Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/TRCSRSLASHEx7_BLR02

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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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(sel₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(from₁(X)) -> FROM₁(s₁(X))
PROPER₁(from₁(X)) -> FROM₁(proper₁(X))
ACTIVE₁(2nd₁(cons₂(X, XS))) -> HEAD₁(XS)
TAKE₂(X1, mark₁(X2)) -> TAKE₂(X1, X2)
ACTIVE₁(from₁(X)) -> FROM₁(active₁(X))
PROPER₁(head₁(X)) -> PROPER₁(X)
S₁(mark₁(X)) -> S₁(X)
PROPER₁(take₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(take₂(s₁(N), cons₂(X, XS))) -> TAKE₂(N, XS)
ACTIVE₁(2nd₁(X)) -> ACTIVE₁(X)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
2ND₁(mark₁(X)) -> 2ND₁(X)
HEAD₁(mark₁(X)) -> HEAD₁(X)
FROM₁(mark₁(X)) -> FROM₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
TAKE₂(ok₁(X1), ok₁(X2)) -> TAKE₂(X1, X2)
2ND₁(ok₁(X)) -> 2ND₁(X)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X2)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)
PROPER₁(take₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(s₁(X)) -> S₁(active₁(X))
PROPER₁(s₁(X)) -> PROPER₁(X)
ACTIVE₁(take₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(take₂(X1, X2)) -> TAKE₂(active₁(X1), X2)
PROPER₁(2nd₁(X)) -> 2ND₁(proper₁(X))
HEAD₁(ok₁(X)) -> HEAD₁(X)
PROPER₁(take₂(X1, X2)) -> TAKE₂(proper₁(X1), proper₁(X2))
TOP₁(mark₁(X)) -> PROPER₁(X)
PROPER₁(sel₂(X1, X2)) -> SEL₂(proper₁(X1), proper₁(X2))
PROPER₁(2nd₁(X)) -> PROPER₁(X)
ACTIVE₁(sel₂(X1, X2)) -> SEL₂(X1, active₁(X2))
ACTIVE₁(head₁(X)) -> ACTIVE₁(X)
ACTIVE₁(2nd₁(X)) -> 2ND₁(active₁(X))
ACTIVE₁(take₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(sel₂(s₁(N), cons₂(X, XS))) -> SEL₂(N, XS)
ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X1)
SEL₂(X1, mark₁(X2)) -> SEL₂(X1, X2)
PROPER₁(s₁(X)) -> S₁(proper₁(X))
S₁(ok₁(X)) -> S₁(X)
ACTIVE₁(head₁(X)) -> HEAD₁(active₁(X))
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
ACTIVE₁(from₁(X)) -> CONS₂(X, from₁(s₁(X)))
CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)
TAKE₂(mark₁(X1), X2) -> TAKE₂(X1, X2)
ACTIVE₁(cons₂(X1, X2)) -> CONS₂(active₁(X1), X2)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
PROPER₁(head₁(X)) -> HEAD₁(proper₁(X))
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X1)
SEL₂(ok₁(X1), ok₁(X2)) -> SEL₂(X1, X2)
ACTIVE₁(from₁(X)) -> S₁(X)
ACTIVE₁(take₂(X1, X2)) -> TAKE₂(X1, active₁(X2))
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
PROPER₁(from₁(X)) -> PROPER₁(X)
ACTIVE₁(sel₂(X1, X2)) -> SEL₂(active₁(X1), X2)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
ACTIVE₁(take₂(s₁(N), cons₂(X, XS))) -> CONS₂(X, take₂(N, XS))
