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

0 QTRS
↳1 QTRSRRRProof (⇔)
↳2 QTRS
↳3 Overlay + Local Confluence (⇔)
↳4 QTRS
↳5 DependencyPairsProof (⇔)
↳6 QDP
↳7 UsableRulesProof (⇔)
↳8 QDP
↳9 QReductionProof (⇔)
↳10 QDP
↳11 NonTerminationProof (⇔)
↳12 NO

(0) Obligation:

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

from(X) → cons(X)
length → 0
length → s(length1)
length1 → length

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(cons(x1)) = x1   
POL(from(x1)) = 1 + x1   
POL(length) = 1   
POL(length1) = 1   
POL(s(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

from(X) → cons(X)
length → 0


(2) Obligation:

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

length → s(length1)
length1 → length

Q is empty.

(3) Overlay + Local Confluence (EQUIVALENT transformation)

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

(4) Obligation:

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

length → s(length1)
length1 → length

The set Q consists of the following terms:

length
length1

(5) DependencyPairsProof (EQUIVALENT transformation)

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

(6) Obligation:

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

LENGTH → LENGTH1
LENGTH1 → LENGTH

The TRS R consists of the following rules:

length → s(length1)
length1 → length

The set Q consists of the following terms:

length
length1

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

(7) 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.

(8) Obligation:

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

LENGTH → LENGTH1
LENGTH1 → LENGTH

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

length
length1

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

(9) 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].

length
length1

(10) Obligation:

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

LENGTH → LENGTH1
LENGTH1 → LENGTH

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

(11) 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 evaluates to t =LENGTH1

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



Rewriting sequence

LENGTH1 → LENGTH
with rule LENGTH1 → LENGTH at position [] and matcher [ ]

LENGTH → LENGTH1
with rule LENGTH → LENGTH1

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.



(12) NO