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(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(after(X1, X2)) → after(active(X1), X2)
active(after(X1, X2)) → after(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
after(mark(X1), X2) → mark(after(X1, X2))
after(X1, mark(X2)) → mark(after(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(after(X1, X2)) → after(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
after(ok(X1), ok(X2)) → ok(after(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(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(after(X1, X2)) → after(active(X1), X2)
active(after(X1, X2)) → after(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
after(mark(X1), X2) → mark(after(X1, X2))
after(X1, mark(X2)) → mark(after(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(after(X1, X2)) → after(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
after(ok(X1), ok(X2)) → ok(after(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)
CONS(mark(X1), X2) → CONS(X1, X2)
FROM(mark(X)) → FROM(X)
ACTIVE(after(X1, X2)) → ACTIVE(X1)
TOP(mark(X)) → PROPER(X)
PROPER(after(X1, X2)) → PROPER(X2)
ACTIVE(after(X1, X2)) → AFTER(X1, active(X2))
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
TOP(ok(X)) → ACTIVE(X)
PROPER(after(X1, X2)) → AFTER(proper(X1), proper(X2))
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
AFTER(mark(X1), X2) → AFTER(X1, X2)
PROPER(from(X)) → PROPER(X)
AFTER(ok(X1), ok(X2)) → AFTER(X1, X2)
ACTIVE(from(X)) → ACTIVE(X)
TOP(ok(X)) → TOP(active(X))
ACTIVE(after(s(N), cons(X, XS))) → AFTER(N, XS)
S(mark(X)) → S(X)
PROPER(after(X1, X2)) → PROPER(X1)
PROPER(from(X)) → FROM(proper(X))
ACTIVE(after(X1, X2)) → ACTIVE(X2)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
PROPER(s(X)) → S(proper(X))
AFTER(X1, mark(X2)) → AFTER(X1, X2)
ACTIVE(from(X)) → S(X)
ACTIVE(s(X)) → ACTIVE(X)
TOP(mark(X)) → TOP(proper(X))
ACTIVE(after(X1, X2)) → AFTER(active(X1), X2)
ACTIVE(from(X)) → FROM(active(X))
ACTIVE(from(X)) → CONS(X, from(s(X)))
FROM(ok(X)) → FROM(X)
ACTIVE(s(X)) → S(active(X))
ACTIVE(from(X)) → FROM(s(X))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(after(X1, X2)) → after(active(X1), X2)
active(after(X1, X2)) → after(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
after(mark(X1), X2) → mark(after(X1, X2))
after(X1, mark(X2)) → mark(after(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(after(X1, X2)) → after(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
after(ok(X1), ok(X2)) → ok(after(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)
CONS(mark(X1), X2) → CONS(X1, X2)
FROM(mark(X)) → FROM(X)
ACTIVE(after(X1, X2)) → ACTIVE(X1)
TOP(mark(X)) → PROPER(X)
PROPER(after(X1, X2)) → PROPER(X2)
ACTIVE(after(X1, X2)) → AFTER(X1, active(X2))
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
TOP(ok(X)) → ACTIVE(X)
PROPER(after(X1, X2)) → AFTER(proper(X1), proper(X2))
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
AFTER(mark(X1), X2) → AFTER(X1, X2)
PROPER(from(X)) → PROPER(X)
AFTER(ok(X1), ok(X2)) → AFTER(X1, X2)
ACTIVE(from(X)) → ACTIVE(X)
TOP(ok(X)) → TOP(active(X))
ACTIVE(after(s(N), cons(X, XS))) → AFTER(N, XS)
S(mark(X)) → S(X)
PROPER(after(X1, X2)) → PROPER(X1)
PROPER(from(X)) → FROM(proper(X))
ACTIVE(after(X1, X2)) → ACTIVE(X2)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
PROPER(s(X)) → S(proper(X))
AFTER(X1, mark(X2)) → AFTER(X1, X2)
ACTIVE(from(X)) → S(X)
ACTIVE(s(X)) → ACTIVE(X)
TOP(mark(X)) → TOP(proper(X))
ACTIVE(after(X1, X2)) → AFTER(active(X1), X2)
ACTIVE(from(X)) → FROM(active(X))
ACTIVE(from(X)) → CONS(X, from(s(X)))
FROM(ok(X)) → FROM(X)
ACTIVE(s(X)) → S(active(X))
ACTIVE(from(X)) → FROM(s(X))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(after(X1, X2)) → after(active(X1), X2)
active(after(X1, X2)) → after(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
after(mark(X1), X2) → mark(after(X1, X2))
after(X1, mark(X2)) → mark(after(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(after(X1, X2)) → after(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
after(ok(X1), ok(X2)) → ok(after(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 7 SCCs with 15 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(after(X1, X2)) → after(active(X1), X2)
active(after(X1, X2)) → after(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
after(mark(X1), X2) → mark(after(X1, X2))
after(X1, mark(X2)) → mark(after(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(after(X1, X2)) → after(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
after(ok(X1), ok(X2)) → ok(after(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

