Termination w.r.t. Q proof of /tmp/tmpHz5U6m/Ex1_GL02a

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

    ↳8 PisEmptyProof (⇔)

    ↳9 TRUE

  ↳10 QDP

(0) Obligation:

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(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(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 8 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1) = x1
n__take(x1, x2) = n__take(x1, x2)
TAKE(x1, x2) = TAKE(x1, x2)
s(x1) = s(x1)
cons(x1, x2) = cons(x1, x2)
n__length(x1) = n__length(x1)
LENGTH(x1) = LENGTH(x1)

Recursive path order with status [RPO].
Quasi-Precedence:

n_{_take₂} > [TAKE₂, LENGTH₁]
s₁ > [TAKE₂, LENGTH₁]
cons₂ > [TAKE₂, LENGTH₁]
n_{_length₁} > [TAKE₂, LENGTH₁]

Status:

n_{_take₂}: multiset
TAKE₂: multiset
s₁: multiset
cons₂: multiset
n_{_length₁}: multiset
LENGTH₁: multiset

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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(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(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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

(8) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(9) TRUE

(10) Obligation:

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(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(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(n__length(X)) → length(X)
activate(X) → X

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