Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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:

PROPER1(from1(X)) -> FROM1(proper1(X))
ACTIVE1(from1(X)) -> FROM1(s1(X))
FIRST2(X1, mark1(X2)) -> FIRST2(X1, X2)
ACTIVE1(first2(s1(X), cons2(Y, Z))) -> FIRST2(X, Z)
TOP1(mark1(X)) -> PROPER1(X)
ACTIVE1(from1(X)) -> FROM1(active1(X))
PROPER1(first2(X1, X2)) -> PROPER1(X1)
PROPER1(first2(X1, X2)) -> PROPER1(X2)
FIRST2(mark1(X1), X2) -> FIRST2(X1, X2)
S1(mark1(X)) -> S1(X)
ACTIVE1(first2(X1, X2)) -> FIRST2(X1, active1(X2))
ACTIVE1(first2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(first2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(first2(X1, X2)) -> FIRST2(active1(X1), X2)
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(s1(X)) -> S1(proper1(X))
S1(ok1(X)) -> S1(X)
ACTIVE1(s1(X)) -> ACTIVE1(X)
ACTIVE1(first2(s1(X), cons2(Y, Z))) -> CONS2(Y, first2(X, Z))
ACTIVE1(from1(X)) -> CONS2(X, from1(s1(X)))
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
ACTIVE1(cons2(X1, X2)) -> CONS2(active1(X1), X2)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
FROM1(mark1(X)) -> FROM1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(first2(X1, X2)) -> FIRST2(proper1(X1), proper1(X2))
TOP1(ok1(X)) -> ACTIVE1(X)
FIRST2(ok1(X1), ok1(X2)) -> FIRST2(X1, X2)
ACTIVE1(from1(X)) -> S1(X)
ACTIVE1(s1(X)) -> S1(active1(X))
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
PROPER1(s1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
PROPER1(from1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
ACTIVE1(from1(X)) -> ACTIVE1(X)
FROM1(ok1(X)) -> FROM1(X)

The TRS R consists of the following rules:

active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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:

PROPER1(from1(X)) -> FROM1(proper1(X))
ACTIVE1(from1(X)) -> FROM1(s1(X))
FIRST2(X1, mark1(X2)) -> FIRST2(X1, X2)
ACTIVE1(first2(s1(X), cons2(Y, Z))) -> FIRST2(X, Z)
TOP1(mark1(X)) -> PROPER1(X)
ACTIVE1(from1(X)) -> FROM1(active1(X))
PROPER1(first2(X1, X2)) -> PROPER1(X1)
PROPER1(first2(X1, X2)) -> PROPER1(X2)
FIRST2(mark1(X1), X2) -> FIRST2(X1, X2)
S1(mark1(X)) -> S1(X)
ACTIVE1(first2(X1, X2)) -> FIRST2(X1, active1(X2))
ACTIVE1(first2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(first2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(first2(X1, X2)) -> FIRST2(active1(X1), X2)
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(s1(X)) -> S1(proper1(X))
S1(ok1(X)) -> S1(X)
ACTIVE1(s1(X)) -> ACTIVE1(X)
ACTIVE1(first2(s1(X), cons2(Y, Z))) -> CONS2(Y, first2(X, Z))
ACTIVE1(from1(X)) -> CONS2(X, from1(s1(X)))
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
ACTIVE1(cons2(X1, X2)) -> CONS2(active1(X1), X2)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
FROM1(mark1(X)) -> FROM1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(first2(X1, X2)) -> FIRST2(proper1(X1), proper1(X2))
TOP1(ok1(X)) -> ACTIVE1(X)
FIRST2(ok1(X1), ok1(X2)) -> FIRST2(X1, X2)
ACTIVE1(from1(X)) -> S1(X)
ACTIVE1(s1(X)) -> S1(active1(X))
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
PROPER1(s1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
PROPER1(from1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
ACTIVE1(from1(X)) -> ACTIVE1(X)
FROM1(ok1(X)) -> FROM1(X)

The TRS R consists of the following rules:

active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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 7 SCCs with 16 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

FROM1(mark1(X)) -> FROM1(X)
FROM1(ok1(X)) -> FROM1(X)

The TRS R consists of the following rules:

active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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.


FROM1(mark1(X)) -> FROM1(X)
FROM1(ok1(X)) -> FROM1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(FROM1(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
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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

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

CONS2(mark1(X1), X2) -> CONS2(X1, X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)

The TRS R consists of the following rules:

active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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.

CONS2(mark1(X1), X2) -> CONS2(X1, X2)
Used ordering: Polynomial interpretation [21]:

POL(CONS2(x1, x2)) = x2   
POL(mark1(x1)) = 0   
POL(ok1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

CONS2(mark1(X1), X2) -> CONS2(X1, X2)

The TRS R consists of the following rules:

active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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(mark1(X1), X2) -> CONS2(X1, X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(CONS2(x1, x2)) = x1   
POL(mark1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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

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

S1(ok1(X)) -> S1(X)
S1(mark1(X)) -> S1(X)

The TRS R consists of the following rules:

active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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)
S1(mark1(X)) -> S1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(S1(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
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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

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

FIRST2(X1, mark1(X2)) -> FIRST2(X1, X2)
FIRST2(ok1(X1), ok1(X2)) -> FIRST2(X1, X2)
FIRST2(mark1(X1), X2) -> FIRST2(X1, X2)

The TRS R consists of the following rules:

active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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.


