Termination w.r.t. Q proof of Transformed_CSR_04_Ex14_AEGL02

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

0 QTRS

↳1 DependencyPairsProof (⇔, 6 ms)

↳2 QDP

↳3 DependencyGraphProof (⇔, 0 ms)

↳4 AND

    ↳5 QDP

        ↳6 UsableRulesProof (⇔, 0 ms)

        ↳7 QDP

        ↳8 QDPSizeChangeProof (⇔, 0 ms)

        ↳9 YES

    ↳10 QDP

        ↳11 TransformationProof (⇔, 0 ms)

        ↳12 QDP

        ↳13 TransformationProof (⇔, 0 ms)

        ↳14 QDP

        ↳15 DependencyGraphProof (⇔, 0 ms)

        ↳16 QDP

        ↳17 QDPOrderProof (⇔, 0 ms)

        ↳18 QDP

        ↳19 NonTerminationLoopProof (⇒, 0 ms)

        ↳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) TransformationProof (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__from(x0)) → LENGTH(from(activate(x0)))
LENGTH1(n__s(x0)) → LENGTH(s(activate(x0))) → LENGTH1(n__s(x0)) → LENGTH(s(activate(x0)))
LENGTH1(n__nil) → LENGTH(nil) → LENGTH1(n__nil) → LENGTH(nil)
LENGTH1(n__cons(x0, x1)) → LENGTH(cons(activate(x0), x1)) → LENGTH1(n__cons(x0, x1)) → LENGTH(cons(activate(x0), x1))
LENGTH1(x0) → LENGTH(x0) → 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) TransformationProof (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) → 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,JAR06].

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

POL( LENGTH(x₁) ) = x₁

POL( cons(x₁, x₂) ) = 2x₂

POL( from(x₁) ) = 0

POL( s(x₁) ) = 2

POL( activate(x₁) ) = x₁

POL( n__from(x₁) ) = 0

POL( n__s(x₁) ) = 2

POL( n__nil ) = 0

POL( nil ) = 0

POL( n__cons(x₁, x₂) ) = 2x₂

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] 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) NonTerminationLoopProof (COMPLETE 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.

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Complex Obligation (AND)

(5) Obligation:

(6) UsableRulesProof (EQUIVALENT transformation)

(7) Obligation:

(8) QDPSizeChangeProof (EQUIVALENT transformation)

(9) YES

(10) Obligation:

(11) TransformationProof (EQUIVALENT transformation)

(12) Obligation:

(13) TransformationProof (EQUIVALENT transformation)

(14) Obligation:

(15) DependencyGraphProof (EQUIVALENT transformation)

(16) Obligation:

(17) QDPOrderProof (EQUIVALENT transformation)

(18) Obligation:

(19) NonTerminationLoopProof (COMPLETE transformation)

(20) NO