Termination w.r.t. Q proof of TRS/TRCSREx2_Luc03b

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.

FST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(X)
ACTIVATE₁(n__fst₂(X1, X2)) -> ACTIVATE₁(X1)
FST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__fst₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__fst₂(X1, X2)) -> FST₂(activate₁(X1), activate₁(X2))

The remaining pairs can at least be oriented weakly.

ACTIVATE₁(n__add₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(activate₁(X1), activate₁(X2))
ACTIVATE₁(n__len₁(X)) -> LEN₁(activate₁(X))
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)

Used ordering: Polynomial interpretation [21]:

POL(0) = 1
POL(ACTIVATE₁(x₁)) = 2·x₁
POL(ADD₂(x₁, x₂)) = 2·x₁
POL(FST₂(x₁, x₂)) = 3 + 2·x₁ + 3·x₂
POL(LEN₁(x₁)) = 2·x₁
POL(activate₁(x₁)) = x₁
POL(add₂(x₁, x₂)) = 3·x₁ + x₂
POL(cons₂(x₁, x₂)) = x₂
POL(from₁(x₁)) = x₁
POL(fst₂(x₁, x₂)) = 2 + 3·x₁ + 2·x₂
POL(len₁(x₁)) = x₁
POL(n__add₂(x₁, x₂)) = 3·x₁ + x₂
POL(n__from₁(x₁)) = x₁
POL(n__fst₂(x₁, x₂)) = 2 + 3·x₁ + 2·x₂
POL(n__len₁(x₁)) = x₁
POL(n__s₁(x₁)) = x₁
POL(nil) = 2
POL(s₁(x₁)) = x₁

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

ACTIVATE₁(n__add₂(X1, X2)) -> 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)
ACTIVATE₁(n__len₁(X)) -> LEN₁(activate₁(X))
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(activate₁(X1), activate₁(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(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__len₁(X)) -> ACTIVATE₁(X)
LEN₁(cons₂(X, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__add₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__len₁(X)) -> LEN₁(activate₁(X))
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(activate₁(X1), activate₁(X2))

The remaining pairs can at least be oriented weakly.

ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)

Used ordering: Polynomial interpretation [21]:

POL(0) = 3
POL(ACTIVATE₁(x₁)) = 2 + 2·x₁
POL(ADD₂(x₁, x₂)) = 2 + 2·x₁
POL(LEN₁(x₁)) = 3 + 2·x₁
POL(activate₁(x₁)) = x₁
POL(add₂(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(cons₂(x₁, x₂)) = x₂
POL(from₁(x₁)) = 3·x₁
POL(fst₂(x₁, x₂)) = 3·x₁
POL(len₁(x₁)) = 2 + x₁
POL(n__add₂(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(n__from₁(x₁)) = 3·x₁
POL(n__fst₂(x₁, x₂)) = 3·x₁
POL(n__len₁(x₁)) = 2 + x₁
POL(n__s₁(x₁)) = x₁
POL(nil) = 1
POL(s₁(x₁)) = x₁

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
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.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ 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 interpretation [21]:

POL(ACTIVATE₁(x₁)) = 3·x₁
POL(n__from₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ 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.