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

0 QTRS
  ↳1 QTRS Reverse (⇔)
    ↳2 QTRS
  ↳3 QTRS Reverse (⇔)
    ↳4 QTRS
  ↳5 QTRSRRRProof (⇔)
    ↳6 QTRS
    ↳7 Overlay + Local Confluence (⇔)
    ↳8 QTRS
    ↳9 DependencyPairsProof (⇔)
    ↳10 QDP
      ↳11 UsableRulesProof (⇔)
        ↳12 QDP
        ↳13 QReductionProof (⇔)
        ↳14 QDP
      ↳15 UsableRulesProof (⇔)
        ↳16 QDP
        ↳17 QReductionProof (⇔)
        ↳18 QDP
        ↳19 NonTerminationProof (⇔)
        ↳20 FALSE

(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) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

from(x) → cons(x)
length'(x) → 0'(x)
length'(x) → length1'(s(x))
length1'(x) → length'(x)

Q is empty.

(3) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(4) Obligation:

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

from(x) → cons(x)
length'(x) → 0'(x)
length'(x) → length1'(s(x))
length1'(x) → length'(x)

Q is empty.

(5) 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


(6) Obligation:

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

length → s(length1)
length1 → length

Q is empty.

(7) Overlay + Local Confluence (EQUIVALENT transformation)

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

(8) 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

(9) DependencyPairsProof (EQUIVALENT transformation)

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

(10) 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.

(11) 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.

(12) 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.

(13) 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

(14) 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.

(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:

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.

(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].

length
length1

(18) 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.

(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 = LENGTH1 evaluates to t =LENGTH1

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



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.



(20) FALSE