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

AFTER(mark(X1), X2) → AFTER(X1, X2)
AFTER(ok(X1), ok(X2)) → AFTER(X1, X2)
AFTER(X1, mark(X2)) → AFTER(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:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ 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(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(after(X1, X2)) → after(active(X1), X2)
active(after(X1, X2)) → after(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
after(mark(X1), X2) → mark(after(X1, X2))
after(X1, mark(X2)) → mark(after(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(after(X1, X2)) → after(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
after(ok(X1), ok(X2)) → ok(after(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
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ 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)

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:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(after(X1, X2)) → after(active(X1), X2)
active(after(X1, X2)) → after(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
after(mark(X1), X2) → mark(after(X1, X2))
after(X1, mark(X2)) → mark(after(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(after(X1, X2)) → after(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
after(ok(X1), ok(X2)) → ok(after(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
          ↳ 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)

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:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

FROM(mark(X)) → FROM(X)
FROM(ok(X)) → FROM(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(after(X1, X2)) → after(active(X1), X2)
active(after(X1, X2)) → after(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
after(mark(X1), X2) → mark(after(X1, X2))
after(X1, mark(X2)) → mark(after(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(after(X1, X2)) → after(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
after(ok(X1), ok(X2)) → ok(after(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
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

FROM(mark(X)) → FROM(X)
FROM(ok(X)) → FROM(X)

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:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP

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

PROPER(after(X1, X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(after(X1, X2)) → PROPER(X2)
PROPER(from(X)) → PROPER(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(after(X1, X2)) → after(active(X1), X2)
active(after(X1, X2)) → after(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
after(mark(X1), X2) → mark(after(X1, X2))
after(X1, mark(X2)) → mark(after(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(after(X1, X2)) → after(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
after(ok(X1), ok(X2)) → ok(after(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
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ QDP
          ↳ QDP

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

PROPER(after(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(after(X1, X2)) → PROPER(X2)
PROPER(from(X)) → PROPER(X)

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:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ UsableRulesProof
          ↳ QDP

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

ACTIVE(after(X1, X2)) → ACTIVE(X1)
ACTIVE(after(X1, X2)) → ACTIVE(X2)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(from(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(after(X1, X2)) → after(active(X1), X2)
active(after(X1, X2)) → after(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
after(mark(X1), X2) → mark(after(X1, X2))
after(X1, mark(X2)) → mark(after(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(after(X1, X2)) → after(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
after(ok(X1), ok(X2)) → ok(after(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
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof
          ↳ QDP

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

ACTIVE(after(X1, X2)) → ACTIVE(X1)
ACTIVE(after(X1, X2)) → ACTIVE(X2)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(from(X)) → ACTIVE(X)

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:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(after(X1, X2)) → after(active(X1), X2)
active(after(X1, X2)) → after(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
after(mark(X1), X2) → mark(after(X1, X2))
after(X1, mark(X2)) → mark(after(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(after(X1, X2)) → after(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
after(ok(X1), ok(X2)) → ok(after(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)) = 2·x1   
POL(active(x1)) = 2·x1   
POL(after(x1, x2)) = 2·x1 + x2   
POL(cons(x1, x2)) = x1 + x2   
POL(from(x1)) = x1   
POL(mark(x1)) = x1   
POL(ok(x1)) = 2·x1   
POL(proper(x1)) = x1   
POL(s(x1)) = x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(after(X1, X2)) → after(active(X1), X2)
active(after(X1, X2)) → after(X1, active(X2))
after(mark(X1), X2) → mark(after(X1, X2))
after(X1, mark(X2)) → mark(after(X1, X2))
after(ok(X1), ok(X2)) → ok(after(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(after(X1, X2)) → after(proper(X1), proper(X2))
proper(0) → ok(0)

