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:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ACTIVE1(from1(X)) -> FROM1(s1(X))
PROPER1(from1(X)) -> FROM1(proper1(X))
TOP1(mark1(X)) -> PROPER1(X)
LENGTH1(ok1(X)) -> LENGTH1(X)
PROPER1(length1(X)) -> PROPER1(X)
LENGTH11(ok1(X)) -> LENGTH11(X)
ACTIVE1(from1(X)) -> FROM1(active1(X))
ACTIVE1(length1(cons2(X, Y))) -> S1(length11(Y))
PROPER1(length11(X)) -> PROPER1(X)
S1(mark1(X)) -> S1(X)
PROPER1(length11(X)) -> LENGTH11(proper1(X))
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(s1(X)) -> S1(proper1(X))
S1(ok1(X)) -> S1(X)
ACTIVE1(s1(X)) -> ACTIVE1(X)
ACTIVE1(from1(X)) -> CONS2(X, from1(s1(X)))
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
ACTIVE1(cons2(X1, X2)) -> CONS2(active1(X1), X2)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(length11(X)) -> LENGTH1(X)
FROM1(mark1(X)) -> FROM1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
ACTIVE1(length1(cons2(X, Y))) -> LENGTH11(Y)
PROPER1(length1(X)) -> LENGTH1(proper1(X))
TOP1(ok1(X)) -> ACTIVE1(X)
ACTIVE1(from1(X)) -> S1(X)
ACTIVE1(s1(X)) -> S1(active1(X))
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
PROPER1(s1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
PROPER1(from1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
ACTIVE1(from1(X)) -> ACTIVE1(X)
FROM1(ok1(X)) -> FROM1(X)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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:

ACTIVE1(from1(X)) -> FROM1(s1(X))
PROPER1(from1(X)) -> FROM1(proper1(X))
TOP1(mark1(X)) -> PROPER1(X)
LENGTH1(ok1(X)) -> LENGTH1(X)
PROPER1(length1(X)) -> PROPER1(X)
LENGTH11(ok1(X)) -> LENGTH11(X)
ACTIVE1(from1(X)) -> FROM1(active1(X))
ACTIVE1(length1(cons2(X, Y))) -> S1(length11(Y))
PROPER1(length11(X)) -> PROPER1(X)
S1(mark1(X)) -> S1(X)
PROPER1(length11(X)) -> LENGTH11(proper1(X))
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(s1(X)) -> S1(proper1(X))
S1(ok1(X)) -> S1(X)
ACTIVE1(s1(X)) -> ACTIVE1(X)
ACTIVE1(from1(X)) -> CONS2(X, from1(s1(X)))
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
ACTIVE1(cons2(X1, X2)) -> CONS2(active1(X1), X2)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(length11(X)) -> LENGTH1(X)
FROM1(mark1(X)) -> FROM1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
ACTIVE1(length1(cons2(X, Y))) -> LENGTH11(Y)
PROPER1(length1(X)) -> LENGTH1(proper1(X))
TOP1(ok1(X)) -> ACTIVE1(X)
ACTIVE1(from1(X)) -> S1(X)
ACTIVE1(s1(X)) -> S1(active1(X))
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
PROPER1(s1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
PROPER1(from1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
ACTIVE1(from1(X)) -> ACTIVE1(X)
FROM1(ok1(X)) -> FROM1(X)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 8 SCCs with 16 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

LENGTH11(ok1(X)) -> LENGTH11(X)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


LENGTH11(ok1(X)) -> LENGTH11(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LENGTH11(x1)  =  LENGTH11(x1)
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
ok1 > LENGTH11


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

LENGTH1(ok1(X)) -> LENGTH1(X)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


LENGTH1(ok1(X)) -> LENGTH1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LENGTH1(x1)  =  LENGTH1(x1)
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
ok1 > LENGTH1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

S1(ok1(X)) -> S1(X)
S1(mark1(X)) -> S1(X)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


S1(mark1(X)) -> S1(X)
The remaining pairs can at least by weakly be oriented.

S1(ok1(X)) -> S1(X)
Used ordering: Combined order from the following AFS and order.
S1(x1)  =  S1(x1)
ok1(x1)  =  x1
mark1(x1)  =  mark1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

S1(ok1(X)) -> S1(X)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


S1(ok1(X)) -> S1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
S1(x1)  =  S1(x1)
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
ok1 > S1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

CONS2(mark1(X1), X2) -> CONS2(X1, X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
The remaining pairs can at least by weakly be oriented.

CONS2(mark1(X1), X2) -> CONS2(X1, X2)
Used ordering: Combined order from the following AFS and order.
CONS2(x1, x2)  =  x2
mark1(x1)  =  mark
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

CONS2(mark1(X1), X2) -> CONS2(X1, X2)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


CONS2(mark1(X1), X2) -> CONS2(X1, X2)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
CONS2(x1, x2)  =  CONS1(x1)
mark1(x1)  =  mark1(x1)

Lexicographic Path Order [19].
Precedence:
[CONS1, mark1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

FROM1(mark1(X)) -> FROM1(X)
FROM1(ok1(X)) -> FROM1(X)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


FROM1(ok1(X)) -> FROM1(X)
The remaining pairs can at least by weakly be oriented.

