Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/TRCSRSLASHEx1_GL02a

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(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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:

TOP₁(mark₁(X)) -> PROPER₁(X)
TAKE₂(X1, mark₁(X2)) -> TAKE₂(X1, X2)
LENGTH₁(ok₁(X)) -> LENGTH₁(X)
PROPER₁(length₁(X)) -> PROPER₁(X)
PROPER₁(eq₂(X1, X2)) -> EQ₂(proper₁(X1), proper₁(X2))
INF₁(ok₁(X)) -> INF₁(X)
ACTIVE₁(length₁(cons₂(X, L))) -> LENGTH₁(L)
PROPER₁(take₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(take₂(s₁(X), cons₂(Y, L))) -> TAKE₂(X, L)
PROPER₁(eq₂(X1, X2)) -> PROPER₁(X2)
LENGTH₁(mark₁(X)) -> LENGTH₁(X)
INF₁(mark₁(X)) -> INF₁(X)
ACTIVE₁(take₂(X1, X2)) -> ACTIVE₁(X2)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
PROPER₁(s₁(X)) -> S₁(proper₁(X))
S₁(ok₁(X)) -> S₁(X)
TAKE₂(mark₁(X1), X2) -> TAKE₂(X1, X2)
ACTIVE₁(inf₁(X)) -> CONS₂(X, inf₁(s₁(X)))
ACTIVE₁(length₁(cons₂(X, L))) -> S₁(length₁(L))
ACTIVE₁(length₁(X)) -> ACTIVE₁(X)
PROPER₁(inf₁(X)) -> PROPER₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(inf₁(X)) -> INF₁(active₁(X))
ACTIVE₁(length₁(X)) -> LENGTH₁(active₁(X))
TAKE₂(ok₁(X1), ok₁(X2)) -> TAKE₂(X1, X2)
EQ₂(ok₁(X1), ok₁(X2)) -> EQ₂(X1, X2)
PROPER₁(length₁(X)) -> LENGTH₁(proper₁(X))
TOP₁(ok₁(X)) -> ACTIVE₁(X)
PROPER₁(eq₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(take₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(inf₁(X)) -> INF₁(proper₁(X))
ACTIVE₁(take₂(X1, X2)) -> TAKE₂(X1, active₁(X2))
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
ACTIVE₁(eq₂(s₁(X), s₁(Y))) -> EQ₂(X, Y)
PROPER₁(s₁(X)) -> PROPER₁(X)
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
ACTIVE₁(inf₁(X)) -> INF₁(s₁(X))
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(take₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(take₂(X1, X2)) -> TAKE₂(active₁(X1), X2)
ACTIVE₁(inf₁(X)) -> ACTIVE₁(X)
ACTIVE₁(inf₁(X)) -> S₁(X)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
ACTIVE₁(take₂(s₁(X), cons₂(Y, L))) -> CONS₂(Y, take₂(X, L))
PROPER₁(take₂(X1, X2)) -> TAKE₂(proper₁(X1), proper₁(X2))

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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:

TOP₁(mark₁(X)) -> PROPER₁(X)
TAKE₂(X1, mark₁(X2)) -> TAKE₂(X1, X2)
LENGTH₁(ok₁(X)) -> LENGTH₁(X)
PROPER₁(length₁(X)) -> PROPER₁(X)
PROPER₁(eq₂(X1, X2)) -> EQ₂(proper₁(X1), proper₁(X2))
INF₁(ok₁(X)) -> INF₁(X)
ACTIVE₁(length₁(cons₂(X, L))) -> LENGTH₁(L)
PROPER₁(take₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(take₂(s₁(X), cons₂(Y, L))) -> TAKE₂(X, L)
PROPER₁(eq₂(X1, X2)) -> PROPER₁(X2)
LENGTH₁(mark₁(X)) -> LENGTH₁(X)
INF₁(mark₁(X)) -> INF₁(X)
ACTIVE₁(take₂(X1, X2)) -> ACTIVE₁(X2)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
PROPER₁(s₁(X)) -> S₁(proper₁(X))
S₁(ok₁(X)) -> S₁(X)
TAKE₂(mark₁(X1), X2) -> TAKE₂(X1, X2)
ACTIVE₁(inf₁(X)) -> CONS₂(X, inf₁(s₁(X)))
ACTIVE₁(length₁(cons₂(X, L))) -> S₁(length₁(L))
ACTIVE₁(length₁(X)) -> ACTIVE₁(X)
PROPER₁(inf₁(X)) -> PROPER₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(inf₁(X)) -> INF₁(active₁(X))
ACTIVE₁(length₁(X)) -> LENGTH₁(active₁(X))
TAKE₂(ok₁(X1), ok₁(X2)) -> TAKE₂(X1, X2)
EQ₂(ok₁(X1), ok₁(X2)) -> EQ₂(X1, X2)
PROPER₁(length₁(X)) -> LENGTH₁(proper₁(X))
TOP₁(ok₁(X)) -> ACTIVE₁(X)
PROPER₁(eq₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(take₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(inf₁(X)) -> INF₁(proper₁(X))
ACTIVE₁(take₂(X1, X2)) -> TAKE₂(X1, active₁(X2))
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
ACTIVE₁(eq₂(s₁(X), s₁(Y))) -> EQ₂(X, Y)
PROPER₁(s₁(X)) -> PROPER₁(X)
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
ACTIVE₁(inf₁(X)) -> INF₁(s₁(X))
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(take₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(take₂(X1, X2)) -> TAKE₂(active₁(X1), X2)
ACTIVE₁(inf₁(X)) -> ACTIVE₁(X)
ACTIVE₁(inf₁(X)) -> S₁(X)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
ACTIVE₁(take₂(s₁(X), cons₂(Y, L))) -> CONS₂(Y, take₂(X, L))
PROPER₁(take₂(X1, X2)) -> TAKE₂(proper₁(X1), proper₁(X2))

