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

(0) Obligation:

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

from(X) → cons(X, from(s(X)))
length(nil) → 0
length(cons(X, Y)) → s(length1(Y))
length1(X) → length(X)

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

from(X) → cons(X, from(s(X)))
length(nil) → 0
length(cons(X, Y)) → s(length1(Y))
length1(X) → length(X)

The set Q consists of the following terms:

from(x0)
length(nil)
length(cons(x0, x1))
length1(x0)

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

FROM(X) → FROM(s(X))
LENGTH(cons(X, Y)) → LENGTH1(Y)
LENGTH1(X) → LENGTH(X)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
length(nil) → 0
length(cons(X, Y)) → s(length1(Y))
length1(X) → length(X)

The set Q consists of the following terms:

from(x0)
length(nil)
length(cons(x0, x1))
length1(x0)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) Obligation:

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

LENGTH1(X) → LENGTH(X)
LENGTH(cons(X, Y)) → LENGTH1(Y)

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
length(nil) → 0
length(cons(X, Y)) → s(length1(Y))
length1(X) → length(X)

The set Q consists of the following terms:

from(x0)
length(nil)
length(cons(x0, x1))
length1(x0)

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

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

LENGTH1(X) → LENGTH(X)
LENGTH(cons(X, Y)) → LENGTH1(Y)

R is empty.
The set Q consists of the following terms:

from(x0)
length(nil)
length(cons(x0, x1))
length1(x0)

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

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

from(x0)
length(nil)
length(cons(x0, x1))
length1(x0)

(11) Obligation:

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

LENGTH1(X) → LENGTH(X)
LENGTH(cons(X, Y)) → LENGTH1(Y)

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

(12) 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:

• LENGTH(cons(X, Y)) → LENGTH1(Y)
  The graph contains the following edges 1 > 1

• LENGTH1(X) → LENGTH(X)
  The graph contains the following edges 1 >= 1

(13) YES

(14) Obligation:

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

FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

from(X) → cons(X, from(s(X)))
length(nil) → 0
length(cons(X, Y)) → s(length1(Y))
length1(X) → length(X)

The set Q consists of the following terms:

from(x0)
length(nil)
length(cons(x0, x1))
length1(x0)

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

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

FROM(X) → FROM(s(X))

R is empty.
The set Q consists of the following terms:

from(x0)
length(nil)
length(cons(x0, x1))
length1(x0)

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

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

from(x0)
length(nil)
length(cons(x0, x1))
length1(x0)

(18) Obligation:

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

FROM(X) → FROM(s(X))

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

(19) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule FROM(X) → FROM(s(X)) we obtained the following new rules [LPAR04]:

FROM(s(z0)) → FROM(s(s(z0)))

(20) Obligation:

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

FROM(s(z0)) → FROM(s(s(z0)))

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

(21) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule FROM(s(z0)) → FROM(s(s(z0))) we obtained the following new rules [LPAR04]:

FROM(s(s(z0))) → FROM(s(s(s(z0))))

(22) Obligation:

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

FROM(s(s(z0))) → FROM(s(s(s(z0))))

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

(23) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = FROM(s(s(z0))) evaluates to t =FROM(s(s(s(z0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

• Matcher: [z0 / s(z0)]
• Semiunifier: [ ]

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Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from FROM(s(s(z0))) to FROM(s(s(s(z0)))).

(24) NO