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:

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, n__fst2(activate1(X), activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(n__add2(activate1(X), Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(n__len1(activate1(Z)))
fst2(X1, X2) -> n__fst2(X1, X2)
from1(X) -> n__from1(X)
add2(X1, X2) -> n__add2(X1, X2)
len1(X) -> n__len1(X)
activate1(n__fst2(X1, X2)) -> fst2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__add2(X1, X2)) -> add2(X1, X2)
activate1(n__len1(X)) -> len1(X)
activate1(X) -> X

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, n__fst2(activate1(X), activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(n__add2(activate1(X), Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(n__len1(activate1(Z)))
fst2(X1, X2) -> n__fst2(X1, X2)
from1(X) -> n__from1(X)
add2(X1, X2) -> n__add2(X1, X2)
len1(X) -> n__len1(X)
activate1(n__fst2(X1, X2)) -> fst2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__add2(X1, X2)) -> add2(X1, X2)
activate1(n__len1(X)) -> len1(X)
activate1(X) -> X

Q is empty.

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

ACTIVATE1(n__fst2(X1, X2)) -> FST2(X1, X2)
FST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(X)
ACTIVATE1(n__len1(X)) -> LEN1(X)
ADD2(s1(X), Y) -> ACTIVATE1(X)
ACTIVATE1(n__add2(X1, X2)) -> ADD2(X1, X2)
ACTIVATE1(n__from1(X)) -> FROM1(X)
LEN1(cons2(X, Z)) -> ACTIVATE1(Z)
FST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)

The TRS R consists of the following rules:

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, n__fst2(activate1(X), activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(n__add2(activate1(X), Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(n__len1(activate1(Z)))
fst2(X1, X2) -> n__fst2(X1, X2)
from1(X) -> n__from1(X)
add2(X1, X2) -> n__add2(X1, X2)
len1(X) -> n__len1(X)
activate1(n__fst2(X1, X2)) -> fst2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__add2(X1, X2)) -> add2(X1, X2)
activate1(n__len1(X)) -> len1(X)
activate1(X) -> 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:

ACTIVATE1(n__fst2(X1, X2)) -> FST2(X1, X2)
FST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(X)
ACTIVATE1(n__len1(X)) -> LEN1(X)
ADD2(s1(X), Y) -> ACTIVATE1(X)
ACTIVATE1(n__add2(X1, X2)) -> ADD2(X1, X2)
ACTIVATE1(n__from1(X)) -> FROM1(X)
LEN1(cons2(X, Z)) -> ACTIVATE1(Z)
FST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)

The TRS R consists of the following rules:

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, n__fst2(activate1(X), activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(n__add2(activate1(X), Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(n__len1(activate1(Z)))
fst2(X1, X2) -> n__fst2(X1, X2)
from1(X) -> n__from1(X)
add2(X1, X2) -> n__add2(X1, X2)
len1(X) -> n__len1(X)
activate1(n__fst2(X1, X2)) -> fst2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__add2(X1, X2)) -> add2(X1, X2)
activate1(n__len1(X)) -> len1(X)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPAfsSolverProof

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

FST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(X)
ACTIVATE1(n__fst2(X1, X2)) -> FST2(X1, X2)
ACTIVATE1(n__len1(X)) -> LEN1(X)
ADD2(s1(X), Y) -> ACTIVATE1(X)
ACTIVATE1(n__add2(X1, X2)) -> ADD2(X1, X2)
FST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
LEN1(cons2(X, Z)) -> ACTIVATE1(Z)

The TRS R consists of the following rules:

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, n__fst2(activate1(X), activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(n__add2(activate1(X), Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(n__len1(activate1(Z)))
fst2(X1, X2) -> n__fst2(X1, X2)
from1(X) -> n__from1(X)
add2(X1, X2) -> n__add2(X1, X2)
len1(X) -> n__len1(X)
activate1(n__fst2(X1, X2)) -> fst2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__add2(X1, X2)) -> add2(X1, X2)
activate1(n__len1(X)) -> len1(X)
activate1(X) -> 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(s1(X), cons2(Y, Z)) -> ACTIVATE1(X)
ACTIVATE1(n__fst2(X1, X2)) -> FST2(X1, X2)
FST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
Used argument filtering: FST2(x1, x2)  =  FST2(x1, x2)
s1(x1)  =  x1
cons2(x1, x2)  =  x2
ACTIVATE1(x1)  =  x1
n__fst2(x1, x2)  =  n__fst2(x1, x2)
n__len1(x1)  =  x1
LEN1(x1)  =  x1
ADD2(x1, x2)  =  x1
n__add2(x1, x2)  =  x1
Used ordering: Quasi Precedence: n__fst_2 > FST_2