FROM₁(ok₁(X)) -> FROM₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(from₁(X)) -> FROM₁(s₁(X))
PROPER₁(from₁(X)) -> FROM₁(proper₁(X))
ACTIVE₁(2nd₁(cons₂(X, XS))) -> HEAD₁(XS)
TAKE₂(X1, mark₁(X2)) -> TAKE₂(X1, X2)
ACTIVE₁(from₁(X)) -> FROM₁(active₁(X))
PROPER₁(head₁(X)) -> PROPER₁(X)
S₁(mark₁(X)) -> S₁(X)
PROPER₁(take₂(X1, X2)) -> PROPER₁(X2)
ACTIVE₁(take₂(s₁(N), cons₂(X, XS))) -> TAKE₂(N, XS)
ACTIVE₁(2nd₁(X)) -> ACTIVE₁(X)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
2ND₁(mark₁(X)) -> 2ND₁(X)
HEAD₁(mark₁(X)) -> HEAD₁(X)
FROM₁(mark₁(X)) -> FROM₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
TAKE₂(ok₁(X1), ok₁(X2)) -> TAKE₂(X1, X2)
2ND₁(ok₁(X)) -> 2ND₁(X)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X2)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)
PROPER₁(take₂(X1, X2)) -> PROPER₁(X1)
ACTIVE₁(s₁(X)) -> S₁(active₁(X))
PROPER₁(s₁(X)) -> PROPER₁(X)
ACTIVE₁(take₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(take₂(X1, X2)) -> TAKE₂(active₁(X1), X2)
PROPER₁(2nd₁(X)) -> 2ND₁(proper₁(X))
HEAD₁(ok₁(X)) -> HEAD₁(X)
PROPER₁(take₂(X1, X2)) -> TAKE₂(proper₁(X1), proper₁(X2))
TOP₁(mark₁(X)) -> PROPER₁(X)
PROPER₁(sel₂(X1, X2)) -> SEL₂(proper₁(X1), proper₁(X2))
PROPER₁(2nd₁(X)) -> PROPER₁(X)
ACTIVE₁(sel₂(X1, X2)) -> SEL₂(X1, active₁(X2))
ACTIVE₁(head₁(X)) -> ACTIVE₁(X)
ACTIVE₁(2nd₁(X)) -> 2ND₁(active₁(X))
ACTIVE₁(take₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(sel₂(s₁(N), cons₂(X, XS))) -> SEL₂(N, XS)
ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X1)
SEL₂(X1, mark₁(X2)) -> SEL₂(X1, X2)
PROPER₁(s₁(X)) -> S₁(proper₁(X))
S₁(ok₁(X)) -> S₁(X)
ACTIVE₁(head₁(X)) -> HEAD₁(active₁(X))
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
ACTIVE₁(from₁(X)) -> CONS₂(X, from₁(s₁(X)))
CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)
TAKE₂(mark₁(X1), X2) -> TAKE₂(X1, X2)
ACTIVE₁(cons₂(X1, X2)) -> CONS₂(active₁(X1), X2)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
PROPER₁(head₁(X)) -> HEAD₁(proper₁(X))
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X1)
SEL₂(ok₁(X1), ok₁(X2)) -> SEL₂(X1, X2)
ACTIVE₁(from₁(X)) -> S₁(X)
ACTIVE₁(take₂(X1, X2)) -> TAKE₂(X1, active₁(X2))
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
PROPER₁(from₁(X)) -> PROPER₁(X)
ACTIVE₁(sel₂(X1, X2)) -> SEL₂(active₁(X1), X2)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
ACTIVE₁(take₂(s₁(N), cons₂(X, XS))) -> CONS₂(X, take₂(N, XS))
FROM₁(ok₁(X)) -> FROM₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 10 SCCs with 25 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL₂(X1, mark₁(X2)) -> SEL₂(X1, X2)
SEL₂(ok₁(X1), ok₁(X2)) -> SEL₂(X1, X2)
SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SEL₂(ok₁(X1), ok₁(X2)) -> SEL₂(X1, X2)

