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(after2(0, XS)) -> mark1(XS)
active1(after2(s1(N), cons2(X, XS))) -> mark1(after2(N, XS))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(after2(X1, X2)) -> after2(active1(X1), X2)
active1(after2(X1, X2)) -> after2(X1, active1(X2))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
after2(mark1(X1), X2) -> mark1(after2(X1, X2))
after2(X1, mark1(X2)) -> mark1(after2(X1, X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(after2(X1, X2)) -> after2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
after2(ok1(X1), ok1(X2)) -> ok1(after2(X1, X2))
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(after2(0, XS)) -> mark1(XS)
active1(after2(s1(N), cons2(X, XS))) -> mark1(after2(N, XS))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(after2(X1, X2)) -> after2(active1(X1), X2)
active1(after2(X1, X2)) -> after2(X1, active1(X2))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
after2(mark1(X1), X2) -> mark1(after2(X1, X2))
after2(X1, mark1(X2)) -> mark1(after2(X1, X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(after2(X1, X2)) -> after2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
after2(ok1(X1), ok1(X2)) -> ok1(after2(X1, X2))
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)
PROPER1(after2(X1, X2)) -> PROPER1(X2)
PROPER1(after2(X1, X2)) -> PROPER1(X1)
ACTIVE1(from1(X)) -> FROM1(active1(X))
ACTIVE1(after2(X1, X2)) -> ACTIVE1(X2)
S1(mark1(X)) -> S1(X)
AFTER2(X1, mark1(X2)) -> AFTER2(X1, X2)
ACTIVE1(after2(s1(N), cons2(X, XS))) -> AFTER2(N, XS)
PROPER1(after2(X1, X2)) -> AFTER2(proper1(X1), proper1(X2))
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(s1(X)) -> S1(proper1(X))
AFTER2(ok1(X1), ok1(X2)) -> AFTER2(X1, X2)
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)
FROM1(mark1(X)) -> FROM1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
ACTIVE1(after2(X1, X2)) -> AFTER2(X1, active1(X2))
ACTIVE1(after2(X1, X2)) -> AFTER2(active1(X1), X2)
AFTER2(mark1(X1), X2) -> AFTER2(X1, X2)
TOP1(ok1(X)) -> ACTIVE1(X)
ACTIVE1(after2(X1, X2)) -> ACTIVE1(X1)
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(after2(0, XS)) -> mark1(XS)
active1(after2(s1(N), cons2(X, XS))) -> mark1(after2(N, XS))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(after2(X1, X2)) -> after2(active1(X1), X2)
active1(after2(X1, X2)) -> after2(X1, active1(X2))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
after2(mark1(X1), X2) -> mark1(after2(X1, X2))
after2(X1, mark1(X2)) -> mark1(after2(X1, X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(after2(X1, X2)) -> after2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
after2(ok1(X1), ok1(X2)) -> ok1(after2(X1, X2))
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)
PROPER1(after2(X1, X2)) -> PROPER1(X2)
PROPER1(after2(X1, X2)) -> PROPER1(X1)
ACTIVE1(from1(X)) -> FROM1(active1(X))
ACTIVE1(after2(X1, X2)) -> ACTIVE1(X2)
S1(mark1(X)) -> S1(X)
AFTER2(X1, mark1(X2)) -> AFTER2(X1, X2)
ACTIVE1(after2(s1(N), cons2(X, XS))) -> AFTER2(N, XS)
PROPER1(after2(X1, X2)) -> AFTER2(proper1(X1), proper1(X2))
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(s1(X)) -> S1(proper1(X))
AFTER2(ok1(X1), ok1(X2)) -> AFTER2(X1, X2)
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)
FROM1(mark1(X)) -> FROM1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
ACTIVE1(after2(X1, X2)) -> AFTER2(X1, active1(X2))
ACTIVE1(after2(X1, X2)) -> AFTER2(active1(X1), X2)
AFTER2(mark1(X1), X2) -> AFTER2(X1, X2)
TOP1(ok1(X)) -> ACTIVE1(X)
ACTIVE1(after2(X1, X2)) -> ACTIVE1(X1)
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(after2(0, XS)) -> mark1(XS)
active1(after2(s1(N), cons2(X, XS))) -> mark1(after2(N, XS))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(after2(X1, X2)) -> after2(active1(X1), X2)
active1(after2(X1, X2)) -> after2(X1, active1(X2))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
after2(mark1(X1), X2) -> mark1(after2(X1, X2))
after2(X1, mark1(X2)) -> mark1(after2(X1, X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(after2(X1, X2)) -> after2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
after2(ok1(X1), ok1(X2)) -> ok1(after2(X1, X2))
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 7 SCCs with 15 less nodes.

