Termination w.r.t. Q of the given QTRS could be disproven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 Narrowing (⇔)
↳6 QDP
↳7 Narrowing (⇔)
↳8 QDP
↳9 DependencyGraphProof (⇔)
↳10 QDP
↳11 Narrowing (⇔)
↳12 QDP
↳13 DependencyGraphProof (⇔)
↳14 QDP
↳15 Narrowing (⇔)
↳16 QDP
↳17 NonTerminationProof (⇔)
↳18 NO

(0) Obligation:

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

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
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:

FROM(X) → CONS(X, n__from(s(X)))
LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH(n__cons(X, Y)) → ACTIVATE(Y)
LENGTH1(X) → LENGTH(activate(X))
LENGTH1(X) → ACTIVATE(X)
ACTIVATE(n__from(X)) → FROM(X)
ACTIVATE(n__nil) → NIL
ACTIVATE(n__cons(X1, X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
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 1 SCC with 6 less nodes.

(4) Obligation:

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

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(X) → LENGTH(activate(X))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(5) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule LENGTH1(X) → LENGTH(activate(X)) at position [0] we obtained the following new rules [LPAR04]:

LENGTH1(n__from(x0)) → LENGTH(from(x0))
LENGTH1(n__nil) → LENGTH(nil)
LENGTH1(n__cons(x0, x1)) → LENGTH(cons(x0, x1))
LENGTH1(x0) → LENGTH(x0)

(6) Obligation:

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

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(n__from(x0)) → LENGTH(from(x0))
LENGTH1(n__nil) → LENGTH(nil)
LENGTH1(n__cons(x0, x1)) → LENGTH(cons(x0, x1))
LENGTH1(x0) → LENGTH(x0)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(7) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule LENGTH1(n__from(x0)) → LENGTH(from(x0)) at position [0] we obtained the following new rules [LPAR04]:

LENGTH1(n__from(x0)) → LENGTH(cons(x0, n__from(s(x0))))
LENGTH1(n__from(x0)) → LENGTH(n__from(x0))

(8) Obligation:

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

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(n__nil) → LENGTH(nil)
LENGTH1(n__cons(x0, x1)) → LENGTH(cons(x0, x1))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__from(x0)) → LENGTH(cons(x0, n__from(s(x0))))
LENGTH1(n__from(x0)) → LENGTH(n__from(x0))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(10) Obligation:

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

LENGTH1(n__nil) → LENGTH(nil)
LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(n__cons(x0, x1)) → LENGTH(cons(x0, x1))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__from(x0)) → LENGTH(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(11) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule LENGTH1(n__nil) → LENGTH(nil) at position [0] we obtained the following new rules [LPAR04]:

LENGTH1(n__nil) → LENGTH(n__nil)

(12) Obligation:

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

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(n__cons(x0, x1)) → LENGTH(cons(x0, x1))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__from(x0)) → LENGTH(cons(x0, n__from(s(x0))))
LENGTH1(n__nil) → LENGTH(n__nil)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(13) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(14) Obligation:

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

LENGTH1(n__cons(x0, x1)) → LENGTH(cons(x0, x1))
LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__from(x0)) → LENGTH(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(15) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule LENGTH1(n__cons(x0, x1)) → LENGTH(cons(x0, x1)) at position [0] we obtained the following new rules [LPAR04]:

LENGTH1(n__cons(x0, x1)) → LENGTH(n__cons(x0, x1))

(16) Obligation:

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

LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__from(x0)) → LENGTH(cons(x0, n__from(s(x0))))
LENGTH1(n__cons(x0, x1)) → LENGTH(n__cons(x0, x1))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
length(n__nil) → 0
length(n__cons(X, Y)) → s(length1(activate(Y)))
length1(X) → length(activate(X))
from(X) → n__from(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(X)
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(X) → X

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

(17) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = LENGTH1(activate(n__from(x0))) evaluates to t =LENGTH1(activate(n__from(s(x0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [x0 / s(x0)]
  • Semiunifier: [ ]



Rewriting sequence

LENGTH1(activate(n__from(x0))) → LENGTH1(n__from(x0))
with rule activate(X) → X at position [0] and matcher [X / n__from(x0)]

LENGTH1(n__from(x0)) → LENGTH(cons(x0, n__from(s(x0))))
with rule LENGTH1(n__from(x0')) → LENGTH(cons(x0', n__from(s(x0')))) at position [] and matcher [x0' / x0]

LENGTH(cons(x0, n__from(s(x0)))) → LENGTH(n__cons(x0, n__from(s(x0))))
with rule cons(X1, X2) → n__cons(X1, X2) at position [0] and matcher [X1 / x0, X2 / n__from(s(x0))]

LENGTH(n__cons(x0, n__from(s(x0)))) → LENGTH1(activate(n__from(s(x0))))
with rule LENGTH(n__cons(X, Y)) → LENGTH1(activate(Y))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.



(18) NO