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 DependencyGraphProof (⇔)
↳8 QDP
↳9 UsableRulesProof (⇔)
↳10 QDP
↳11 QReductionProof (⇔)
↳12 QDP
↳13 Narrowing (⇔)
↳14 QDP
↳15 UsableRulesProof (⇔)
↳16 QDP
↳17 QReductionProof (⇔)
↳18 QDP
↳19 Rewriting (⇔)
↳20 QDP
↳21 UsableRulesProof (⇔)
↳22 QDP
↳23 QReductionProof (⇔)
↳24 QDP
↳25 Instantiation (⇔)
↳26 QDP
↳27 NonTerminationProof (⇔)
↳28 FALSE

(0) Obligation:

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

zeros → cons(0, n__zeros)
and(tt, X) → activate(X)
length(nil) → 0
length(cons(N, L)) → s(length(activate(L)))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x1)) = 1 + x1   
POL(and(x1, x2)) = 2 + x1 + x2   
POL(cons(x1, x2)) = 1 + x1 + x2   
POL(length(x1)) = x1   
POL(n__zeros) = 0   
POL(nil) = 1   
POL(s(x1)) = x1   
POL(tt) = 0   
POL(zeros) = 1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

and(tt, X) → activate(X)
length(nil) → 0
zeros → n__zeros
activate(X) → X


(2) Obligation:

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

zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
activate(n__zeros) → zeros

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:

zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
activate(n__zeros) → zeros

The set Q consists of the following terms:

zeros
length(cons(x0, x1))
activate(n__zeros)

(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(cons(N, L)) → LENGTH(activate(L))
LENGTH(cons(N, L)) → ACTIVATE(L)
ACTIVATE(n__zeros) → ZEROS

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
activate(n__zeros) → zeros

The set Q consists of the following terms:

zeros
length(cons(x0, x1))
activate(n__zeros)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(8) Obligation:

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

LENGTH(cons(N, L)) → LENGTH(activate(L))

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
activate(n__zeros) → zeros

The set Q consists of the following terms:

zeros
length(cons(x0, x1))
activate(n__zeros)

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

(9) 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.

(10) Obligation:

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

LENGTH(cons(N, L)) → LENGTH(activate(L))

The TRS R consists of the following rules:

activate(n__zeros) → zeros
zeros → cons(0, n__zeros)

The set Q consists of the following terms:

zeros
length(cons(x0, x1))
activate(n__zeros)

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

(11) 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(cons(x0, x1))

(12) Obligation:

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

LENGTH(cons(N, L)) → LENGTH(activate(L))

The TRS R consists of the following rules:

activate(n__zeros) → zeros
zeros → cons(0, n__zeros)

The set Q consists of the following terms:

zeros
activate(n__zeros)

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

(13) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule LENGTH(cons(N, L)) → LENGTH(activate(L)) at position [0] we obtained the following new rules [LPAR04]:

LENGTH(cons(y0, n__zeros)) → LENGTH(zeros)

(14) Obligation:

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

LENGTH(cons(y0, n__zeros)) → LENGTH(zeros)

The TRS R consists of the following rules:

activate(n__zeros) → zeros
zeros → cons(0, n__zeros)

The set Q consists of the following terms:

zeros
activate(n__zeros)

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(cons(y0, n__zeros)) → LENGTH(zeros)

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)

The set Q consists of the following terms:

zeros
activate(n__zeros)

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].

activate(n__zeros)

(18) Obligation:

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

LENGTH(cons(y0, n__zeros)) → LENGTH(zeros)

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)

The set Q consists of the following terms:

zeros

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

(19) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LENGTH(cons(y0, n__zeros)) → LENGTH(zeros) at position [0] we obtained the following new rules [LPAR04]:

LENGTH(cons(y0, n__zeros)) → LENGTH(cons(0, n__zeros))

(20) Obligation:

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

LENGTH(cons(y0, n__zeros)) → LENGTH(cons(0, n__zeros))

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)

The set Q consists of the following terms:

zeros

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

(21) 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.

(22) Obligation:

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

LENGTH(cons(y0, n__zeros)) → LENGTH(cons(0, n__zeros))

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

zeros

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

(23) 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].

zeros

(24) Obligation:

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

LENGTH(cons(y0, n__zeros)) → LENGTH(cons(0, n__zeros))

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

(25) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule LENGTH(cons(y0, n__zeros)) → LENGTH(cons(0, n__zeros)) we obtained the following new rules [LPAR04]:

LENGTH(cons(0, n__zeros)) → LENGTH(cons(0, n__zeros))

(26) Obligation:

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

LENGTH(cons(0, n__zeros)) → LENGTH(cons(0, n__zeros))

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

(27) 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 = LENGTH(cons(0, n__zeros)) evaluates to t =LENGTH(cons(0, n__zeros))

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



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from LENGTH(cons(0, n__zeros)) to LENGTH(cons(0, n__zeros)).



(28) FALSE