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(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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:

ACTIVE1(from1(X)) -> FROM1(s1(X))
ACTIVE1(fst2(X1, X2)) -> ACTIVE1(X1)
PROPER1(from1(X)) -> FROM1(proper1(X))
ACTIVE1(fst2(X1, X2)) -> FST2(active1(X1), X2)
PROPER1(len1(X)) -> LEN1(proper1(X))
ACTIVE1(fst2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(from1(X)) -> FROM1(active1(X))
ACTIVE1(add2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(fst2(s1(X), cons2(Y, Z))) -> FST2(X, Z)
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(len1(X)) -> PROPER1(X)
ACTIVE1(fst2(s1(X), cons2(Y, Z))) -> CONS2(Y, fst2(X, Z))
FROM1(mark1(X)) -> FROM1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
ACTIVE1(add2(X1, X2)) -> ADD2(active1(X1), X2)
ACTIVE1(len1(X)) -> LEN1(active1(X))
TOP1(ok1(X)) -> ACTIVE1(X)
LEN1(ok1(X)) -> LEN1(X)
ACTIVE1(add2(X1, X2)) -> ADD2(X1, active1(X2))
PROPER1(s1(X)) -> PROPER1(X)
ACTIVE1(fst2(X1, X2)) -> FST2(X1, active1(X2))
ACTIVE1(add2(s1(X), Y)) -> ADD2(X, Y)
FST2(X1, mark1(X2)) -> FST2(X1, X2)
TOP1(mark1(X)) -> PROPER1(X)
ADD2(mark1(X1), X2) -> ADD2(X1, X2)
PROPER1(fst2(X1, X2)) -> PROPER1(X2)
ACTIVE1(add2(X1, X2)) -> ACTIVE1(X1)
PROPER1(fst2(X1, X2)) -> FST2(proper1(X1), proper1(X2))
PROPER1(add2(X1, X2)) -> PROPER1(X1)
PROPER1(add2(X1, X2)) -> ADD2(proper1(X1), proper1(X2))
PROPER1(s1(X)) -> S1(proper1(X))
S1(ok1(X)) -> S1(X)
PROPER1(fst2(X1, X2)) -> PROPER1(X1)
ACTIVE1(len1(cons2(X, Z))) -> S1(len1(Z))
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
ACTIVE1(from1(X)) -> CONS2(X, from1(s1(X)))
ACTIVE1(cons2(X1, X2)) -> CONS2(active1(X1), X2)
LEN1(mark1(X)) -> LEN1(X)
ADD2(ok1(X1), ok1(X2)) -> ADD2(X1, X2)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(len1(cons2(X, Z))) -> LEN1(Z)
FST2(ok1(X1), ok1(X2)) -> FST2(X1, X2)
PROPER1(add2(X1, X2)) -> PROPER1(X2)
ACTIVE1(len1(X)) -> ACTIVE1(X)
ACTIVE1(from1(X)) -> S1(X)
ADD2(X1, mark1(X2)) -> ADD2(X1, X2)
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
TOP1(ok1(X)) -> TOP1(active1(X))
PROPER1(from1(X)) -> PROPER1(X)
FST2(mark1(X1), X2) -> FST2(X1, X2)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
ACTIVE1(from1(X)) -> ACTIVE1(X)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
FROM1(ok1(X)) -> FROM1(X)
ACTIVE1(add2(s1(X), Y)) -> S1(add2(X, Y))

The TRS R consists of the following rules:

active1(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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:

