Termination w.r.t. Q proof of /tmp/tmpsQg44s/Ex1_GL02a

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:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(n__s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__length(X)) → length(activate(X))
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(n__s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__length(X)) → length(activate(X))
activate(X) → X

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ACTIVATE(n__inf(X)) → ACTIVATE(X)
LENGTH(cons(X, L)) → S(n__length(activate(L)))
TAKE(s(X), cons(Y, L)) → ACTIVATE(X)
LENGTH(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
TAKE(s(X), cons(Y, L)) → ACTIVATE(Y)
ACTIVATE(n__length(X)) → LENGTH(activate(X))
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__s(X)) → S(X)
EQ(n__s(X), n__s(Y)) → ACTIVATE(X)
ACTIVATE(n__inf(X)) → INF(activate(X))
EQ(n__s(X), n__s(Y)) → EQ(activate(X), activate(Y))
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
TAKE(s(X), cons(Y, L)) → ACTIVATE(L)
ACTIVATE(n__0) → 0¹
EQ(n__s(X), n__s(Y)) → ACTIVATE(Y)
ACTIVATE(n__length(X)) → ACTIVATE(X)
LENGTH(nil) → 0¹

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(n__s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__length(X)) → length(activate(X))
activate(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:

ACTIVATE(n__inf(X)) → ACTIVATE(X)
LENGTH(cons(X, L)) → S(n__length(activate(L)))
TAKE(s(X), cons(Y, L)) → ACTIVATE(X)
LENGTH(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
TAKE(s(X), cons(Y, L)) → ACTIVATE(Y)
ACTIVATE(n__length(X)) → LENGTH(activate(X))
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__s(X)) → S(X)
EQ(n__s(X), n__s(Y)) → ACTIVATE(X)
ACTIVATE(n__inf(X)) → INF(activate(X))
EQ(n__s(X), n__s(Y)) → EQ(activate(X), activate(Y))
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
TAKE(s(X), cons(Y, L)) → ACTIVATE(L)
ACTIVATE(n__0) → 0¹
EQ(n__s(X), n__s(Y)) → ACTIVATE(Y)
ACTIVATE(n__length(X)) → ACTIVATE(X)
LENGTH(nil) → 0¹

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(n__s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__length(X)) → length(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

ACTIVATE(n__inf(X)) → ACTIVATE(X)
TAKE(s(X), cons(Y, L)) → ACTIVATE(X)
LENGTH(cons(X, L)) → ACTIVATE(L)
TAKE(s(X), cons(Y, L)) → ACTIVATE(Y)
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
ACTIVATE(n__length(X)) → LENGTH(activate(X))
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__length(X)) → ACTIVATE(X)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
TAKE(s(X), cons(Y, L)) → ACTIVATE(L)

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(n__s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__length(X)) → length(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

LENGTH(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__length(X)) → ACTIVATE(X)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)

The remaining pairs can at least be oriented weakly.

ACTIVATE(n__inf(X)) → ACTIVATE(X)
TAKE(s(X), cons(Y, L)) → ACTIVATE(X)
TAKE(s(X), cons(Y, L)) → ACTIVATE(Y)
ACTIVATE(n__length(X)) → LENGTH(activate(X))
TAKE(s(X), cons(Y, L)) → ACTIVATE(L)

Used ordering: Polynomial interpretation [25,35]:

POL(LENGTH(x₁)) = 1 + x_1
POL(n__length(x₁)) = 4 + (4)x_1
POL(n__inf(x₁)) = (4)x_1
POL(activate(x₁)) = x_1
POL(n__s(x₁)) = (1/2)x_1
POL(take(x₁, x₂)) = 1/4 + (2)x_1 + (13/4)x_2
POL(0) = 2
POL(TAKE(x₁, x₂)) = (1/2)x_1 + (1/2)x_2
POL(cons(x₁, x₂)) = (3)x_1 + (1/2)x_2
POL(inf(x₁)) = (4)x_1
POL(n__0) = 2
POL(n__take(x₁, x₂)) = 1/4 + (2)x_1 + (13/4)x_2
POL(s(x₁)) = (1/2)x_1
POL(length(x₁)) = 4 + (4)x_1
POL(ACTIVATE(x₁)) = (1/4)x_1
POL(nil) = 3/2

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

activate(n__inf(X)) → inf(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__length(X)) → length(activate(X))
activate(X) → X
take(0, X) → nil
inf(X) → cons(X, n__inf(n__s(X)))
length(nil) → 0
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
0 → n__0
length(cons(X, L)) → s(n__length(activate(L)))
inf(X) → n__inf(X)
s(X) → n__s(X)
length(X) → n__length(X)
take(X1, X2) → n__take(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

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

ACTIVATE(n__inf(X)) → ACTIVATE(X)
TAKE(s(X), cons(Y, L)) → ACTIVATE(X)
TAKE(s(X), cons(Y, L)) → ACTIVATE(Y)
ACTIVATE(n__length(X)) → LENGTH(activate(X))
TAKE(s(X), cons(Y, L)) → ACTIVATE(L)

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(n__s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__length(X)) → length(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

          ↳ QDP

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

ACTIVATE(n__inf(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(n__s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__length(X)) → length(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

ACTIVATE(n__inf(X)) → ACTIVATE(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(n__inf(x₁)) = 1/4 + (7/2)x_1
POL(ACTIVATE(x₁)) = (2)x_1

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(n__s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__length(X)) → length(activate(X))
activate(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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

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

EQ(n__s(X), n__s(Y)) → EQ(activate(X), activate(Y))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(n__s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__length(X)) → length(activate(X))
activate(X) → X

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