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

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

(0) Obligation:

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

from(X) → cons(X, n__from(n__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)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(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(n__s(X)))
LENGTH(n__cons(X, Y)) → S(length1(activate(Y)))
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(activate(X))
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__nil) → NIL
ACTIVATE(n__cons(X1, X2)) → CONS(activate(X1), X2)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__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)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(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 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__s(X)) → ACTIVATE(X)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__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)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

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

(6) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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

ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)

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

(8) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • ACTIVATE(n__s(X)) → ACTIVATE(X)
    The graph contains the following edges 1 > 1

  • ACTIVATE(n__from(X)) → ACTIVATE(X)
    The graph contains the following edges 1 > 1

  • ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
    The graph contains the following edges 1 > 1

(9) YES

(10) 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(n__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)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(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(X) → LENGTH(activate(X)) at position [0] we obtained the following new rules [LPAR04]:

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

(12) 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(activate(x0)))
LENGTH1(n__s(x0)) → LENGTH(s(activate(x0)))
LENGTH1(n__nil) → LENGTH(nil)
LENGTH1(n__cons(x0, x1)) → LENGTH(cons(activate(x0), x1))
LENGTH1(x0) → LENGTH(x0)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__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)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

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

(13) 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)

(14) 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(activate(x0)))
LENGTH1(n__s(x0)) → LENGTH(s(activate(x0)))
LENGTH1(n__cons(x0, x1)) → LENGTH(cons(activate(x0), x1))
LENGTH1(x0) → LENGTH(x0)
LENGTH1(n__nil) → LENGTH(n__nil)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__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)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

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

(15) DependencyGraphProof (EQUIVALENT transformation)

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

(16) Obligation:

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

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

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__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)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


LENGTH1(n__s(x0)) → LENGTH(s(activate(x0)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( LENGTH1(x1) ) = 2x1 + 1


POL( LENGTH(x1) ) = x1 + 1


POL( cons(x1, x2) ) = 2x2


POL( from(x1) ) = max{0, -2}


POL( s(x1) ) = x1 + 2


POL( activate(x1) ) = x1


POL( n__from(x1) ) = max{0, -2}


POL( n__s(x1) ) = x1 + 2


POL( n__nil ) = 2


POL( nil ) = 2


POL( n__cons(x1, x2) ) = 2x2



The following usable rules [FROCOS05] were oriented:

activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X
from(X) → cons(X, n__from(n__s(X)))
from(X) → n__from(X)
s(X) → n__s(X)
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil

(18) Obligation:

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

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

The TRS R consists of the following rules:

from(X) → cons(X, n__from(n__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)
s(X) → n__s(X)
nil → n__nil
cons(X1, X2) → n__cons(X1, X2)
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__nil) → nil
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(X) → X

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

(19) 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 = LENGTH(from(X)) evaluates to t =LENGTH(from(activate(n__s(X))))

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



Rewriting sequence

LENGTH(from(X)) → LENGTH(cons(X, n__from(n__s(X))))
with rule from(X') → cons(X', n__from(n__s(X'))) at position [0] and matcher [X' / X]

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

LENGTH(n__cons(X, n__from(n__s(X)))) → LENGTH1(activate(n__from(n__s(X))))
with rule LENGTH(n__cons(X', Y)) → LENGTH1(activate(Y)) at position [] and matcher [X' / X, Y / n__from(n__s(X))]

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

LENGTH1(n__from(n__s(X))) → LENGTH(from(activate(n__s(X))))
with rule LENGTH1(n__from(x0)) → LENGTH(from(activate(x0)))

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.



(20) NO