X)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(f(X)) → ACTIVE(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))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

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

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

ACTIVE(g(X)) → ACTIVE(X)
The graph contains the following edges 1 > 1
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
The graph contains the following edges 1 > 1
ACTIVE(s(X)) → ACTIVE(X)
The graph contains the following edges 1 > 1
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
The graph contains the following edges 1 > 1
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
The graph contains the following edges 1 > 1
ACTIVE(f(X)) → ACTIVE(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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(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))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
g(mark(X)) → mark(g(X))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
g(ok(X)) → ok(g(X))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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(0) = 0
POL(TOP(x₁)) = x₁
POL(active(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = x₁ + x₂
POL(f(x₁)) = x₁
POL(g(x₁)) = x₁
POL(mark(x₁)) = x₁
POL(ok(x₁)) = 2·x₁
POL(proper(x₁)) = x₁
POL(s(x₁)) = x₁
POL(sel(x₁, x₂)) = x₁ + 2·x₂

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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:

proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
g(mark(X)) → mark(g(X))
g(ok(X)) → ok(g(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
f(mark(X)) → mark(f(X))
f(ok(X)) → ok(f(X))
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))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))

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(g(x0))) → TOP(g(proper(x0)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(f(x0))) → TOP(f(proper(x0)))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(0)) → TOP(ok(0))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ Narrowing

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

TOP(mark(g(x0))) → TOP(g(proper(x0)))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(f(x0))) → TOP(f(proper(x0)))
TOP(mark(0)) → TOP(ok(0))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
g(mark(X)) → mark(g(X))
g(ok(X)) → ok(g(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
f(mark(X)) → mark(f(X))
f(ok(X)) → ok(f(X))
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))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))

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(g(s(x0)))) → TOP(mark(s(s(g(x0)))))
TOP(ok(sel(0, cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(g(x0))) → TOP(g(active(x0)))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(ok(f(x0))) → TOP(f(active(x0)))
TOP(ok(g(0))) → TOP(mark(s(0)))
TOP(ok(f(x0))) → TOP(mark(cons(x0, f(g(x0)))))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

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

TOP(mark(g(x0))) → TOP(g(proper(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(f(x0))) → TOP(f(proper(x0)))
TOP(ok(g(0))) → TOP(mark(s(0)))
TOP(mark(0)) → TOP(ok(0))
TOP(ok(g(s(x0)))) → TOP(mark(s(s(g(x0)))))
TOP(ok(sel(0, cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(ok(g(x0))) → TOP(g(active(x0)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(ok(f(x0))) → TOP(f(active(x0)))
TOP(ok(f(x0))) → TOP(mark(cons(x0, f(g(x0)))))

The TRS R consists of the following rules:

proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
g(mark(X)) → mark(g(X))
g(ok(X)) → ok(g(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
f(mark(X)) → mark(f(X))
f(ok(X)) → ok(f(X))
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))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))

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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

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

TOP(mark(g(x0))) → TOP(g(proper(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(f(x0))) → TOP(f(proper(x0)))
TOP(ok(g(0))) → TOP(mark(s(0)))
TOP(ok(g(s(x0)))) → TOP(mark(s(s(g(x0)))))
TOP(ok(sel(0, cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(ok(g(x0))) → TOP(g(active(x0)))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(ok(f(x0))) → TOP(f(active(x0)))
TOP(ok(f(x0))) → TOP(mark(cons(x0, f(g(x0)))))

The TRS R consists of the following rules:

proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
g(mark(X)) → mark(g(X))
g(ok(X)) → ok(g(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
f(mark(X)) → mark(f(X))
f(ok(X)) → ok(f(X))
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))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(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.

TOP(ok(sel(0, cons(x0, x1)))) → TOP(mark(x0))

The remaining pairs can at least be oriented weakly.