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

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

AFTER2(ok1(X1), ok1(X2)) -> AFTER2(X1, X2)
AFTER2(mark1(X1), X2) -> AFTER2(X1, X2)
AFTER2(X1, mark1(X2)) -> AFTER2(X1, X2)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(after2(0, XS)) -> mark1(XS)
active1(after2(s1(N), cons2(X, XS))) -> mark1(after2(N, XS))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(after2(X1, X2)) -> after2(active1(X1), X2)
active1(after2(X1, X2)) -> after2(X1, active1(X2))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
after2(mark1(X1), X2) -> mark1(after2(X1, X2))
after2(X1, mark1(X2)) -> mark1(after2(X1, X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(after2(X1, X2)) -> after2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
after2(ok1(X1), ok1(X2)) -> ok1(after2(X1, X2))
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 oriented strictly and are deleted.


AFTER2(ok1(X1), ok1(X2)) -> AFTER2(X1, X2)
AFTER2(mark1(X1), X2) -> AFTER2(X1, X2)
AFTER2(X1, mark1(X2)) -> AFTER2(X1, X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( AFTER2(x1, x2) ) = 3x1 + x2 + 1


POL( mark1(x1) ) = 3x1 + 3


POL( ok1(x1) ) = 2x1 + 3



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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(after2(0, XS)) -> mark1(XS)
active1(after2(s1(N), cons2(X, XS))) -> mark1(after2(N, XS))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(after2(X1, X2)) -> after2(active1(X1), X2)
active1(after2(X1, X2)) -> after2(X1, active1(X2))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
after2(mark1(X1), X2) -> mark1(after2(X1, X2))
after2(X1, mark1(X2)) -> mark1(after2(X1, X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(after2(X1, X2)) -> after2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
after2(ok1(X1), ok1(X2)) -> ok1(after2(X1, X2))
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

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(after2(0, XS)) -> mark1(XS)
active1(after2(s1(N), cons2(X, XS))) -> mark1(after2(N, XS))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(after2(X1, X2)) -> after2(active1(X1), X2)
active1(after2(X1, X2)) -> after2(X1, active1(X2))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
after2(mark1(X1), X2) -> mark1(after2(X1, X2))
after2(X1, mark1(X2)) -> mark1(after2(X1, X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(after2(X1, X2)) -> after2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
after2(ok1(X1), ok1(X2)) -> ok1(after2(X1, X2))
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 oriented strictly and are deleted.


S1(ok1(X)) -> S1(X)
S1(mark1(X)) -> S1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( S1(x1) ) = 2x1


POL( mark1(x1) ) = 2x1 + 2


POL( ok1(x1) ) = 3x1 + 1



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ 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(after2(0, XS)) -> mark1(XS)
active1(after2(s1(N), cons2(X, XS))) -> mark1(after2(N, XS))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(after2(X1, X2)) -> after2(active1(X1), X2)
active1(after2(X1, X2)) -> after2(X1, active1(X2))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
after2(mark1(X1), X2) -> mark1(after2(X1, X2))
after2(X1, mark1(X2)) -> mark1(after2(X1, X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(after2(X1, X2)) -> after2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
after2(ok1(X1), ok1(X2)) -> ok1(after2(X1, X2))
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

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(after2(0, XS)) -> mark1(XS)
active1(after2(s1(N), cons2(X, XS))) -> mark1(after2(N, XS))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(after2(X1, X2)) -> after2(active1(X1), X2)
active1(after2(X1, X2)) -> after2(X1, active1(X2))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
after2(mark1(X1), X2) -> mark1(after2(X1, X2))
after2(X1, mark1(X2)) -> mark1(after2(X1, X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(after2(X1, X2)) -> after2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
after2(ok1(X1), ok1(X2)) -> ok1(after2(X1, X2))
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 oriented strictly and are deleted.


CONS2(mark1(X1), X2) -> CONS2(X1, X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( CONS2(x1, x2) ) = max{0, 2x1 + 2x2 - 1}


POL( mark1(x1) ) = 2x1 + 1


POL( ok1(x1) ) = x1 + 3



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ 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(after2(0, XS)) -> mark1(XS)
active1(after2(s1(N), cons2(X, XS))) -> mark1(after2(N, XS))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(after2(X1, X2)) -> after2(active1(X1), X2)
active1(after2(X1, X2)) -> after2(X1, active1(X2))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
after2(mark1(X1), X2) -> mark1(after2(X1, X2))
after2(X1, mark1(X2)) -> mark1(after2(X1, X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(after2(X1, X2)) -> after2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
after2(ok1(X1), ok1(X2)) -> ok1(after2(X1, X2))
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

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(after2(0, XS)) -> mark1(XS)
active1(after2(s1(N), cons2(X, XS))) -> mark1(after2(N, XS))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(after2(X1, X2)) -> after2(active1(X1), X2)
active1(after2(X1, X2)) -> after2(X1, active1(X2))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
after2(mark1(X1), X2) -> mark1(after2(X1, X2))
after2(X1, mark1(X2)) -> mark1(after2(X1, X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(after2(X1, X2)) -> after2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
after2(ok1(X1), ok1(X2)) -> ok1(after2(X1, X2))
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 oriented strictly and are deleted.