ACTIVE1(from1(X)) -> FROM1(s1(X))
ACTIVE1(fst2(X1, X2)) -> ACTIVE1(X1)
PROPER1(from1(X)) -> FROM1(proper1(X))
ACTIVE1(fst2(X1, X2)) -> FST2(active1(X1), X2)
PROPER1(len1(X)) -> LEN1(proper1(X))
ACTIVE1(fst2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(from1(X)) -> FROM1(active1(X))
ACTIVE1(add2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(fst2(s1(X), cons2(Y, Z))) -> FST2(X, Z)
TOP1(mark1(X)) -> TOP1(proper1(X))
PROPER1(len1(X)) -> PROPER1(X)
ACTIVE1(fst2(s1(X), cons2(Y, Z))) -> CONS2(Y, fst2(X, Z))
FROM1(mark1(X)) -> FROM1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
ACTIVE1(add2(X1, X2)) -> ADD2(active1(X1), X2)
ACTIVE1(len1(X)) -> LEN1(active1(X))
TOP1(ok1(X)) -> ACTIVE1(X)
LEN1(ok1(X)) -> LEN1(X)
ACTIVE1(add2(X1, X2)) -> ADD2(X1, active1(X2))
PROPER1(s1(X)) -> PROPER1(X)
ACTIVE1(fst2(X1, X2)) -> FST2(X1, active1(X2))
ACTIVE1(add2(s1(X), Y)) -> ADD2(X, Y)
FST2(X1, mark1(X2)) -> FST2(X1, X2)
TOP1(mark1(X)) -> PROPER1(X)
ADD2(mark1(X1), X2) -> ADD2(X1, X2)
PROPER1(fst2(X1, X2)) -> PROPER1(X2)
ACTIVE1(add2(X1, X2)) -> ACTIVE1(X1)
PROPER1(fst2(X1, X2)) -> FST2(proper1(X1), proper1(X2))
PROPER1(add2(X1, X2)) -> PROPER1(X1)
PROPER1(add2(X1, X2)) -> ADD2(proper1(X1), proper1(X2))
PROPER1(s1(X)) -> S1(proper1(X))
S1(ok1(X)) -> S1(X)
PROPER1(fst2(X1, X2)) -> PROPER1(X1)
ACTIVE1(len1(cons2(X, Z))) -> S1(len1(Z))
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
ACTIVE1(from1(X)) -> CONS2(X, from1(s1(X)))
ACTIVE1(cons2(X1, X2)) -> CONS2(active1(X1), X2)
LEN1(mark1(X)) -> LEN1(X)
ADD2(ok1(X1), ok1(X2)) -> ADD2(X1, X2)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(len1(cons2(X, Z))) -> LEN1(Z)
FST2(ok1(X1), ok1(X2)) -> FST2(X1, X2)
PROPER1(add2(X1, X2)) -> PROPER1(X2)
ACTIVE1(len1(X)) -> ACTIVE1(X)
ACTIVE1(from1(X)) -> S1(X)
ADD2(X1, mark1(X2)) -> ADD2(X1, X2)
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
TOP1(ok1(X)) -> TOP1(active1(X))
PROPER1(from1(X)) -> PROPER1(X)
FST2(mark1(X1), X2) -> FST2(X1, X2)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
ACTIVE1(from1(X)) -> ACTIVE1(X)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
FROM1(ok1(X)) -> FROM1(X)
ACTIVE1(add2(s1(X), Y)) -> S1(add2(X, Y))

The TRS R consists of the following rules:

active1(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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 24 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:

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

The TRS R consists of the following rules:

active1(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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)) = 3·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
          ↳ QDP
          ↳ QDP

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

active1(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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:

LEN1(mark1(X)) -> LEN1(X)
LEN1(ok1(X)) -> LEN1(X)

The TRS R consists of the following rules:

active1(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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.


LEN1(mark1(X)) -> LEN1(X)
LEN1(ok1(X)) -> LEN1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(LEN1(x1)) = 3·x1   
POL(mark1(x1)) = 3 + x1   
POL(ok1(x1)) = 3 + 2·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(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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:

ADD2(X1, mark1(X2)) -> ADD2(X1, X2)
ADD2(mark1(X1), X2) -> ADD2(X1, X2)
ADD2(ok1(X1), ok1(X2)) -> ADD2(X1, X2)

The TRS R consists of the following rules:

active1(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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.


