Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/TRCSRSLASHEx1_GL02a_Z.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

eq₂(n__0, n__0) -> true
eq₂(n__s₁(X), n__s₁(Y)) -> eq₂(activate₁(X), activate₁(Y))
eq₂(X, Y) -> false
inf₁(X) -> cons₂(X, n__inf₁(s₁(X)))
take₂(0, X) -> nil
take₂(s₁(X), cons₂(Y, L)) -> cons₂(activate₁(Y), n__take₂(activate₁(X), activate₁(L)))
length₁(nil) -> 0
length₁(cons₂(X, L)) -> s₁(n__length₁(activate₁(L)))
0 -> n__0
s₁(X) -> n__s₁(X)
inf₁(X) -> n__inf₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
length₁(X) -> n__length₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__inf₁(X)) -> inf₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(X1, X2)
activate₁(n__length₁(X)) -> length₁(X)
activate₁(X) -> X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

eq₂(n__0, n__0) -> true
eq₂(n__s₁(X), n__s₁(Y)) -> eq₂(activate₁(X), activate₁(Y))
eq₂(X, Y) -> false
inf₁(X) -> cons₂(X, n__inf₁(s₁(X)))
take₂(0, X) -> nil
take₂(s₁(X), cons₂(Y, L)) -> cons₂(activate₁(Y), n__take₂(activate₁(X), activate₁(L)))
length₁(nil) -> 0
length₁(cons₂(X, L)) -> s₁(n__length₁(activate₁(L)))
0 -> n__0
s₁(X) -> n__s₁(X)
inf₁(X) -> n__inf₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
length₁(X) -> n__length₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__inf₁(X)) -> inf₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(X1, X2)
activate₁(n__length₁(X)) -> length₁(X)
activate₁(X) -> X

Q is empty.

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

ACTIVATE₁(n__take₂(X1, X2)) -> TAKE₂(X1, X2)
TAKE₂(s₁(X), cons₂(Y, L)) -> ACTIVATE₁(L)
LENGTH₁(cons₂(X, L)) -> S₁(n__length₁(activate₁(L)))
ACTIVATE₁(n__0) -> 0¹
ACTIVATE₁(n__s₁(X)) -> S₁(X)
TAKE₂(s₁(X), cons₂(Y, L)) -> ACTIVATE₁(X)
EQ₂(n__s₁(X), n__s₁(Y)) -> ACTIVATE₁(X)
EQ₂(n__s₁(X), n__s₁(Y)) -> EQ₂(activate₁(X), activate₁(Y))
LENGTH₁(cons₂(X, L)) -> ACTIVATE₁(L)
INF₁(X) -> S₁(X)
ACTIVATE₁(n__length₁(X)) -> LENGTH₁(X)
TAKE₂(s₁(X), cons₂(Y, L)) -> ACTIVATE₁(Y)
ACTIVATE₁(n__inf₁(X)) -> INF₁(X)
EQ₂(n__s₁(X), n__s₁(Y)) -> ACTIVATE₁(Y)
LENGTH₁(nil) -> 0¹

The TRS R consists of the following rules:

eq₂(n__0, n__0) -> true
eq₂(n__s₁(X), n__s₁(Y)) -> eq₂(activate₁(X), activate₁(Y))
eq₂(X, Y) -> false
inf₁(X) -> cons₂(X, n__inf₁(s₁(X)))
take₂(0, X) -> nil
take₂(s₁(X), cons₂(Y, L)) -> cons₂(activate₁(Y), n__take₂(activate₁(X), activate₁(L)))
length₁(nil) -> 0
length₁(cons₂(X, L)) -> s₁(n__length₁(activate₁(L)))
0 -> n__0
s₁(X) -> n__s₁(X)
inf₁(X) -> n__inf₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
length₁(X) -> n__length₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__inf₁(X)) -> inf₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(X1, X2)
activate₁(n__length₁(X)) -> length₁(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__take₂(X1, X2)) -> TAKE₂(X1, X2)
TAKE₂(s₁(X), cons₂(Y, L)) -> ACTIVATE₁(L)
LENGTH₁(cons₂(X, L)) -> S₁(n__length₁(activate₁(L)))
ACTIVATE₁(n__0) -> 0¹
ACTIVATE₁(n__s₁(X)) -> S₁(X)
TAKE₂(s₁(X), cons₂(Y, L)) -> ACTIVATE₁(X)
EQ₂(n__s₁(X), n__s₁(Y)) -> ACTIVATE₁(X)
EQ₂(n__s₁(X), n__s₁(Y)) -> EQ₂(activate₁(X), activate₁(Y))
LENGTH₁(cons₂(X, L)) -> ACTIVATE₁(L)
INF₁(X) -> S₁(X)
ACTIVATE₁(n__length₁(X)) -> LENGTH₁(X)
TAKE₂(s₁(X), cons₂(Y, L)) -> ACTIVATE₁(Y)
ACTIVATE₁(n__inf₁(X)) -> INF₁(X)
EQ₂(n__s₁(X), n__s₁(Y)) -> ACTIVATE₁(Y)
LENGTH₁(nil) -> 0¹