Used argument filtering: SEL₂(x1, x2) = x2
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL₂(X1, mark₁(X2)) -> SEL₂(X1, X2)
SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SEL₂(X1, mark₁(X2)) -> SEL₂(X1, X2)

Used argument filtering: SEL₂(x1, x2) = x2
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)

Used argument filtering: SEL₂(x1, x2) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ 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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

2ND₁(mark₁(X)) -> 2ND₁(X)
2ND₁(ok₁(X)) -> 2ND₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

2ND₁(ok₁(X)) -> 2ND₁(X)

Used argument filtering: 2ND₁(x1) = x1
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

2ND₁(mark₁(X)) -> 2ND₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

2ND₁(mark₁(X)) -> 2ND₁(X)

Used argument filtering: 2ND₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ 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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

HEAD₁(ok₁(X)) -> HEAD₁(X)
HEAD₁(mark₁(X)) -> HEAD₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

HEAD₁(mark₁(X)) -> HEAD₁(X)

Used argument filtering: HEAD₁(x1) = x1
ok₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

HEAD₁(ok₁(X)) -> HEAD₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

HEAD₁(ok₁(X)) -> HEAD₁(X)

Used argument filtering: HEAD₁(x1) = x1
ok₁(x1) = ok₁(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

          ↳ 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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ 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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

          ↳ 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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

          ↳ 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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ 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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

          ↳ 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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

          ↳ 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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ 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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

          ↳ 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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

          ↳ 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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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(s₁(X)) -> PROPER₁(X)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(from₁(X)) -> PROPER₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(head₁(X)) -> PROPER₁(X)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(2nd₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(sel₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X2)

Used argument filtering: PROPER₁(x1) = x1
take₂(x1, x2) = take₂(x1, x2)
s₁(x1) = x1
sel₂(x1, x2) = sel₂(x1, x2)
from₁(x1) = x1
cons₂(x1, x2) = cons₂(x1, x2)
head₁(x1) = x1
2nd₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ 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(from₁(X)) -> PROPER₁(X)
PROPER₁(head₁(X)) -> PROPER₁(X)
PROPER₁(2nd₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(2nd₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = x1
from₁(x1) = x1
head₁(x1) = x1
2nd₁(x1) = 2nd₁(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

          ↳ QDP

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

PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(from₁(X)) -> PROPER₁(X)
PROPER₁(head₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(head₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
s₁(x1) = x1
from₁(x1) = x1
head₁(x1) = head₁(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

                        ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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
s₁(x1) = x1
from₁(x1) = from₁(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

                        ↳ 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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

          ↳ 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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(2nd₁(X)) -> ACTIVE₁(X)
ACTIVE₁(head₁(X)) -> ACTIVE₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(take₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(take₂(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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(take₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(take₂(X1, X2)) -> ACTIVE₁(X1)

Used argument filtering: ACTIVE₁(x1) = x1
sel₂(x1, x2) = sel₂(x1, x2)
2nd₁(x1) = x1
head₁(x1) = x1
cons₂(x1, x2) = x1
take₂(x1, x2) = take₂(x1, x2)
from₁(x1) = x1
s₁(x1) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVE₁(head₁(X)) -> ACTIVE₁(X)
ACTIVE₁(2nd₁(X)) -> ACTIVE₁(X)
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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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
head₁(x1) = x1
2nd₁(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

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVE₁(2nd₁(X)) -> ACTIVE₁(X)
ACTIVE₁(head₁(X)) -> ACTIVE₁(X)
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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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
2nd₁(x1) = x1
head₁(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

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVE₁(head₁(X)) -> ACTIVE₁(X)
ACTIVE₁(2nd₁(X)) -> ACTIVE₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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
head₁(x1) = x1
2nd₁(x1) = x1
cons₂(x1, x2) = cons₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVE₁(2nd₁(X)) -> ACTIVE₁(X)
ACTIVE₁(head₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(head₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
2nd₁(x1) = x1
head₁(x1) = head₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ QDP

                        ↳ QDPAfsSolverProof

                          ↳ QDP

                            ↳ QDPAfsSolverProof

                              ↳ QDP

                                ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVE₁(2nd₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(2nd₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
2nd₁(x1) = 2nd₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ 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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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(head₁(cons₂(X, XS))) -> mark₁(X)
active₁(2nd₁(cons₂(X, XS))) -> mark₁(head₁(XS))
active₁(take₂(0, XS)) -> mark₁(nil)
active₁(take₂(s₁(N), cons₂(X, XS))) -> mark₁(cons₂(X, take₂(N, XS)))
active₁(sel₂(0, cons₂(X, XS))) -> mark₁(X)
active₁(sel₂(s₁(N), cons₂(X, XS))) -> mark₁(sel₂(N, XS))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(head₁(X)) -> head₁(active₁(X))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
head₁(mark₁(X)) -> mark₁(head₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(head₁(X)) -> head₁(proper₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(nil) -> ok₁(nil)
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
head₁(ok₁(X)) -> ok₁(head₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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