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

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)) = 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(mark1(X)) -> LENGTH1(X)
LENGTH1(ok1(X)) -> LENGTH1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

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

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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(X1, mark1(X2)) -> TAKE2(X1, X2)
TAKE2(ok1(X1), ok1(X2)) -> TAKE2(X1, X2)
The remaining pairs can at least be oriented weakly.

TAKE2(mark1(X1), X2) -> TAKE2(X1, X2)
Used ordering: Polynomial interpretation [21]:

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

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.
none
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
                    ↳ 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(mark1(X)) -> INF1(X)
INF1(ok1(X)) -> INF1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

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

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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(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 remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PROPER1(x1)) = x1   
POL(cons2(x1, x2)) = 1 + x1 + x2   
POL(eq2(x1, x2)) = 1 + x1 + x2   
POL(inf1(x1)) = 1 + x1   
POL(length1(x1)) = 1 + x1   
POL(s1(x1)) = 1 + 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
            ↳ 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(length1(X)) -> ACTIVE1(X)
ACTIVE1(take2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(take2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(inf1(X)) -> ACTIVE1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVE1(x1)) = x1   
POL(inf1(x1)) = 1 + x1   
POL(length1(x1)) = 1 + 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
                ↳ 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.