Termination w.r.t. Q proof of TRS/TRCSREx1_GL02a

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 9 SCCs with 20 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(CONS₂(x₁, x₂)) = x₁·x₂
POL(ok₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S₁(ok₁(X)) -> S₁(X)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

S₁(ok₁(X)) -> S₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(S₁(x₁)) = x₁²
POL(ok₁(x₁)) = 1 + x₁²

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

EQ₂(ok₁(X1), ok₁(X2)) -> EQ₂(X1, X2)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

EQ₂(ok₁(X1), ok₁(X2)) -> EQ₂(X1, X2)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(EQ₂(x₁, x₂)) = x₁·x₂
POL(ok₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

LENGTH₁(ok₁(X)) -> LENGTH₁(X)

The remaining pairs can at least be oriented weakly.

LENGTH₁(mark₁(X)) -> LENGTH₁(X)

Used ordering: Polynomial interpretation [21]:

POL(LENGTH₁(x₁)) = x₁²
POL(mark₁(x₁)) = x₁
POL(ok₁(x₁)) = 1 + x₁²

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

LENGTH₁(mark₁(X)) -> LENGTH₁(X)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

LENGTH₁(mark₁(X)) -> LENGTH₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(LENGTH₁(x₁)) = x₁²
POL(mark₁(x₁)) = 1 + x₁²

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(TAKE₂(x₁, x₂)) = x₁·x₂
POL(mark₁(x₁)) = x₁
POL(ok₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

TAKE₂(mark₁(X1), X2) -> TAKE₂(X1, X2)

The remaining pairs can at least be oriented weakly.

TAKE₂(X1, mark₁(X2)) -> TAKE₂(X1, X2)

Used ordering: Polynomial interpretation [21]:

POL(TAKE₂(x₁, x₂)) = x₁
POL(mark₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

TAKE₂(X1, mark₁(X2)) -> TAKE₂(X1, X2)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

TAKE₂(X1, mark₁(X2)) -> TAKE₂(X1, X2)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(TAKE₂(x₁, x₂)) = x₂
POL(mark₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

INF₁(ok₁(X)) -> INF₁(X)

The remaining pairs can at least be oriented weakly.

INF₁(mark₁(X)) -> INF₁(X)

Used ordering: Polynomial interpretation [21]:

POL(INF₁(x₁)) = x₁²
POL(mark₁(x₁)) = x₁
POL(ok₁(x₁)) = 1 + x₁²

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

INF₁(mark₁(X)) -> INF₁(X)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

INF₁(mark₁(X)) -> INF₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(INF₁(x₁)) = x₁²
POL(mark₁(x₁)) = 1 + x₁²

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

PROPER₁(eq₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(eq₂(X1, X2)) -> PROPER₁(X1)

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(PROPER₁(x₁)) = x₁²
POL(cons₂(x₁, x₂)) = x₁ + x₂
POL(eq₂(x₁, x₂)) = 1 + x₁ + x₂
POL(inf₁(x₁)) = x₁
POL(length₁(x₁)) = x₁
POL(s₁(x₁)) = x₁
POL(take₂(x₁, x₂)) = x₁ + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

PROPER₁(inf₁(X)) -> PROPER₁(X)

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(PROPER₁(x₁)) = x₁²
POL(cons₂(x₁, x₂)) = x₁ + x₂
POL(inf₁(x₁)) = 1 + x₁
POL(length₁(x₁)) = x₁
POL(s₁(x₁)) = x₁
POL(take₂(x₁, x₂)) = x₁ + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

          ↳ 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(length₁(X)) -> PROPER₁(X)
PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

PROPER₁(length₁(X)) -> PROPER₁(X)

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(PROPER₁(x₁)) = x₁²
POL(cons₂(x₁, x₂)) = x₁ + x₂
POL(length₁(x₁)) = 1 + x₁
POL(s₁(x₁)) = x₁
POL(take₂(x₁, x₂)) = x₁ + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

PROPER₁(s₁(X)) -> PROPER₁(X)

The remaining pairs can at least be oriented weakly.