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(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(0)) → TOP(ok(0))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(after(x0, x1))) → TOP(after(proper(x0), proper(x1)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(0)) → TOP(ok(0))
TOP(ok(X)) → TOP(active(X))
TOP(mark(after(x0, x1))) → TOP(after(proper(x0), proper(x1)))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(after(X1, X2)) → after(active(X1), X2)
active(after(X1, X2)) → after(X1, active(X2))
after(mark(X1), X2) → mark(after(X1, X2))
after(X1, mark(X2)) → mark(after(X1, X2))
after(ok(X1), ok(X2)) → ok(after(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(after(X1, X2)) → after(proper(X1), proper(X2))
proper(0) → ok(0)

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(after(0, x0))) → TOP(mark(x0))
TOP(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(after(s(x0), cons(x1, x2)))) → TOP(mark(after(x0, x2)))
TOP(ok(after(x0, x1))) → TOP(after(active(x0), x1))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(ok(after(x0, x1))) → TOP(after(x0, active(x1)))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(after(s(x0), cons(x1, x2)))) → TOP(mark(after(x0, x2)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(mark(0)) → TOP(ok(0))
TOP(ok(after(0, x0))) → TOP(mark(x0))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(after(x0, x1))) → TOP(after(active(x0), x1))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(ok(after(x0, x1))) → TOP(after(x0, active(x1)))
TOP(mark(after(x0, x1))) → TOP(after(proper(x0), proper(x1)))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(after(X1, X2)) → after(active(X1), X2)
active(after(X1, X2)) → after(X1, active(X2))
after(mark(X1), X2) → mark(after(X1, X2))
after(X1, mark(X2)) → mark(after(X1, X2))
after(ok(X1), ok(X2)) → ok(after(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(after(X1, X2)) → after(proper(X1), proper(X2))
proper(0) → ok(0)

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
            ↳ UsableRulesReductionPairsProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ Narrowing
                      ↳ QDP
                        ↳ DependencyGraphProof
QDP
                            ↳ QDPOrderProof

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

TOP(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(after(0, x0))) → TOP(mark(x0))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(after(s(x0), cons(x1, x2)))) → TOP(mark(after(x0, x2)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(ok(after(x0, x1))) → TOP(after(active(x0), x1))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(ok(after(x0, x1))) → TOP(after(x0, active(x1)))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(mark(after(x0, x1))) → TOP(after(proper(x0), proper(x1)))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(after(X1, X2)) → after(active(X1), X2)
active(after(X1, X2)) → after(X1, active(X2))
after(mark(X1), X2) → mark(after(X1, X2))
after(X1, mark(X2)) → mark(after(X1, X2))
after(ok(X1), ok(X2)) → ok(after(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(after(X1, X2)) → after(proper(X1), proper(X2))
proper(0) → ok(0)

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(after(0, x0))) → TOP(mark(x0))
The remaining pairs can at least be oriented weakly.

TOP(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(after(s(x0), cons(x1, x2)))) → TOP(mark(after(x0, x2)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(ok(after(x0, x1))) → TOP(after(active(x0), x1))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(ok(after(x0, x1))) → TOP(after(x0, active(x1)))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(mark(after(x0, x1))) → TOP(after(proper(x0), proper(x1)))
Used ordering: Polynomial interpretation [25]:

POL(0) = 1   
POL(TOP(x1)) = x1   
POL(active(x1)) = x1   
POL(after(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = x2   
POL(from(x1)) = 0   
POL(mark(x1)) = x1   
POL(ok(x1)) = x1   
POL(proper(x1)) = x1   
POL(s(x1)) = x1   

The following usable rules [17] were oriented:

active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(after(X1, X2)) → after(X1, active(X2))
active(after(X1, X2)) → after(active(X1), X2)
after(X1, mark(X2)) → mark(after(X1, X2))
after(mark(X1), X2) → mark(after(X1, X2))
s(mark(X)) → mark(s(X))
after(ok(X1), ok(X2)) → ok(after(X1, X2))
active(after(0, XS)) → mark(XS)
active(from(X)) → mark(cons(X, from(s(X))))
active(from(X)) → from(active(X))
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
proper(after(X1, X2)) → after(proper(X1), proper(X2))
proper(0) → ok(0)
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(after(s(x0), cons(x1, x2)))) → TOP(mark(after(x0, x2)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(after(x0, x1))) → TOP(after(active(x0), x1))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(ok(after(x0, x1))) → TOP(after(x0, active(x1)))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(mark(after(x0, x1))) → TOP(after(proper(x0), proper(x1)))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(after(X1, X2)) → after(active(X1), X2)
active(after(X1, X2)) → after(X1, active(X2))
after(mark(X1), X2) → mark(after(X1, X2))
after(X1, mark(X2)) → mark(after(X1, X2))
after(ok(X1), ok(X2)) → ok(after(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(after(X1, X2)) → after(proper(X1), proper(X2))
proper(0) → ok(0)

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(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
The remaining pairs can at least be oriented weakly.

