Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/TRCSRSLASHEx14_AEGL02

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.

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 contains 8 SCCs with 16 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

LENGTH1₁(ok₁(X)) -> LENGTH1₁(X)

Used argument filtering: LENGTH1₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

LENGTH₁(ok₁(X)) -> LENGTH₁(X)

Used argument filtering: LENGTH₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

S₁(mark₁(X)) -> S₁(X)

Used argument filtering: S₁(x1) = x1
ok₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

S₁(ok₁(X)) -> S₁(X)

Used argument filtering: S₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
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

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
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

          ↳ QDP

          ↳ 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(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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

FROM₁(ok₁(X)) -> FROM₁(X)

Used argument filtering: FROM₁(x1) = x1
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

FROM₁(mark₁(X)) -> FROM₁(X)

Used argument filtering: FROM₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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(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

            ↳ QDPAfsSolverProof

          ↳ 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.
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
length₁(x1) = x1
s₁(x1) = x1
from₁(x1) = x1
cons₂(x1, x2) = cons₂(x1, x2)
length1₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(length1₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = x1
length₁(x1) = x1
from₁(x1) = x1
length1₁(x1) = length1₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ 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)

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(from₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
length₁(x1) = x1
s₁(x1) = x1
from₁(x1) = from₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(length₁(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(length₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = x1
length₁(x1) = length₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(s₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

                              ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
cons₂(x1, x2) = x1
from₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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)
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(from₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
cons₂(x1, x2) = x1
from₁(x1) = from₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ 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(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.
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

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ 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.