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(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
length1(ok(X)) → ok(length1(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
length1(ok(X)) → ok(length1(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

S(ok(X)) → S(X)
CONS(mark(X1), X2) → CONS(X1, X2)
FROM(mark(X)) → FROM(X)
LENGTH(ok(X)) → LENGTH(X)
TOP(mark(X)) → PROPER(X)
PROPER(length1(X)) → PROPER(X)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(length(cons(X, Y))) → LENGTH1(Y)
TOP(ok(X)) → ACTIVE(X)
PROPER(length1(X)) → LENGTH1(proper(X))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(length(X)) → LENGTH(proper(X))
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
PROPER(from(X)) → PROPER(X)
ACTIVE(from(X)) → ACTIVE(X)
TOP(ok(X)) → TOP(active(X))
LENGTH1(ok(X)) → LENGTH1(X)
ACTIVE(length(cons(X, Y))) → S(length1(Y))
S(mark(X)) → S(X)
PROPER(from(X)) → FROM(proper(X))
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
PROPER(length(X)) → PROPER(X)
PROPER(s(X)) → S(proper(X))
ACTIVE(from(X)) → S(X)
ACTIVE(s(X)) → ACTIVE(X)
TOP(mark(X)) → TOP(proper(X))
ACTIVE(length1(X)) → LENGTH(X)
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(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
length1(ok(X)) → ok(length1(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

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

S(ok(X)) → S(X)
CONS(mark(X1), X2) → CONS(X1, X2)
FROM(mark(X)) → FROM(X)
LENGTH(ok(X)) → LENGTH(X)
TOP(mark(X)) → PROPER(X)
PROPER(length1(X)) → PROPER(X)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(length(cons(X, Y))) → LENGTH1(Y)
TOP(ok(X)) → ACTIVE(X)
PROPER(length1(X)) → LENGTH1(proper(X))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(length(X)) → LENGTH(proper(X))
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
PROPER(from(X)) → PROPER(X)
ACTIVE(from(X)) → ACTIVE(X)
TOP(ok(X)) → TOP(active(X))
LENGTH1(ok(X)) → LENGTH1(X)
ACTIVE(length(cons(X, Y))) → S(length1(Y))
S(mark(X)) → S(X)
PROPER(from(X)) → FROM(proper(X))
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
PROPER(length(X)) → PROPER(X)
PROPER(s(X)) → S(proper(X))
ACTIVE(from(X)) → S(X)
ACTIVE(s(X)) → ACTIVE(X)
TOP(mark(X)) → TOP(proper(X))
ACTIVE(length1(X)) → LENGTH(X)
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(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
length1(ok(X)) → ok(length1(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 8 SCCs with 16 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:

LENGTH1(ok(X)) → LENGTH1(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
length1(ok(X)) → ok(length1(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

LENGTH1(ok(X)) → LENGTH1(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
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

LENGTH(ok(X)) → LENGTH(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
length1(ok(X)) → ok(length1(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

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

LENGTH(ok(X)) → LENGTH(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
          ↳ 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(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
length1(ok(X)) → ok(length1(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We 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
          ↳ 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
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(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
length1(ok(X)) → ok(length1(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We 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
          ↳ 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
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(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
length1(ok(X)) → ok(length1(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

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

PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(length(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(from(X)) → PROPER(X)
PROPER(length1(X)) → PROPER(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
length1(ok(X)) → ok(length1(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

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

PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(length(X)) → PROPER(X)
PROPER(from(X)) → PROPER(X)
PROPER(length1(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
QDP
            ↳ UsableRulesProof
          ↳ QDP

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

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(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
length1(ok(X)) → ok(length1(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

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

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
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(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
length(ok(X)) → ok(length(X))
length1(ok(X)) → ok(length1(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(cons(x1, x2)) = x1 + x2   
POL(from(x1)) = 2·x1   
POL(length(x1)) = 2·x1   
POL(length1(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
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
          ↳ 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(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
length1(ok(X)) → ok(length1(X))
length(ok(X)) → ok(length(X))
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))
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))