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

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

ACTIVATE1(n__len1(X)) -> LEN1(X)
ADD2(s1(X), Y) -> ACTIVATE1(X)
ACTIVATE1(n__add2(X1, X2)) -> ADD2(X1, X2)
LEN1(cons2(X, Z)) -> ACTIVATE1(Z)

The TRS R consists of the following rules:

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, n__fst2(activate1(X), activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(n__add2(activate1(X), Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(n__len1(activate1(Z)))
fst2(X1, X2) -> n__fst2(X1, X2)
from1(X) -> n__from1(X)
add2(X1, X2) -> n__add2(X1, X2)
len1(X) -> n__len1(X)
activate1(n__fst2(X1, X2)) -> fst2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__add2(X1, X2)) -> add2(X1, X2)
activate1(n__len1(X)) -> len1(X)
activate1(X) -> 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(cons2(X, Z)) -> ACTIVATE1(Z)
Used argument filtering: ACTIVATE1(x1)  =  x1
n__len1(x1)  =  x1
LEN1(x1)  =  x1
ADD2(x1, x2)  =  x1
s1(x1)  =  x1
n__add2(x1, x2)  =  x1
cons2(x1, x2)  =  cons1(x2)
Used ordering: Quasi Precedence: trivial


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

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

ACTIVATE1(n__len1(X)) -> LEN1(X)
ADD2(s1(X), Y) -> ACTIVATE1(X)
ACTIVATE1(n__add2(X1, X2)) -> ADD2(X1, X2)

The TRS R consists of the following rules:

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, n__fst2(activate1(X), activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(n__add2(activate1(X), Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(n__len1(activate1(Z)))
fst2(X1, X2) -> n__fst2(X1, X2)
from1(X) -> n__from1(X)
add2(X1, X2) -> n__add2(X1, X2)
len1(X) -> n__len1(X)
activate1(n__fst2(X1, X2)) -> fst2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__add2(X1, X2)) -> add2(X1, X2)
activate1(n__len1(X)) -> len1(X)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

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

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

ADD2(s1(X), Y) -> ACTIVATE1(X)
ACTIVATE1(n__add2(X1, X2)) -> ADD2(X1, X2)

The TRS R consists of the following rules:

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, n__fst2(activate1(X), activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(n__add2(activate1(X), Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(n__len1(activate1(Z)))
fst2(X1, X2) -> n__fst2(X1, X2)
from1(X) -> n__from1(X)
add2(X1, X2) -> n__add2(X1, X2)
len1(X) -> n__len1(X)
activate1(n__fst2(X1, X2)) -> fst2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__add2(X1, X2)) -> add2(X1, X2)
activate1(n__len1(X)) -> len1(X)
activate1(X) -> 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.

ACTIVATE1(n__add2(X1, X2)) -> ADD2(X1, X2)
Used argument filtering: ADD2(x1, x2)  =  x1
s1(x1)  =  x1
ACTIVATE1(x1)  =  x1
n__add2(x1, x2)  =  n__add1(x1)
Used ordering: Quasi Precedence: trivial


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

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

ADD2(s1(X), Y) -> ACTIVATE1(X)

The TRS R consists of the following rules:

fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, n__fst2(activate1(X), activate1(Z)))
from1(X) -> cons2(X, n__from1(s1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(n__add2(activate1(X), Y))
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(n__len1(activate1(Z)))
fst2(X1, X2) -> n__fst2(X1, X2)
from1(X) -> n__from1(X)
add2(X1, X2) -> n__add2(X1, X2)
len1(X) -> n__len1(X)
activate1(n__fst2(X1, X2)) -> fst2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__add2(X1, X2)) -> add2(X1, X2)
activate1(n__len1(X)) -> len1(X)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.