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(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)
FST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
LEN₁(cons₂(X, Z)) -> ACTIVATE₁(Z)

The remaining pairs can at least be oriented weakly.

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)

Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE₁(x₁)) = x₁
POL(ADD₂(x₁, x₂)) = x₁
POL(FST₂(x₁, x₂)) = x₁ + x₂
POL(LEN₁(x₁)) = x₁
POL(cons₂(x₁, x₂)) = 1 + x₂
POL(n__add₂(x₁, x₂)) = x₁
POL(n__fst₂(x₁, x₂)) = x₁ + x₂
POL(n__len₁(x₁)) = 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)
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 [13,14,18] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(X1, X2)

The remaining pairs can at least be oriented weakly.

ADD₂(s₁(X), Y) -> ACTIVATE₁(X)

Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE₁(x₁)) = x₁
POL(ADD₂(x₁, x₂)) = x₁
POL(n__add₂(x₁, x₂)) = 1 + x₁
POL(s₁(x₁)) = x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ 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 [13,14,18] contains 0 SCCs with 1 less node.