FROM1(mark1(X)) -> FROM1(X)
FROM1(ok1(X)) -> FROM1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( mark1(x1) ) = 3x1 + 1


POL( FROM1(x1) ) = 2x1


POL( ok1(x1) ) = 2x1 + 2



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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(after2(0, XS)) -> mark1(XS)
active1(after2(s1(N), cons2(X, XS))) -> mark1(after2(N, XS))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(after2(X1, X2)) -> after2(active1(X1), X2)
active1(after2(X1, X2)) -> after2(X1, active1(X2))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
after2(mark1(X1), X2) -> mark1(after2(X1, X2))
after2(X1, mark1(X2)) -> mark1(after2(X1, X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(after2(X1, X2)) -> after2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
after2(ok1(X1), ok1(X2)) -> ok1(after2(X1, X2))
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

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

PROPER1(after2(X1, X2)) -> PROPER1(X2)
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(from1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(after2(X1, X2)) -> PROPER1(X1)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(after2(0, XS)) -> mark1(XS)
active1(after2(s1(N), cons2(X, XS))) -> mark1(after2(N, XS))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(after2(X1, X2)) -> after2(active1(X1), X2)
active1(after2(X1, X2)) -> after2(X1, active1(X2))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
after2(mark1(X1), X2) -> mark1(after2(X1, X2))
after2(X1, mark1(X2)) -> mark1(after2(X1, X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(after2(X1, X2)) -> after2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
after2(ok1(X1), ok1(X2)) -> ok1(after2(X1, X2))
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 oriented strictly and are deleted.


PROPER1(after2(X1, X2)) -> PROPER1(X2)
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(from1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(after2(X1, X2)) -> PROPER1(X1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( after2(x1, x2) ) = 2x1 + 2x2 + 1


POL( from1(x1) ) = 2x1 + 1


POL( s1(x1) ) = x1 + 1


POL( cons2(x1, x2) ) = x1 + 2x2 + 1


POL( PROPER1(x1) ) = x1 + 3



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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(after2(0, XS)) -> mark1(XS)
active1(after2(s1(N), cons2(X, XS))) -> mark1(after2(N, XS))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(after2(X1, X2)) -> after2(active1(X1), X2)
active1(after2(X1, X2)) -> after2(X1, active1(X2))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
after2(mark1(X1), X2) -> mark1(after2(X1, X2))
after2(X1, mark1(X2)) -> mark1(after2(X1, X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(after2(X1, X2)) -> after2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
after2(ok1(X1), ok1(X2)) -> ok1(after2(X1, X2))
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

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)
ACTIVE1(after2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(after2(X1, X2)) -> ACTIVE1(X1)

The TRS R consists of the following rules:

active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(after2(0, XS)) -> mark1(XS)
active1(after2(s1(N), cons2(X, XS))) -> mark1(after2(N, XS))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(after2(X1, X2)) -> after2(active1(X1), X2)
active1(after2(X1, X2)) -> after2(X1, active1(X2))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
after2(mark1(X1), X2) -> mark1(after2(X1, X2))
after2(X1, mark1(X2)) -> mark1(after2(X1, X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(after2(X1, X2)) -> after2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
after2(ok1(X1), ok1(X2)) -> ok1(after2(X1, X2))
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 oriented strictly and are deleted.


ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(from1(X)) -> ACTIVE1(X)
ACTIVE1(s1(X)) -> ACTIVE1(X)
ACTIVE1(after2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(after2(X1, X2)) -> ACTIVE1(X1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( after2(x1, x2) ) = 2x1 + 2x2 + 3


POL( from1(x1) ) = 3x1 + 3


POL( s1(x1) ) = 2x1 + 3


POL( ACTIVE1(x1) ) = x1 + 2


POL( cons2(x1, x2) ) = 2x1 + 3x2 + 3



The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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(after2(0, XS)) -> mark1(XS)
active1(after2(s1(N), cons2(X, XS))) -> mark1(after2(N, XS))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(after2(X1, X2)) -> after2(active1(X1), X2)
active1(after2(X1, X2)) -> after2(X1, active1(X2))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
after2(mark1(X1), X2) -> mark1(after2(X1, X2))
after2(X1, mark1(X2)) -> mark1(after2(X1, X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(after2(X1, X2)) -> after2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
after2(ok1(X1), ok1(X2)) -> ok1(after2(X1, X2))
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

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(after2(0, XS)) -> mark1(XS)
active1(after2(s1(N), cons2(X, XS))) -> mark1(after2(N, XS))
active1(from1(X)) -> from1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(after2(X1, X2)) -> after2(active1(X1), X2)
active1(after2(X1, X2)) -> after2(X1, active1(X2))
from1(mark1(X)) -> mark1(from1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
after2(mark1(X1), X2) -> mark1(after2(X1, X2))
after2(X1, mark1(X2)) -> mark1(after2(X1, X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(after2(X1, X2)) -> after2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
from1(ok1(X)) -> ok1(from1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
after2(ok1(X1), ok1(X2)) -> ok1(after2(X1, X2))
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.