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(length(x0))) → TOP(length(proper(x0)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(length1(x0))) → TOP(length1(proper(x0)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(nil)) → TOP(ok(nil))
TOP(mark(0)) → TOP(ok(0))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesReductionPairsProof
              ↳ QDP
                ↳ Narrowing
QDP
                    ↳ DependencyGraphProof

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

TOP(mark(length(x0))) → TOP(length(proper(x0)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(length1(x0))) → TOP(length1(proper(x0)))
TOP(mark(nil)) → TOP(ok(nil))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(0)) → TOP(ok(0))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
length1(ok(X)) → ok(length1(X))
length(ok(X)) → ok(length(X))
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))
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(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 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesReductionPairsProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ DependencyGraphProof
QDP
                        ↳ Narrowing

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

TOP(mark(length(x0))) → TOP(length(proper(x0)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(length1(x0))) → TOP(length1(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))

The TRS R consists of the following rules:

proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
length1(ok(X)) → ok(length1(X))
length(ok(X)) → ok(length(X))
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))
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))

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(from(x0))) → TOP(from(active(x0)))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(length(cons(x0, x1)))) → TOP(mark(s(length1(x1))))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(ok(length(nil))) → TOP(mark(0))
TOP(ok(length1(x0))) → TOP(mark(length(x0)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesReductionPairsProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ QDP
                        ↳ Narrowing
QDP
                            ↳ DependencyGraphProof

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

TOP(ok(from(x0))) → TOP(from(active(x0)))
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(length(nil))) → TOP(mark(0))
TOP(ok(length1(x0))) → TOP(mark(length(x0)))
TOP(mark(length(x0))) → TOP(length(proper(x0)))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(length(cons(x0, x1)))) → TOP(mark(s(length1(x1))))
TOP(mark(length1(x0))) → TOP(length1(proper(x0)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))

The TRS R consists of the following rules:

proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
length1(ok(X)) → ok(length1(X))
length(ok(X)) → ok(length(X))
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))
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(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 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesReductionPairsProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ 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(s(x0))) → TOP(s(active(x0)))
TOP(mark(length(x0))) → TOP(length(proper(x0)))
TOP(ok(length(cons(x0, x1)))) → TOP(mark(s(length1(x1))))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(length1(x0))) → TOP(length1(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(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(ok(length1(x0))) → TOP(mark(length(x0)))

The TRS R consists of the following rules:

proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
length1(ok(X)) → ok(length1(X))
length(ok(X)) → ok(length(X))
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))
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


TOP(ok(length1(x0))) → TOP(mark(length(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(mark(length(x0))) → TOP(length(proper(x0)))
TOP(ok(length(cons(x0, x1)))) → TOP(mark(s(length1(x1))))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(length1(x0))) → TOP(length1(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(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
Used ordering: Polynomial interpretation [25]:

POL(0) = 0   
POL(TOP(x1)) = x1   
POL(active(x1)) = 0   
POL(cons(x1, x2)) = 0   
POL(from(x1)) = 0   
POL(length(x1)) = 0   
POL(length1(x1)) = 1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(ok(x1)) = x1   
POL(proper(x1)) = 0   
POL(s(x1)) = 0   

The following usable rules [17] were oriented:

from(mark(X)) → mark(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
length1(ok(X)) → ok(length1(X))
s(mark(X)) → mark(s(X))
length(ok(X)) → ok(length(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(ok(X)) → ok(s(X))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesReductionPairsProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ 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(mark(length(x0))) → TOP(length(proper(x0)))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(ok(length(cons(x0, x1)))) → TOP(mark(s(length1(x1))))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(length1(x0))) → TOP(length1(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(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))

The TRS R consists of the following rules:

proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
length1(ok(X)) → ok(length1(X))
length(ok(X)) → ok(length(X))
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))
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


TOP(ok(length(cons(x0, x1)))) → TOP(mark(s(length1(x1))))
The remaining pairs can at least be oriented weakly.

