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

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

active1(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.

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 contains 9 SCCs with 24 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

S1(ok1(X)) -> S1(X)
Used argument filtering: S1(x1)  =  x1
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

LEN1(ok1(X)) -> LEN1(X)
Used argument filtering: LEN1(x1)  =  x1
mark1(x1)  =  x1
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

LEN1(mark1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

LEN1(mark1(X)) -> LEN1(X)
Used argument filtering: LEN1(x1)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ADD2(ok1(X1), ok1(X2)) -> ADD2(X1, X2)
Used argument filtering: ADD2(x1, x2)  =  x2
mark1(x1)  =  x1
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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)

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ADD2(X1, mark1(X2)) -> ADD2(X1, X2)
Used argument filtering: ADD2(x1, x2)  =  x2
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

ADD2(mark1(X1), 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ADD2(mark1(X1), X2) -> ADD2(X1, X2)
Used argument filtering: ADD2(x1, x2)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
QDP
                        ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

FROM1(ok1(X)) -> FROM1(X)
Used argument filtering: FROM1(x1)  =  x1
mark1(x1)  =  x1
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

The TRS R consists of the following rules:

active1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

FROM1(mark1(X)) -> FROM1(X)
Used argument filtering: FROM1(x1)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

FST2(X1, mark1(X2)) -> FST2(X1, X2)
Used argument filtering: FST2(x1, x2)  =  x2
ok1(x1)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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)

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

FST2(ok1(X1), ok1(X2)) -> FST2(X1, X2)
Used argument filtering: FST2(x1, x2)  =  x2
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

FST2(mark1(X1), 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

FST2(mark1(X1), X2) -> FST2(X1, X2)
Used argument filtering: FST2(x1, x2)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
QDP
                        ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

active1(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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
Used argument filtering: CONS2(x1, x2)  =  x2
ok1(x1)  =  ok1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CONS2(mark1(X1), X2) -> CONS2(X1, X2)
Used argument filtering: CONS2(x1, x2)  =  x1
mark1(x1)  =  mark1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER1(add2(X1, X2)) -> PROPER1(X1)
PROPER1(fst2(X1, X2)) -> PROPER1(X2)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(fst2(X1, X2)) -> PROPER1(X1)
PROPER1(add2(X1, X2)) -> PROPER1(X2)
Used argument filtering: PROPER1(x1)  =  x1
add2(x1, x2)  =  add2(x1, x2)
fst2(x1, x2)  =  fst2(x1, x2)
s1(x1)  =  x1
from1(x1)  =  x1
cons2(x1, x2)  =  cons2(x1, x2)
len1(x1)  =  x1
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

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

PROPER1(s1(X)) -> PROPER1(X)
PROPER1(from1(X)) -> PROPER1(X)
PROPER1(len1(X)) -> 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER1(len1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1)  =  x1
s1(x1)  =  x1
from1(x1)  =  x1
len1(x1)  =  len1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

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

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

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER1(from1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1)  =  x1
s1(x1)  =  x1
from1(x1)  =  from1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
QDP
                        ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP

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

PROPER1(s1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER1(s1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
                      ↳ QDP
                        ↳ QDPAfsSolverProof
QDP
                            ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

active1(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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE1(fst2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(add2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(fst2(X1, X2)) -> ACTIVE1(X2)
ACTIVE1(add2(X1, X2)) -> ACTIVE1(X2)
Used argument filtering: ACTIVE1(x1)  =  x1
len1(x1)  =  x1
fst2(x1, x2)  =  fst2(x1, x2)
cons2(x1, x2)  =  x1
add2(x1, x2)  =  add2(x1, x2)
from1(x1)  =  x1
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof
          ↳ QDP

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

ACTIVE1(len1(X)) -> ACTIVE1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(from1(X)) -> 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE1(from1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1)  =  x1
len1(x1)  =  x1
cons2(x1, x2)  =  x1
from1(x1)  =  from1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ QDPAfsSolverProof
          ↳ QDP

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

ACTIVE1(len1(X)) -> ACTIVE1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
Used argument filtering: ACTIVE1(x1)  =  x1
len1(x1)  =  x1
cons2(x1, x2)  =  cons1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
QDP
                        ↳ QDPAfsSolverProof
          ↳ QDP

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

ACTIVE1(len1(X)) -> 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE1(len1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1)  =  x1
len1(x1)  =  len1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ QDPAfsSolverProof
                      ↳ QDP
                        ↳ QDPAfsSolverProof
QDP
                            ↳ 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

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.