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(zeros) -> mark1(cons2(0, zeros))
active1(tail1(cons2(X, XS))) -> mark1(XS)
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(tail1(X)) -> tail1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(tail1(X)) -> tail1(proper1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
tail1(ok1(X)) -> ok1(tail1(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(zeros) -> mark1(cons2(0, zeros))
active1(tail1(cons2(X, XS))) -> mark1(XS)
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(tail1(X)) -> tail1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(tail1(X)) -> tail1(proper1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.

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

CONS2(mark1(X1), X2) -> CONS2(X1, X2)
TOP1(mark1(X)) -> PROPER1(X)
ACTIVE1(zeros) -> CONS2(0, zeros)
ACTIVE1(cons2(X1, X2)) -> CONS2(active1(X1), X2)
TAIL1(mark1(X)) -> TAIL1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
TAIL1(ok1(X)) -> TAIL1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(tail1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> ACTIVE1(X)
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
TOP1(ok1(X)) -> TOP1(active1(X))
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(tail1(X)) -> TAIL1(proper1(X))
ACTIVE1(tail1(X)) -> TAIL1(active1(X))
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
ACTIVE1(tail1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(zeros) -> mark1(cons2(0, zeros))
active1(tail1(cons2(X, XS))) -> mark1(XS)
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(tail1(X)) -> tail1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(tail1(X)) -> tail1(proper1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
tail1(ok1(X)) -> ok1(tail1(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:

CONS2(mark1(X1), X2) -> CONS2(X1, X2)
TOP1(mark1(X)) -> PROPER1(X)
ACTIVE1(zeros) -> CONS2(0, zeros)
ACTIVE1(cons2(X1, X2)) -> CONS2(active1(X1), X2)
TAIL1(mark1(X)) -> TAIL1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
TAIL1(ok1(X)) -> TAIL1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(tail1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> ACTIVE1(X)
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
TOP1(ok1(X)) -> TOP1(active1(X))
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(tail1(X)) -> TAIL1(proper1(X))
ACTIVE1(tail1(X)) -> TAIL1(active1(X))
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
ACTIVE1(tail1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(zeros) -> mark1(cons2(0, zeros))
active1(tail1(cons2(X, XS))) -> mark1(XS)
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(tail1(X)) -> tail1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(tail1(X)) -> tail1(proper1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
tail1(ok1(X)) -> ok1(tail1(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 contains 5 SCCs with 7 less nodes.

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

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

TAIL1(mark1(X)) -> TAIL1(X)
TAIL1(ok1(X)) -> TAIL1(X)

The TRS R consists of the following rules:

active1(zeros) -> mark1(cons2(0, zeros))
active1(tail1(cons2(X, XS))) -> mark1(XS)
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(tail1(X)) -> tail1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(tail1(X)) -> tail1(proper1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
tail1(ok1(X)) -> ok1(tail1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TAIL1(ok1(X)) -> TAIL1(X)
Used argument filtering: TAIL1(x1)  =  x1
mark1(x1)  =  x1
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


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

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

TAIL1(mark1(X)) -> TAIL1(X)

The TRS R consists of the following rules:

active1(zeros) -> mark1(cons2(0, zeros))
active1(tail1(cons2(X, XS))) -> mark1(XS)
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(tail1(X)) -> tail1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(tail1(X)) -> tail1(proper1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
tail1(ok1(X)) -> ok1(tail1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TAIL1(mark1(X)) -> TAIL1(X)
Used argument filtering: TAIL1(x1)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(zeros) -> mark1(cons2(0, zeros))
active1(tail1(cons2(X, XS))) -> mark1(XS)
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(tail1(X)) -> tail1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(tail1(X)) -> tail1(proper1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
tail1(ok1(X)) -> ok1(tail1(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
            ↳ QDPAfsSolverProof
          ↳ 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(zeros) -> mark1(cons2(0, zeros))
active1(tail1(cons2(X, XS))) -> mark1(XS)
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(tail1(X)) -> tail1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(tail1(X)) -> tail1(proper1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
tail1(ok1(X)) -> ok1(tail1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
Used argument filtering: CONS2(x1, x2)  =  x2
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ 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(zeros) -> mark1(cons2(0, zeros))
active1(tail1(cons2(X, XS))) -> mark1(XS)
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(tail1(X)) -> tail1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(tail1(X)) -> tail1(proper1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
tail1(ok1(X)) -> ok1(tail1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CONS2(mark1(X1), X2) -> CONS2(X1, X2)
Used argument filtering: CONS2(x1, x2)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(zeros) -> mark1(cons2(0, zeros))
active1(tail1(cons2(X, XS))) -> mark1(XS)
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(tail1(X)) -> tail1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(tail1(X)) -> tail1(proper1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
tail1(ok1(X)) -> ok1(tail1(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
            ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

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

PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(tail1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(zeros) -> mark1(cons2(0, zeros))
active1(tail1(cons2(X, XS))) -> mark1(XS)
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(tail1(X)) -> tail1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(tail1(X)) -> tail1(proper1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
tail1(ok1(X)) -> ok1(tail1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
Used argument filtering: PROPER1(x1)  =  x1
cons2(x1, x2)  =  cons2(x1, x2)
tail1(x1)  =  x1
Used ordering: Quasi Precedence: trivial


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

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

PROPER1(tail1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(zeros) -> mark1(cons2(0, zeros))
active1(tail1(cons2(X, XS))) -> mark1(XS)
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(tail1(X)) -> tail1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(tail1(X)) -> tail1(proper1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
tail1(ok1(X)) -> ok1(tail1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER1(tail1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1)  =  x1
tail1(x1)  =  tail1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

active1(zeros) -> mark1(cons2(0, zeros))
active1(tail1(cons2(X, XS))) -> mark1(XS)
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(tail1(X)) -> tail1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(tail1(X)) -> tail1(proper1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
tail1(ok1(X)) -> ok1(tail1(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
            ↳ QDPAfsSolverProof
          ↳ QDP

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

ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(tail1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(zeros) -> mark1(cons2(0, zeros))
active1(tail1(cons2(X, XS))) -> mark1(XS)
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(tail1(X)) -> tail1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(tail1(X)) -> tail1(proper1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
tail1(ok1(X)) -> ok1(tail1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE1(tail1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1)  =  x1
cons2(x1, x2)  =  x1
tail1(x1)  =  tail1(x1)
Used ordering: Quasi Precedence: trivial


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

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

ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)

The TRS R consists of the following rules:

active1(zeros) -> mark1(cons2(0, zeros))
active1(tail1(cons2(X, XS))) -> mark1(XS)
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(tail1(X)) -> tail1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(tail1(X)) -> tail1(proper1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
tail1(ok1(X)) -> ok1(tail1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
Used argument filtering: ACTIVE1(x1)  =  x1
cons2(x1, x2)  =  cons1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP

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

active1(zeros) -> mark1(cons2(0, zeros))
active1(tail1(cons2(X, XS))) -> mark1(XS)
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(tail1(X)) -> tail1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(tail1(X)) -> tail1(proper1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
tail1(ok1(X)) -> ok1(tail1(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

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(zeros) -> mark1(cons2(0, zeros))
active1(tail1(cons2(X, XS))) -> mark1(XS)
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(tail1(X)) -> tail1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(zeros) -> ok1(zeros)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(tail1(X)) -> tail1(proper1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
tail1(ok1(X)) -> ok1(tail1(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.