TOP(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(after(s(x0), cons(x1, x2)))) → TOP(mark(after(x0, x2)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(after(x0, x1))) → TOP(after(active(x0), x1))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(ok(after(x0, x1))) → TOP(after(x0, active(x1)))
TOP(mark(after(x0, x1))) → TOP(after(proper(x0), proper(x1)))
Used ordering: Polynomial interpretation with max and min functions [25]:

POL(0) = 0   
POL(TOP(x1)) = x1   
POL(active(x1)) = 0   
POL(after(x1, x2)) = 0   
POL(cons(x1, x2)) = 0   
POL(from(x1)) = 1   
POL(mark(x1)) = x1   
POL(ok(x1)) = x1   
POL(proper(x1)) = 0   
POL(s(x1)) = 0   

The following usable rules [17] were oriented:

after(X1, mark(X2)) → mark(after(X1, X2))
after(mark(X1), X2) → mark(after(X1, X2))
s(mark(X)) → mark(s(X))
after(ok(X1), ok(X2)) → ok(after(X1, X2))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(after(s(x0), cons(x1, x2)))) → TOP(mark(after(x0, x2)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(ok(after(x0, x1))) → TOP(after(active(x0), x1))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(ok(after(x0, x1))) → TOP(after(x0, active(x1)))
TOP(mark(after(x0, x1))) → TOP(after(proper(x0), proper(x1)))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(after(X1, X2)) → after(active(X1), X2)
active(after(X1, X2)) → after(X1, active(X2))
after(mark(X1), X2) → mark(after(X1, X2))
after(X1, mark(X2)) → mark(after(X1, X2))
after(ok(X1), ok(X2)) → ok(after(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(after(X1, X2)) → after(proper(X1), proper(X2))
proper(0) → ok(0)

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(after(s(x0), cons(x1, x2)))) → TOP(mark(after(x0, x2)))
The remaining pairs can at least be oriented weakly.

TOP(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(ok(after(x0, x1))) → TOP(after(active(x0), x1))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(ok(after(x0, x1))) → TOP(after(x0, active(x1)))
TOP(mark(after(x0, x1))) → TOP(after(proper(x0), proper(x1)))
Used ordering: Polynomial interpretation with max and min functions [25]:

POL(0) = 0   
POL(TOP(x1)) = x1   
POL(active(x1)) = x1   
POL(after(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = x2   
POL(from(x1)) = 0   
POL(mark(x1)) = x1   
POL(ok(x1)) = x1   
POL(proper(x1)) = x1   
POL(s(x1)) = 1 + x1   

The following usable rules [17] were oriented:

active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(after(X1, X2)) → after(X1, active(X2))
active(after(X1, X2)) → after(active(X1), X2)
after(X1, mark(X2)) → mark(after(X1, X2))
after(mark(X1), X2) → mark(after(X1, X2))
s(mark(X)) → mark(s(X))
after(ok(X1), ok(X2)) → ok(after(X1, X2))
active(after(0, XS)) → mark(XS)
active(from(X)) → mark(cons(X, from(s(X))))
active(from(X)) → from(active(X))
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
proper(after(X1, X2)) → after(proper(X1), proper(X2))
proper(0) → ok(0)
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(after(x0, x1))) → TOP(after(active(x0), x1))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(ok(after(x0, x1))) → TOP(after(x0, active(x1)))
TOP(mark(after(x0, x1))) → TOP(after(proper(x0), proper(x1)))

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(after(X1, X2)) → after(active(X1), X2)
active(after(X1, X2)) → after(X1, active(X2))
after(mark(X1), X2) → mark(after(X1, X2))
after(X1, mark(X2)) → mark(after(X1, X2))
after(ok(X1), ok(X2)) → ok(after(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(after(X1, X2)) → after(proper(X1), proper(X2))
proper(0) → ok(0)

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