ADD2(X1, mark1(X2)) -> ADD2(X1, X2)
ADD2(mark1(X1), X2) -> ADD2(X1, X2)
ADD2(ok1(X1), ok1(X2)) -> ADD2(X1, X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ADD2(x1, x2)) = 3·x1 + 3·x2   
POL(mark1(x1)) = 3 + x1   
POL(ok1(x1)) = 3 + 2·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(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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:

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

The TRS R consists of the following rules:

active1(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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)) = 3·x1   
POL(mark1(x1)) = 3 + x1   
POL(ok1(x1)) = 3 + 2·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(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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:

FST2(ok1(X1), ok1(X2)) -> FST2(X1, X2)
FST2(mark1(X1), X2) -> FST2(X1, X2)
FST2(X1, mark1(X2)) -> FST2(X1, X2)

The TRS R consists of the following rules:

active1(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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.


FST2(ok1(X1), ok1(X2)) -> FST2(X1, X2)
FST2(mark1(X1), X2) -> FST2(X1, X2)
FST2(X1, mark1(X2)) -> FST2(X1, X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(FST2(x1, x2)) = 3·x1 + 3·x2   
POL(mark1(x1)) = 3 + 2·x1   
POL(ok1(x1)) = 3 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ 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(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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:

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

The TRS R consists of the following rules:

active1(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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)
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)) = 3·x1 + 3·x2   
POL(mark1(x1)) = 3 + x1   
POL(ok1(x1)) = 3 + 2·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(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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(add2(X1, X2)) -> PROPER1(X1)
PROPER1(fst2(X1, X2)) -> PROPER1(X2)
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(from1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(len1(X)) -> PROPER1(X)
PROPER1(fst2(X1, X2)) -> PROPER1(X1)
PROPER1(add2(X1, X2)) -> PROPER1(X2)

The TRS R consists of the following rules:

active1(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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(add2(X1, X2)) -> PROPER1(X1)
PROPER1(fst2(X1, X2)) -> PROPER1(X2)
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(from1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(len1(X)) -> PROPER1(X)
PROPER1(fst2(X1, X2)) -> PROPER1(X1)
PROPER1(add2(X1, X2)) -> PROPER1(X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PROPER1(x1)) = 3·x1   
POL(add2(x1, x2)) = 3 + x1 + 2·x2   
POL(cons2(x1, x2)) = 3 + 2·x1 + 2·x2   
POL(from1(x1)) = 3 + 2·x1   
POL(fst2(x1, x2)) = 3 + 2·x1 + 2·x2   
POL(len1(x1)) = 3 + 2·x1   
POL(s1(x1)) = 3 + 2·x1   

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(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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(len1(X)) -> ACTIVE1(X)
ACTIVE1(fst2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(add2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(fst2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(from1(X)) -> ACTIVE1(X)
ACTIVE1(add2(X1, X2)) -> ACTIVE1(X2)

The TRS R consists of the following rules:

active1(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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(len1(X)) -> ACTIVE1(X)
ACTIVE1(fst2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(add2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(fst2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(from1(X)) -> ACTIVE1(X)
ACTIVE1(add2(X1, X2)) -> ACTIVE1(X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVE1(x1)) = 3·x1   
POL(add2(x1, x2)) = 3 + 2·x1 + 2·x2   
POL(cons2(x1, x2)) = 3 + 2·x1   
POL(from1(x1)) = 3 + 2·x1   
POL(fst2(x1, x2)) = 3 + 2·x1 + 2·x2   
POL(len1(x1)) = 3 + x1   

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(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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
            ↳ 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(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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) = 3   
POL(TOP1(x1)) = x1   
POL(active1(x1)) = x1   
POL(add2(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(cons2(x1, x2)) = x1   
POL(from1(x1)) = 2 + x1   
POL(fst2(x1, x2)) = 2 + 3·x1 + 3·x2   
POL(len1(x1)) = 3 + x1   
POL(mark1(x1)) = 1 + x1   
POL(nil) = 2   
POL(ok1(x1)) = x1   
POL(proper1(x1)) = x1   
POL(s1(x1)) = 1   

