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(n__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)
s1(X) -> n__s1(X)
add2(X1, X2) -> n__add2(X1, X2)
len1(X) -> n__len1(X)
activate1(n__fst2(X1, X2)) -> fst2(activate1(X1), activate1(X2))
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(X)
activate1(n__add2(X1, X2)) -> add2(activate1(X1), activate1(X2))
activate1(n__len1(X)) -> len1(activate1(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(n__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)
s1(X) -> n__s1(X)
add2(X1, X2) -> n__add2(X1, X2)
len1(X) -> n__len1(X)
activate1(n__fst2(X1, X2)) -> fst2(activate1(X1), activate1(X2))
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(X)
activate1(n__add2(X1, X2)) -> add2(activate1(X1), activate1(X2))
activate1(n__len1(X)) -> len1(activate1(X))
activate1(X) -> 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:

FST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(X)
ACTIVATE1(n__add2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__s1(X)) -> S1(X)
ACTIVATE1(n__fst2(X1, X2)) -> ACTIVATE1(X1)
FST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__len1(X)) -> LEN1(activate1(X))
ACTIVATE1(n__add2(X1, X2)) -> ADD2(activate1(X1), activate1(X2))
LEN1(cons2(X, Z)) -> S1(n__len1(activate1(Z)))
ACTIVATE1(n__from1(X)) -> FROM1(activate1(X))
ACTIVATE1(n__fst2(X1, X2)) -> ACTIVATE1(X2)
ADD2(s1(X), Y) -> ACTIVATE1(X)
ACTIVATE1(n__fst2(X1, X2)) -> FST2(activate1(X1), activate1(X2))
ACTIVATE1(n__len1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ADD2(s1(X), Y) -> S1(n__add2(activate1(X), Y))
LEN1(cons2(X, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__add2(X1, X2)) -> ACTIVATE1(X1)

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(n__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)
s1(X) -> n__s1(X)
add2(X1, X2) -> n__add2(X1, X2)
len1(X) -> n__len1(X)
activate1(n__fst2(X1, X2)) -> fst2(activate1(X1), activate1(X2))
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(X)
activate1(n__add2(X1, X2)) -> add2(activate1(X1), activate1(X2))
activate1(n__len1(X)) -> len1(activate1(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:

FST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(X)
ACTIVATE1(n__add2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__s1(X)) -> S1(X)
ACTIVATE1(n__fst2(X1, X2)) -> ACTIVATE1(X1)
FST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__len1(X)) -> LEN1(activate1(X))
ACTIVATE1(n__add2(X1, X2)) -> ADD2(activate1(X1), activate1(X2))
LEN1(cons2(X, Z)) -> S1(n__len1(activate1(Z)))
ACTIVATE1(n__from1(X)) -> FROM1(activate1(X))
ACTIVATE1(n__fst2(X1, X2)) -> ACTIVATE1(X2)
ADD2(s1(X), Y) -> ACTIVATE1(X)
ACTIVATE1(n__fst2(X1, X2)) -> FST2(activate1(X1), activate1(X2))
ACTIVATE1(n__len1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ADD2(s1(X), Y) -> S1(n__add2(activate1(X), Y))
LEN1(cons2(X, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__add2(X1, X2)) -> ACTIVATE1(X1)

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(n__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)
s1(X) -> n__s1(X)
add2(X1, X2) -> n__add2(X1, X2)
len1(X) -> n__len1(X)
activate1(n__fst2(X1, X2)) -> fst2(activate1(X1), activate1(X2))
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(X)
activate1(n__add2(X1, X2)) -> add2(activate1(X1), activate1(X2))
activate1(n__len1(X)) -> len1(activate1(X))
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPOrderProof

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

FST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(X)
ACTIVATE1(n__add2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__fst2(X1, X2)) -> ACTIVATE1(X1)
FST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__add2(X1, X2)) -> ADD2(activate1(X1), activate1(X2))
ACTIVATE1(n__len1(X)) -> LEN1(activate1(X))
ACTIVATE1(n__fst2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__fst2(X1, X2)) -> FST2(activate1(X1), activate1(X2))
ADD2(s1(X), Y) -> ACTIVATE1(X)
ACTIVATE1(n__len1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
LEN1(cons2(X, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__add2(X1, X2)) -> ACTIVATE1(X1)