TOP(mark(g(x0))) → TOP(g(proper(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(f(x0))) → TOP(f(proper(x0)))
TOP(ok(g(0))) → TOP(mark(s(0)))
TOP(ok(g(s(x0)))) → TOP(mark(s(s(g(x0)))))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(ok(g(x0))) → TOP(g(active(x0)))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(ok(f(x0))) → TOP(f(active(x0)))
TOP(ok(f(x0))) → TOP(mark(cons(x0, f(g(x0)))))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(TOP(x₁)) = x₁
POL(active(x₁)) = x₁
POL(cons(x₁, x₂)) = x₁ + x₂
POL(f(x₁)) = x₁
POL(g(x₁)) = 0
POL(mark(x₁)) = x₁
POL(ok(x₁)) = x₁
POL(proper(x₁)) = x₁
POL(s(x₁)) = 0
POL(sel(x₁, x₂)) = 1 + x₂

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

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

TOP(mark(g(x0))) → TOP(g(proper(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(f(x0))) → TOP(f(proper(x0)))
TOP(ok(g(0))) → TOP(mark(s(0)))
TOP(ok(g(s(x0)))) → TOP(mark(s(s(g(x0)))))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(ok(g(x0))) → TOP(g(active(x0)))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(ok(f(x0))) → TOP(f(active(x0)))
TOP(ok(f(x0))) → TOP(mark(cons(x0, f(g(x0)))))

The TRS R consists of the following rules:

proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
g(mark(X)) → mark(g(X))
g(ok(X)) → ok(g(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
f(mark(X)) → mark(f(X))
f(ok(X)) → ok(f(X))
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))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(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.

TOP(ok(f(x0))) → TOP(mark(cons(x0, f(g(x0)))))

The remaining pairs can at least be oriented weakly.

TOP(mark(g(x0))) → TOP(g(proper(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(f(x0))) → TOP(f(proper(x0)))
TOP(ok(g(0))) → TOP(mark(s(0)))
TOP(ok(g(s(x0)))) → TOP(mark(s(s(g(x0)))))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(ok(g(x0))) → TOP(g(active(x0)))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(ok(f(x0))) → TOP(f(active(x0)))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(TOP(x₁)) = x₁
POL(active(x₁)) = 0
POL(cons(x₁, x₂)) = 0
POL(f(x₁)) = 1
POL(g(x₁)) = 0
POL(mark(x₁)) = x₁
POL(ok(x₁)) = x₁
POL(proper(x₁)) = 0
POL(s(x₁)) = 0
POL(sel(x₁, x₂)) = 0

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

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

TOP(mark(g(x0))) → TOP(g(proper(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(f(x0))) → TOP(f(proper(x0)))
TOP(ok(g(0))) → TOP(mark(s(0)))
TOP(ok(g(s(x0)))) → TOP(mark(s(s(g(x0)))))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(ok(g(x0))) → TOP(g(active(x0)))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(ok(f(x0))) → TOP(f(active(x0)))

The TRS R consists of the following rules:

proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
g(mark(X)) → mark(g(X))
g(ok(X)) → ok(g(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
f(mark(X)) → mark(f(X))
f(ok(X)) → ok(f(X))
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))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(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.

TOP(ok(g(0))) → TOP(mark(s(0)))
TOP(ok(g(s(x0)))) → TOP(mark(s(s(g(x0)))))

The remaining pairs can at least be oriented weakly.

TOP(mark(g(x0))) → TOP(g(proper(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(f(x0))) → TOP(f(proper(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(ok(g(x0))) → TOP(g(active(x0)))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(ok(f(x0))) → TOP(f(active(x0)))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(TOP(x₁)) = x₁
POL(active(x₁)) = 0
POL(cons(x₁, x₂)) = 0
POL(f(x₁)) = 0
POL(g(x₁)) = 1
POL(mark(x₁)) = x₁
POL(ok(x₁)) = x₁
POL(proper(x₁)) = 0
POL(s(x₁)) = 0
POL(sel(x₁, x₂)) = 0

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesReductionPairsProof

              ↳ QDP

                ↳ Narrowing

                  ↳ QDP

                    ↳ Narrowing

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

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

TOP(mark(g(x0))) → TOP(g(proper(x0)))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(g(x0))) → TOP(g(active(x0)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(f(x0))) → TOP(f(proper(x0)))
TOP(ok(f(x0))) → TOP(f(active(x0)))

The TRS R consists of the following rules:

proper(f(X)) → f(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
g(mark(X)) → mark(g(X))
g(ok(X)) → ok(g(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
f(mark(X)) → mark(f(X))
f(ok(X)) → ok(f(X))
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))
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(g(X)) → g(active(X))
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))

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