Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/TRCSRSLASHEx2_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.

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 contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: FST₂(x1, x2) = FST₂(x1, x2)
s₁(x1) = x1
cons₂(x1, x2) = x2
ACTIVATE₁(x1) = x1
n__fst₂(x1, x2) = n__fst₂(x1, x2)
n__len₁(x1) = x1
LEN₁(x1) = x1
ADD₂(x1, x2) = x1
n__add₂(x1, x2) = x1
Used ordering: Precedence:

n_{_fst₂} > FST₂

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ QDPAfsSolverProof

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

ACTIVATE₁(n__len₁(X)) -> LEN₁(X)
ADD₂(s₁(X), Y) -> ACTIVATE₁(X)
ACTIVATE₁(n__add₂(X1, X2)) -> ADD₂(X1, X2)
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

LEN₁(cons₂(X, Z)) -> ACTIVATE₁(Z)

Used argument filtering: ACTIVATE₁(x1) = x1
n__len₁(x1) = x1
LEN₁(x1) = x1
ADD₂(x1, x2) = x1
s₁(x1) = x1
n__add₂(x1, x2) = x1
cons₂(x1, x2) = cons₁(x2)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ QDPAfsSolverProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

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 contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ QDPAfsSolverProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: ADD₂(x1, x2) = x1
s₁(x1) = x1
ACTIVATE₁(x1) = x1
n__add₂(x1, x2) = n__add₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ QDPAfsSolverProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ QDPAfsSolverProof

                        ↳ 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 contains 0 SCCs with 1 less node.