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(n__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)
s1(X) -> n__s1(X)
add2(X1, X2) -> n__add2(X1, X2)
len1(X) -> n__len1(X)
activate1(n__fst2(X1, X2)) -> fst2(activate1(X1), activate1(X2))
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(X)
activate1(n__add2(X1, X2)) -> add2(activate1(X1), activate1(X2))
activate1(n__len1(X)) -> len1(activate1(X))
activate1(X) -> 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 strictly oriented and are deleted.


FST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(X)
ACTIVATE1(n__add2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__fst2(X1, X2)) -> ACTIVATE1(X1)
FST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__add2(X1, X2)) -> ADD2(activate1(X1), activate1(X2))
ACTIVATE1(n__len1(X)) -> LEN1(activate1(X))
ACTIVATE1(n__fst2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__fst2(X1, X2)) -> FST2(activate1(X1), activate1(X2))
ACTIVATE1(n__len1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__add2(X1, X2)) -> ACTIVATE1(X1)
The remaining pairs can at least by weakly be oriented.

ADD2(s1(X), Y) -> ACTIVATE1(X)
LEN1(cons2(X, Z)) -> ACTIVATE1(Z)
Used ordering: Combined order from the following AFS and order.
FST2(x1, x2)  =  FST2(x1, x2)
s1(x1)  =  x1
cons2(x1, x2)  =  x2
ACTIVATE1(x1)  =  x1
n__add2(x1, x2)  =  n__add2(x1, x2)
n__fst2(x1, x2)  =  n__fst2(x1, x2)
ADD2(x1, x2)  =  x1
activate1(x1)  =  x1
n__len1(x1)  =  n__len1(x1)
LEN1(x1)  =  x1
n__from1(x1)  =  n__from1(x1)
fst2(x1, x2)  =  fst2(x1, x2)
from1(x1)  =  from1(x1)
n__s1(x1)  =  x1
add2(x1, x2)  =  add2(x1, x2)
len1(x1)  =  len1(x1)
nil  =  nil
0  =  0

Lexicographic Path Order [19].
Precedence:
[nadd2, add2] > [nil, 0]
[nfst2, fst2] > FST2 > [nil, 0]
[nlen1, len1] > [nil, 0]
[nfrom1, from1] > [nil, 0]


The following usable rules [14] were oriented:

activate1(n__fst2(X1, X2)) -> fst2(activate1(X1), activate1(X2))
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(X)
activate1(n__add2(X1, X2)) -> add2(activate1(X1), activate1(X2))
activate1(n__len1(X)) -> len1(activate1(X))
activate1(X) -> X
len1(nil) -> 0
len1(cons2(X, Z)) -> s1(n__len1(activate1(Z)))
len1(X) -> n__len1(X)
s1(X) -> n__s1(X)
add2(0, X) -> X
add2(s1(X), Y) -> s1(n__add2(activate1(X), Y))
add2(X1, X2) -> n__add2(X1, X2)
from1(X) -> cons2(X, n__from1(n__s1(X)))
from1(X) -> n__from1(X)
fst2(0, Z) -> nil
fst2(s1(X), cons2(Y, Z)) -> cons2(Y, n__fst2(activate1(X), activate1(Z)))
fst2(X1, X2) -> n__fst2(X1, X2)



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

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

ADD2(s1(X), Y) -> ACTIVATE1(X)
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(n__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)
s1(X) -> n__s1(X)
add2(X1, X2) -> n__add2(X1, X2)
len1(X) -> n__len1(X)
activate1(n__fst2(X1, X2)) -> fst2(activate1(X1), activate1(X2))
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(X)
activate1(n__add2(X1, X2)) -> add2(activate1(X1), activate1(X2))
activate1(n__len1(X)) -> len1(activate1(X))
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.