The following usable rules [14] were oriented:

cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(len1(nil)) -> mark1(0)
active1(add2(0, X)) -> mark1(X)
proper1(0) -> ok1(0)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(len1(X)) -> len1(active1(X))
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
proper1(s1(X)) -> s1(proper1(X))
from1(ok1(X)) -> ok1(from1(X))
active1(add2(X1, X2)) -> add2(X1, active1(X2))
proper1(nil) -> ok1(nil)
active1(fst2(0, Z)) -> mark1(nil)
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
active1(from1(X)) -> from1(active1(X))
len1(mark1(X)) -> mark1(len1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(X))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
from1(mark1(X)) -> mark1(from1(X))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
s1(ok1(X)) -> ok1(s1(X))
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
proper1(from1(X)) -> from1(proper1(X))
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
proper1(len1(X)) -> len1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ 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(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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) = 3   
POL(TOP1(x1)) = x1   
POL(active1(x1)) = 2·x1   
POL(add2(x1, x2)) = 1 + 3·x2   
POL(cons2(x1, x2)) = 1 + 3·x1 + x2   
POL(from1(x1)) = 3·x1   
POL(fst2(x1, x2)) = 1 + 2·x1   
POL(len1(x1)) = x1   
POL(mark1(x1)) = 0   
POL(nil) = 2   
POL(ok1(x1)) = 1 + 2·x1   
POL(s1(x1)) = 1 + 3·x1   

The following usable rules [14] were oriented:

active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(len1(nil)) -> mark1(0)
active1(add2(0, X)) -> mark1(X)
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(len1(X)) -> len1(active1(X))
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
from1(ok1(X)) -> ok1(from1(X))
active1(fst2(0, Z)) -> mark1(nil)
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
active1(from1(X)) -> from1(active1(X))
len1(mark1(X)) -> mark1(len1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(X))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
from1(mark1(X)) -> mark1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ 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(fst2(0, Z)) -> mark1(nil)
active1(fst2(s1(X), cons2(Y, Z))) -> mark1(cons2(Y, fst2(X, Z)))
active1(from1(X)) -> mark1(cons2(X, from1(s1(X))))
active1(add2(0, X)) -> mark1(X)
active1(add2(s1(X), Y)) -> mark1(s1(add2(X, Y)))
active1(len1(nil)) -> mark1(0)
active1(len1(cons2(X, Z))) -> mark1(s1(len1(Z)))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(active1(X1), X2)
active1(fst2(X1, X2)) -> fst2(X1, active1(X2))
active1(from1(X)) -> from1(active1(X))
active1(add2(X1, X2)) -> add2(active1(X1), X2)
active1(add2(X1, X2)) -> add2(X1, active1(X2))
active1(len1(X)) -> len1(active1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
fst2(mark1(X1), X2) -> mark1(fst2(X1, X2))
fst2(X1, mark1(X2)) -> mark1(fst2(X1, X2))
from1(mark1(X)) -> mark1(from1(X))
add2(mark1(X1), X2) -> mark1(add2(X1, X2))
add2(X1, mark1(X2)) -> mark1(add2(X1, X2))
len1(mark1(X)) -> mark1(len1(X))
proper1(0) -> ok1(0)
proper1(s1(X)) -> s1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(fst2(X1, X2)) -> fst2(proper1(X1), proper1(X2))
proper1(from1(X)) -> from1(proper1(X))
proper1(add2(X1, X2)) -> add2(proper1(X1), proper1(X2))
proper1(len1(X)) -> len1(proper1(X))
s1(ok1(X)) -> ok1(s1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
fst2(ok1(X1), ok1(X2)) -> ok1(fst2(X1, X2))
from1(ok1(X)) -> ok1(from1(X))
add2(ok1(X1), ok1(X2)) -> ok1(add2(X1, X2))
len1(ok1(X)) -> ok1(len1(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.