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 oriented strictly and are deleted.


ACTIVATE1(n__add2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__fst2(X1, X2)) -> ACTIVATE1(X1)
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__add2(X1, X2)) -> ACTIVATE1(X1)
The remaining pairs can at least be oriented weakly.

FST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(X)
FST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
ADD2(s1(X), Y) -> ACTIVATE1(X)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
LEN1(cons2(X, Z)) -> ACTIVATE1(Z)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( from1(x1) ) = 2x1


POL( fst2(x1, x2) ) = x1 + 2x2 + 1


POL( n__from1(x1) ) = 2x1


POL( n__s1(x1) ) = x1


POL( 0 ) = max{0, -2}


POL( ADD2(x1, x2) ) = 2x1 + 2x2 + 2


POL( LEN1(x1) ) = 2x1 + 2


POL( nil ) = 1


POL( cons2(x1, x2) ) = x2


POL( activate1(x1) ) = x1


POL( n__add2(x1, x2) ) = 2x1 + 2x2 + 2


POL( len1(x1) ) = x1 + 2


POL( ACTIVATE1(x1) ) = 2x1 + 2


POL( add2(x1, x2) ) = 2x1 + 2x2 + 2


POL( n__fst2(x1, x2) ) = x1 + 2x2 + 1


POL( FST2(x1, x2) ) = 2x1 + 2x2 + 2


POL( s1(x1) ) = x1


POL( n__len1(x1) ) = x1 + 2



The following usable rules [14] were oriented:

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



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

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

FST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(X)
ADD2(s1(X), Y) -> ACTIVATE1(X)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(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(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
            ↳ QDP
              ↳ DependencyGraphProof
QDP
                  ↳ QDPOrderProof

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

ACTIVATE1(n__from1(X)) -> 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(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 oriented strictly and are deleted.


ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( ACTIVATE1(x1) ) = x1 + 2


POL( n__from1(x1) ) = 2x1 + 1



The following usable rules [14] were oriented: none



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

Q DP problem:
P is empty.
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 TRS P is empty. Hence, there is no (P,Q,R) chain.