FIRST2(ok1(X1), ok1(X2)) -> FIRST2(X1, X2)
FIRST2(mark1(X1), X2) -> FIRST2(X1, X2)
The remaining pairs can at least be oriented weakly.

FIRST2(X1, mark1(X2)) -> FIRST2(X1, X2)
Used ordering: Polynomial interpretation [21]:

POL(FIRST2(x1, x2)) = 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
                ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

FIRST2(X1, mark1(X2)) -> FIRST2(X1, X2)

The TRS R consists of the following rules:

active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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.


FIRST2(X1, mark1(X2)) -> FIRST2(X1, X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(FIRST2(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
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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

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

PROPER1(s1(X)) -> PROPER1(X)
PROPER1(from1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(first2(X1, X2)) -> PROPER1(X1)
PROPER1(first2(X1, X2)) -> PROPER1(X2)

The TRS R consists of the following rules:

active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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)
PROPER1(from1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(first2(X1, X2)) -> PROPER1(X1)
PROPER1(first2(X1, X2)) -> PROPER1(X2)
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(first2(x1, x2)) = 1 + x1 + x2   
POL(from1(x1)) = 1 + x1   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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

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

ACTIVE1(first2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(first2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(from1(X)) -> ACTIVE1(X)
ACTIVE1(s1(X)) -> ACTIVE1(X)

The TRS R consists of the following rules:

active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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(first2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(first2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(from1(X)) -> ACTIVE1(X)
ACTIVE1(s1(X)) -> ACTIVE1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVE1(x1)) = x1   
POL(cons2(x1, x2)) = 1 + x1   
POL(first2(x1, x2)) = 1 + x1 + x2   
POL(from1(x1)) = 1 + x1   
POL(s1(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

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

active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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

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(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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.


TOP1(mark1(X)) -> TOP1(proper1(X))
The remaining pairs can at least be oriented weakly.

TOP1(ok1(X)) -> TOP1(active1(X))
Used ordering: Polynomial interpretation [21]:

POL(0) = 1   
POL(TOP1(x1)) = x1   
POL(active1(x1)) = x1   
POL(cons2(x1, x2)) = x1   
POL(first2(x1, x2)) = x1 + x2   
POL(from1(x1)) = 1 + x1   
POL(mark1(x1)) = 1 + x1   
POL(nil) = 0   
POL(ok1(x1)) = x1   
POL(proper1(x1)) = x1   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented:

from1(ok1(X)) -> ok1(from1(X))
from1(mark1(X)) -> mark1(from1(X))
active1(first2(0, X)) -> mark1(nil)
active1(from1(X)) -> from1(active1(X))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
proper1(from1(X)) -> from1(proper1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
active1(first2(X1, X2)) -> first2(X1, active1(X2))
proper1(nil) -> ok1(nil)
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
active1(s1(X)) -> s1(active1(X))
proper1(0) -> ok1(0)
s1(ok1(X)) -> ok1(s1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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.


TOP1(ok1(X)) -> TOP1(active1(X))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(TOP1(x1)) = x1   
POL(active1(x1)) = x1   
POL(cons2(x1, x2)) = x1 + x2   
POL(first2(x1, x2)) = x1   
POL(from1(x1)) = x1   
POL(mark1(x1)) = 0   
POL(nil) = 0   
POL(ok1(x1)) = 1 + x1   
POL(s1(x1)) = x1   

The following usable rules [14] were oriented:

from1(ok1(X)) -> ok1(from1(X))
from1(mark1(X)) -> mark1(from1(X))
active1(first2(0, X)) -> mark1(nil)
active1(from1(X)) -> from1(active1(X))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
active1(first2(X1, X2)) -> first2(X1, active1(X2))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
active1(s1(X)) -> s1(active1(X))
s1(ok1(X)) -> ok1(s1(X))
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof

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

active1(first2(0, X)) -> mark1(nil)
active1(first2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, first2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(first2(X1, X2)) -> first2(active1(X1), X2)
active1(first2(X1, X2)) -> first2(X1, active1(X2))
active1(s1(X)) -> s1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(from1(X)) -> from1(active1(X))
first2(mark1(X1), X2) -> mark1(first2(X1, X2))
first2(X1, mark1(X2)) -> mark1(first2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
proper1(first2(X1, X2)) -> first2(proper1(X1), proper1(X2))
proper1(0) -> ok1(0)
proper1(nil) -> ok1(nil)
proper1(s1(X)) -> s1(proper1(X))
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
first2(ok1(X1), ok1(X2)) -> ok1(first2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
from1(ok1(X)) -> ok1(from1(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.