TOP(ok(from(x0))) → TOP(from(active(x0)))
TOP(mark(length(x0))) → TOP(length(proper(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(mark(length1(x0))) → TOP(length1(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(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
Used ordering: Polynomial interpretation [25]:

POL(0) = 0   
POL(TOP(x1)) = x1   
POL(active(x1)) = 0   
POL(cons(x1, x2)) = 1   
POL(from(x1)) = 1   
POL(length(x1)) = 1 + x1   
POL(length1(x1)) = 0   
POL(mark(x1)) = x1   
POL(nil) = 1   
POL(ok(x1)) = x1   
POL(proper(x1)) = x1   
POL(s(x1)) = 0   

The following usable rules [17] were oriented:

from(mark(X)) → mark(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
proper(0) → ok(0)
proper(nil) → ok(nil)
length1(ok(X)) → ok(length1(X))
proper(length1(X)) → length1(proper(X))
s(mark(X)) → mark(s(X))
length(ok(X)) → ok(length(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(ok(X)) → ok(s(X))
proper(length(X)) → length(proper(X))
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesReductionPairsProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ 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(mark(length(x0))) → TOP(length(proper(x0)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(length1(x0))) → TOP(length1(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(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))

The TRS R consists of the following rules:

proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
length1(ok(X)) → ok(length1(X))
length(ok(X)) → ok(length(X))
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))
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


TOP(mark(length(x0))) → TOP(length(proper(x0)))
TOP(mark(length1(x0))) → TOP(length1(proper(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(mark(from(x0))) → TOP(from(proper(x0)))
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(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
Used ordering: Polynomial interpretation [25]:

POL(0) = 0   
POL(TOP(x1)) = x1   
POL(active(x1)) = 0   
POL(cons(x1, x2)) = 1   
POL(from(x1)) = 1   
POL(length(x1)) = 0   
POL(length1(x1)) = 0   
POL(mark(x1)) = 1   
POL(nil) = 0   
POL(ok(x1)) = x1   
POL(proper(x1)) = 0   
POL(s(x1)) = 1   

The following usable rules [17] were oriented:

from(mark(X)) → mark(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
length1(ok(X)) → ok(length1(X))
s(mark(X)) → mark(s(X))
length(ok(X)) → ok(length(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(ok(X)) → ok(s(X))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesReductionPairsProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ QDPOrderProof
                                  ↳ 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(mark(from(x0))) → TOP(from(proper(x0)))
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(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))

The TRS R consists of the following rules:

proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
length1(ok(X)) → ok(length1(X))
length(ok(X)) → ok(length(X))
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))
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


TOP(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(mark(from(x0))) → TOP(from(proper(x0)))
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)))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( from(x1) ) =
/1\
\0/
+
/00\
\00/
·x1

M( cons(x1, x2) ) =
/0\
\0/
+
/00\
\00/
·x1+
/00\
\00/
·x2

M( active(x1) ) =
/0\
\0/
+
/00\
\00/
·x1

M( length1(x1) ) =
/0\
\0/
+
/00\
\00/
·x1

M( mark(x1) ) =
/0\
\0/
+
/10\
\00/
·x1

M( ok(x1) ) =
/0\
\0/
+
/10\
\00/
·x1

M( s(x1) ) =
/0\
\0/
+
/00\
\00/
·x1

M( proper(x1) ) =
/0\
\0/
+
/00\
\00/
·x1

M( length(x1) ) =
/0\
\0/
+
/00\
\00/
·x1

M( 0 ) =
/0\
\0/

M( nil ) =
/0\
\0/

Tuple symbols:
M( TOP(x1) ) = 0+
[1,0]
·x1


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesReductionPairsProof
              ↳ QDP
                ↳ Narrowing
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ QDP
                        ↳ Narrowing
                          ↳ QDP
                            ↳ DependencyGraphProof
                              ↳ QDP
                                ↳ QDPOrderProof
                                  ↳ 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(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))

The TRS R consists of the following rules:

proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(length(X)) → length(proper(X))
proper(nil) → ok(nil)
proper(0) → ok(0)
proper(length1(X)) → length1(proper(X))
length1(ok(X)) → ok(length1(X))
length(ok(X)) → ok(length(X))
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))
active(from(X)) → mark(cons(X, from(s(X))))
active(length(nil)) → mark(0)
active(length(cons(X, Y))) → mark(s(length1(Y)))
active(length1(X)) → mark(length(X))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))

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