The TRS R consists of the following rules:

eq₂(n__0, n__0) -> true
eq₂(n__s₁(X), n__s₁(Y)) -> eq₂(activate₁(X), activate₁(Y))
eq₂(X, Y) -> false
inf₁(X) -> cons₂(X, n__inf₁(s₁(X)))
take₂(0, X) -> nil
take₂(s₁(X), cons₂(Y, L)) -> cons₂(activate₁(Y), n__take₂(activate₁(X), activate₁(L)))
length₁(nil) -> 0
length₁(cons₂(X, L)) -> s₁(n__length₁(activate₁(L)))
0 -> n__0
s₁(X) -> n__s₁(X)
inf₁(X) -> n__inf₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
length₁(X) -> n__length₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__inf₁(X)) -> inf₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(X1, X2)
activate₁(n__length₁(X)) -> length₁(X)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 8 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVATE₁(n__take₂(X1, X2)) -> TAKE₂(X1, X2)
ACTIVATE₁(n__length₁(X)) -> LENGTH₁(X)
TAKE₂(s₁(X), cons₂(Y, L)) -> ACTIVATE₁(L)
TAKE₂(s₁(X), cons₂(Y, L)) -> ACTIVATE₁(Y)
TAKE₂(s₁(X), cons₂(Y, L)) -> ACTIVATE₁(X)
LENGTH₁(cons₂(X, L)) -> ACTIVATE₁(L)

The TRS R consists of the following rules:

eq₂(n__0, n__0) -> true
eq₂(n__s₁(X), n__s₁(Y)) -> eq₂(activate₁(X), activate₁(Y))
eq₂(X, Y) -> false
inf₁(X) -> cons₂(X, n__inf₁(s₁(X)))
take₂(0, X) -> nil
take₂(s₁(X), cons₂(Y, L)) -> cons₂(activate₁(Y), n__take₂(activate₁(X), activate₁(L)))
length₁(nil) -> 0
length₁(cons₂(X, L)) -> s₁(n__length₁(activate₁(L)))
0 -> n__0
s₁(X) -> n__s₁(X)
inf₁(X) -> n__inf₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
length₁(X) -> n__length₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__inf₁(X)) -> inf₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(X1, X2)
activate₁(n__length₁(X)) -> length₁(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__take₂(X1, X2)) -> TAKE₂(X1, X2)
TAKE₂(s₁(X), cons₂(Y, L)) -> ACTIVATE₁(L)
TAKE₂(s₁(X), cons₂(Y, L)) -> ACTIVATE₁(Y)
TAKE₂(s₁(X), cons₂(Y, L)) -> ACTIVATE₁(X)
LENGTH₁(cons₂(X, L)) -> ACTIVATE₁(L)

Used argument filtering: ACTIVATE₁(x1) = x1
n__take₂(x1, x2) = n__take₂(x1, x2)
TAKE₂(x1, x2) = TAKE₂(x1, x2)
n__length₁(x1) = x1
LENGTH₁(x1) = x1
s₁(x1) = x1
cons₂(x1, x2) = cons₂(x1, x2)
Used ordering: Quasi Precedence: n__take_2 > TAKE_2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

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

ACTIVATE₁(n__length₁(X)) -> LENGTH₁(X)

The TRS R consists of the following rules:

eq₂(n__0, n__0) -> true
eq₂(n__s₁(X), n__s₁(Y)) -> eq₂(activate₁(X), activate₁(Y))
eq₂(X, Y) -> false
inf₁(X) -> cons₂(X, n__inf₁(s₁(X)))
take₂(0, X) -> nil
take₂(s₁(X), cons₂(Y, L)) -> cons₂(activate₁(Y), n__take₂(activate₁(X), activate₁(L)))
length₁(nil) -> 0
length₁(cons₂(X, L)) -> s₁(n__length₁(activate₁(L)))
0 -> n__0
s₁(X) -> n__s₁(X)
inf₁(X) -> n__inf₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
length₁(X) -> n__length₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__inf₁(X)) -> inf₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(X1, X2)
activate₁(n__length₁(X)) -> length₁(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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

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

EQ₂(n__s₁(X), n__s₁(Y)) -> EQ₂(activate₁(X), activate₁(Y))

The TRS R consists of the following rules:

eq₂(n__0, n__0) -> true
eq₂(n__s₁(X), n__s₁(Y)) -> eq₂(activate₁(X), activate₁(Y))
eq₂(X, Y) -> false
inf₁(X) -> cons₂(X, n__inf₁(s₁(X)))
take₂(0, X) -> nil
take₂(s₁(X), cons₂(Y, L)) -> cons₂(activate₁(Y), n__take₂(activate₁(X), activate₁(L)))
length₁(nil) -> 0
length₁(cons₂(X, L)) -> s₁(n__length₁(activate₁(L)))
0 -> n__0
s₁(X) -> n__s₁(X)
inf₁(X) -> n__inf₁(X)
take₂(X1, X2) -> n__take₂(X1, X2)
length₁(X) -> n__length₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__inf₁(X)) -> inf₁(X)
activate₁(n__take₂(X1, X2)) -> take₂(X1, X2)
activate₁(n__length₁(X)) -> length₁(X)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.