PROPER₁(take₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(take₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)

Used ordering: Polynomial interpretation [21]:

POL(PROPER₁(x₁)) = x₁²
POL(cons₂(x₁, x₂)) = x₁ + x₂
POL(s₁(x₁)) = 1 + x₁
POL(take₂(x₁, x₂)) = x₁ + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

          ↳ 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(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

PROPER₁(take₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(take₂(X1, X2)) -> PROPER₁(X1)

Used ordering: Polynomial interpretation [21]:

POL(PROPER₁(x₁)) = x₁²
POL(cons₂(x₁, x₂)) = 1 + x₁ + x₂
POL(take₂(x₁, x₂)) = x₁ + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

PROPER₁(take₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(take₂(X1, X2)) -> PROPER₁(X2)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

PROPER₁(take₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(take₂(X1, X2)) -> PROPER₁(X2)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PROPER₁(x₁)) = x₁²
POL(take₂(x₁, x₂)) = 1 + x₁ + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

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

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(ACTIVE₁(x₁)) = x₁²
POL(inf₁(x₁)) = x₁
POL(length₁(x₁)) = x₁
POL(take₂(x₁, x₂)) = 1 + x₁ + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

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

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

ACTIVE₁(inf₁(X)) -> ACTIVE₁(X)

The remaining pairs can at least be oriented weakly.

ACTIVE₁(length₁(X)) -> ACTIVE₁(X)

Used ordering: Polynomial interpretation [21]:

POL(ACTIVE₁(x₁)) = x₁²
POL(inf₁(x₁)) = 1 + x₁²
POL(length₁(x₁)) = x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

          ↳ QDP

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

ACTIVE₁(length₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

ACTIVE₁(length₁(X)) -> ACTIVE₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVE₁(x₁)) = x₁²
POL(length₁(x₁)) = 1 + x₁²

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

          ↳ QDP

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

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

TOP₁(ok₁(X)) -> TOP₁(active₁(X))
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))

The TRS R consists of the following rules:

active₁(eq₂(0, 0)) -> mark₁(true)
active₁(eq₂(s₁(X), s₁(Y))) -> mark₁(eq₂(X, Y))
active₁(eq₂(X, Y)) -> mark₁(false)
active₁(inf₁(X)) -> mark₁(cons₂(X, inf₁(s₁(X))))
active₁(take₂(0, X)) -> mark₁(nil)
active₁(take₂(s₁(X), cons₂(Y, L))) -> mark₁(cons₂(Y, take₂(X, L)))
active₁(length₁(nil)) -> mark₁(0)
active₁(length₁(cons₂(X, L))) -> mark₁(s₁(length₁(L)))
active₁(inf₁(X)) -> inf₁(active₁(X))
active₁(take₂(X1, X2)) -> take₂(active₁(X1), X2)
active₁(take₂(X1, X2)) -> take₂(X1, active₁(X2))
active₁(length₁(X)) -> length₁(active₁(X))
inf₁(mark₁(X)) -> mark₁(inf₁(X))
take₂(mark₁(X1), X2) -> mark₁(take₂(X1, X2))
take₂(X1, mark₁(X2)) -> mark₁(take₂(X1, X2))
length₁(mark₁(X)) -> mark₁(length₁(X))
proper₁(eq₂(X1, X2)) -> eq₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(true) -> ok₁(true)
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(false) -> ok₁(false)
proper₁(inf₁(X)) -> inf₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(take₂(X1, X2)) -> take₂(proper₁(X1), proper₁(X2))
proper₁(nil) -> ok₁(nil)
proper₁(length₁(X)) -> length₁(proper₁(X))
eq₂(ok₁(X1), ok₁(X2)) -> ok₁(eq₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
inf₁(ok₁(X)) -> ok₁(inf₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
take₂(ok₁(X1), ok₁(X2)) -> ok₁(take₂(X1, X2))
length₁(ok₁(X)) -> ok₁(length₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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