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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 1 SCC with 1 less node.

↳ 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__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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


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

ACTIVATE1(n__fst2(X1, X2)) -> FST2(X1, X2)
ADD2(s1(X), Y) -> ACTIVATE1(X)
LEN1(cons2(X, Z)) -> ACTIVATE1(Z)
Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE1(x1)) = x1   
POL(ADD2(x1, x2)) = 2·x1   
POL(FST2(x1, x2)) = 1 + 2·x1 + 3·x2   
POL(LEN1(x1)) = 3·x1   
POL(cons2(x1, x2)) = x2   
POL(n__add2(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(n__fst2(x1, x2)) = 1 + 2·x1 + 3·x2   
POL(n__len1(x1)) = 2 + 3·x1   
POL(s1(x1)) = x1   

The following usable rules [14] were oriented: none



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

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

ACTIVATE1(n__fst2(X1, X2)) -> FST2(X1, X2)
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(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 [13,14,18] contains 0 SCCs with 3 less nodes.