Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/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(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.

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 contains 1 SCC with 4 less nodes.

↳ 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__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.
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__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__fst₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__add₂(X1, X2)) -> ACTIVATE₁(X1)

Used argument filtering: FST₂(x1, x2) = FST₂(x1, x2)
s₁(x1) = x1
cons₂(x1, x2) = x2
ACTIVATE₁(x1) = x1
n__add₂(x1, x2) = n__add₂(x1, x2)
n__fst₂(x1, x2) = n__fst₂(x1, x2)
ADD₂(x1, x2) = x1
activate₁(x1) = x1
n__len₁(x1) = x1
LEN₁(x1) = x1
n__from₁(x1) = x1
fst₂(x1, x2) = fst₂(x1, x2)
from₁(x1) = x1
n__s₁(x1) = x1
add₂(x1, x2) = add₂(x1, x2)
len₁(x1) = x1
nil = nil
0 = 0
Used ordering: Quasi Precedence: [FST_2, n__fst_2, fst_2] [n__add_2, add_2] [nil, 0]

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ DependencyGraphProof

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

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)
LEN₁(cons₂(X, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__len₁(X)) -> LEN₁(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 contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPAfsSolverProof

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

ACTIVATE₁(n__len₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)
LEN₁(cons₂(X, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__len₁(X)) -> LEN₁(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVATE₁(n__from₁(X)) -> ACTIVATE₁(X)

Used argument filtering: ACTIVATE₁(x1) = x1
n__len₁(x1) = x1
n__from₁(x1) = n__from₁(x1)
LEN₁(x1) = x1
cons₂(x1, x2) = x2
activate₁(x1) = x1
n__fst₂(x1, x2) = n__fst
fst₂(x1, x2) = fst
from₁(x1) = from₁(x1)
n__s₁(x1) = n__s
s₁(x1) = s
n__add₂(x1, x2) = x2
add₂(x1, x2) = x2
len₁(x1) = x1
nil = nil
0 = 0
Used ordering: Quasi Precedence: [n__from_1, from_1] > [n__s, s] [n__fst, fst] > nil > 0 > [n__s, s]

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPAfsSolverProof

                    ↳ QDP

                      ↳ QDPAfsSolverProof

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

ACTIVATE₁(n__len₁(X)) -> ACTIVATE₁(X)
LEN₁(cons₂(X, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__len₁(X)) -> LEN₁(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVATE₁(n__len₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__len₁(X)) -> LEN₁(activate₁(X))

Used argument filtering: ACTIVATE₁(x1) = x1
n__len₁(x1) = n__len₁(x1)
LEN₁(x1) = x1
cons₂(x1, x2) = x2
activate₁(x1) = x1
n__fst₂(x1, x2) = n__fst
fst₂(x1, x2) = fst
n__from₁(x1) = n__from
from₁(x1) = from
n__s₁(x1) = n__s
s₁(x1) = s
n__add₂(x1, x2) = x2
add₂(x1, x2) = x2
len₁(x1) = len₁(x1)
nil = nil
0 = 0
Used ordering: Quasi Precedence: [n__len_1, len_1] > [n__s, s] [n__fst, fst, nil, 0] > [n__s, s] [n__from, from] > [n__s, s]

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPAfsSolverProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPAfsSolverProof

                    ↳ QDP

                      ↳ QDPAfsSolverProof

                        ↳ QDP

                          ↳ DependencyGraphProof

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

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(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 contains 0 SCCs with 1 less node.