FROM1(mark1(X)) -> FROM1(X)
Used ordering: Combined order from the following AFS and order.
FROM1(x1)  =  FROM1(x1)
mark1(x1)  =  x1
ok1(x1)  =  ok1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

FROM1(mark1(X)) -> FROM1(X)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


FROM1(mark1(X)) -> FROM1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
FROM1(x1)  =  FROM1(x1)
mark1(x1)  =  mark1(x1)

Lexicographic Path Order [19].
Precedence:
mark1 > FROM1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

PROPER1(length1(X)) -> PROPER1(X)
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(from1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(length11(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
The remaining pairs can at least by weakly be oriented.

PROPER1(length1(X)) -> PROPER1(X)
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(from1(X)) -> PROPER1(X)
PROPER1(length11(X)) -> PROPER1(X)
Used ordering: Combined order from the following AFS and order.
PROPER1(x1)  =  PROPER1(x1)
length1(x1)  =  x1
s1(x1)  =  x1
from1(x1)  =  x1
cons2(x1, x2)  =  cons2(x1, x2)
length11(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
cons2 > PROPER1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

PROPER1(s1(X)) -> PROPER1(X)
PROPER1(length1(X)) -> PROPER1(X)
PROPER1(from1(X)) -> PROPER1(X)
PROPER1(length11(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


PROPER1(length1(X)) -> PROPER1(X)
The remaining pairs can at least by weakly be oriented.

PROPER1(s1(X)) -> PROPER1(X)
PROPER1(from1(X)) -> PROPER1(X)
PROPER1(length11(X)) -> PROPER1(X)
Used ordering: Combined order from the following AFS and order.
PROPER1(x1)  =  PROPER1(x1)
s1(x1)  =  x1
length1(x1)  =  length1(x1)
from1(x1)  =  x1
length11(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

PROPER1(s1(X)) -> PROPER1(X)
PROPER1(from1(X)) -> PROPER1(X)
PROPER1(length11(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


PROPER1(from1(X)) -> PROPER1(X)
The remaining pairs can at least by weakly be oriented.

PROPER1(s1(X)) -> PROPER1(X)
PROPER1(length11(X)) -> PROPER1(X)
Used ordering: Combined order from the following AFS and order.
PROPER1(x1)  =  PROPER1(x1)
s1(x1)  =  x1
from1(x1)  =  from1(x1)
length11(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
[PROPER1, from1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

PROPER1(s1(X)) -> PROPER1(X)
PROPER1(length11(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


PROPER1(length11(X)) -> PROPER1(X)
The remaining pairs can at least by weakly be oriented.

PROPER1(s1(X)) -> PROPER1(X)
Used ordering: Combined order from the following AFS and order.
PROPER1(x1)  =  PROPER1(x1)
s1(x1)  =  x1
length11(x1)  =  length11(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
QDP
                            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

PROPER1(s1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


PROPER1(s1(X)) -> PROPER1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
PROPER1(x1)  =  PROPER1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
s1 > PROPER1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ QDPOrderProof
                          ↳ QDP
                            ↳ QDPOrderProof
QDP
                                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
QDP
            ↳ QDPOrderProof
          ↳ QDP

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

ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(from1(X)) -> ACTIVE1(X)
ACTIVE1(s1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
The remaining pairs can at least by weakly be oriented.

ACTIVE1(from1(X)) -> ACTIVE1(X)
ACTIVE1(s1(X)) -> ACTIVE1(X)
Used ordering: Combined order from the following AFS and order.
ACTIVE1(x1)  =  ACTIVE1(x1)
cons2(x1, x2)  =  cons1(x1)
from1(x1)  =  x1
s1(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
[ACTIVE1, cons1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ QDPOrderProof
          ↳ QDP

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

ACTIVE1(from1(X)) -> ACTIVE1(X)
ACTIVE1(s1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ACTIVE1(s1(X)) -> ACTIVE1(X)
The remaining pairs can at least by weakly be oriented.

ACTIVE1(from1(X)) -> ACTIVE1(X)
Used ordering: Combined order from the following AFS and order.
ACTIVE1(x1)  =  ACTIVE1(x1)
from1(x1)  =  x1
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ QDPOrderProof
          ↳ QDP

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

ACTIVE1(from1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ACTIVE1(from1(X)) -> ACTIVE1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ACTIVE1(x1)  =  ACTIVE1(x1)
from1(x1)  =  from1(x1)

Lexicographic Path Order [19].
Precedence:
from1 > ACTIVE1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
          ↳ QDP

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

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

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

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

TOP1(ok1(X)) -> TOP1(active1(X))
TOP1(mark1(X)) -> TOP1(proper1(X))

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(length1(nil)) -> mark1(0)
active1(length1(cons2(X, Y))) -> mark1(s1(length11(Y)))
active1(length11(X)) -> mark1(length1(X))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(length1(X)) -> length1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(0) -> ok1(0)
proper1(length11(X)) -> length11(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
length1(ok1(X)) -> ok1(length1(X))
length11(ok1(X)) -> ok1(length11(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

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