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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 9 SCCs with 20 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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

          ↳ 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(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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

            ↳ 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(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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

          ↳ QDP

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

EQ₂(ok₁(X1), ok₁(X2)) -> EQ₂(X1, X2)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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.

EQ₂(ok₁(X1), ok₁(X2)) -> EQ₂(X1, X2)

Used argument filtering: EQ₂(x1, x2) = x2
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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

          ↳ QDP

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

LENGTH₁(mark₁(X)) -> LENGTH₁(X)
LENGTH₁(ok₁(X)) -> LENGTH₁(X)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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
mark₁(x1) = x1
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

          ↳ QDP

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

LENGTH₁(mark₁(X)) -> LENGTH₁(X)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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(mark₁(X)) -> LENGTH₁(X)

Used argument filtering: LENGTH₁(x1) = 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

          ↳ QDP

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

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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

          ↳ QDP

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

TAKE₂(mark₁(X1), X2) -> TAKE₂(X1, X2)
TAKE₂(X1, mark₁(X2)) -> TAKE₂(X1, X2)
TAKE₂(ok₁(X1), ok₁(X2)) -> TAKE₂(X1, X2)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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.

TAKE₂(ok₁(X1), ok₁(X2)) -> TAKE₂(X1, X2)

Used argument filtering: TAKE₂(x1, x2) = x2
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

          ↳ QDP

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

TAKE₂(mark₁(X1), X2) -> TAKE₂(X1, X2)
TAKE₂(X1, mark₁(X2)) -> TAKE₂(X1, X2)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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.

TAKE₂(X1, mark₁(X2)) -> TAKE₂(X1, X2)

Used argument filtering: TAKE₂(x1, x2) = x2
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

TAKE₂(mark₁(X1), X2) -> TAKE₂(X1, X2)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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.

TAKE₂(mark₁(X1), X2) -> TAKE₂(X1, X2)

Used argument filtering: TAKE₂(x1, x2) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ 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(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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

          ↳ QDP

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

INF₁(mark₁(X)) -> INF₁(X)
INF₁(ok₁(X)) -> INF₁(X)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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.

INF₁(ok₁(X)) -> INF₁(X)

Used argument filtering: INF₁(x1) = x1
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

INF₁(mark₁(X)) -> INF₁(X)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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.

INF₁(mark₁(X)) -> INF₁(X)

Used argument filtering: INF₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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

          ↳ QDP

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

PROPER₁(take₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(take₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(eq₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(inf₁(X)) -> PROPER₁(X)
PROPER₁(length₁(X)) -> PROPER₁(X)
PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(eq₂(X1, X2)) -> PROPER₁(X1)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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(take₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(take₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(eq₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(eq₂(X1, X2)) -> PROPER₁(X1)

Used argument filtering: PROPER₁(x1) = x1
take₂(x1, x2) = take₂(x1, x2)
eq₂(x1, x2) = eq₂(x1, x2)
inf₁(x1) = x1
length₁(x1) = x1
s₁(x1) = x1
cons₂(x1, x2) = cons₂(x1, x2)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

PROPER₁(inf₁(X)) -> PROPER₁(X)
PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(length₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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
inf₁(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

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

PROPER₁(inf₁(X)) -> PROPER₁(X)
PROPER₁(s₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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
inf₁(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

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

PROPER₁(inf₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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(inf₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
inf₁(x1) = inf₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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

            ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVE₁(length₁(X)) -> ACTIVE₁(X)
ACTIVE₁(take₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(take₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(inf₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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(take₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(take₂(X1, X2)) -> ACTIVE₁(X1)

Used argument filtering: ACTIVE₁(x1) = x1
length₁(x1) = x1
take₂(x1, x2) = take₂(x1, x2)
inf₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVE₁(length₁(X)) -> ACTIVE₁(X)
ACTIVE₁(inf₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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(inf₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
length₁(x1) = x1
inf₁(x1) = inf₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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(length₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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(length₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
length₁(x1) = length₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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

          ↳ 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(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(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.