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

0 QTRS
↳1 QTRSRRRProof (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 UsableRulesReductionPairsProof (⇔)
    ↳11 QDP
    ↳12 DependencyGraphProof (⇔)
    ↳13 QDP
    ↳14 UsableRulesProof (⇔)
    ↳15 QDP
    ↳16 QDPSizeChangeProof (⇔)
    ↳17 TRUE
  ↳18 QDP
    ↳19 UsableRulesProof (⇔)
    ↳20 QDP
    ↳21 UsableRulesReductionPairsProof (⇔)
    ↳22 QDP
    ↳23 MRRProof (⇔)
    ↳24 QDP
    ↳25 MNOCProof (⇔)
    ↳26 QDP
    ↳27 MRRProof (⇔)
    ↳28 QDP
    ↳29 UsableRulesProof (⇔)
    ↳30 QDP
    ↳31 QReductionProof (⇔)
    ↳32 QDP
    ↳33 Narrowing (⇔)
    ↳34 QDP
    ↳35 UsableRulesProof (⇔)
    ↳36 QDP
    ↳37 QReductionProof (⇔)
    ↳38 QDP
    ↳39 Rewriting (⇔)
    ↳40 QDP
    ↳41 UsableRulesProof (⇔)
    ↳42 QDP
    ↳43 QReductionProof (⇔)
    ↳44 QDP
    ↳45 Instantiation (⇔)
    ↳46 QDP
    ↳47 NonTerminationProof (⇔)
    ↳48 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)))
take(0, IL) → nil
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x1)) = x1   
POL(and(x1, x2)) = 1 + x1 + x2   
POL(cons(x1, x2)) = x1 + x2   
POL(length(x1)) = 1 + x1   
POL(n__take(x1, x2)) = 1 + x1 + x2   
POL(n__zeros) = 0   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(take(x1, x2)) = 1 + x1 + x2   
POL(tt) = 0   
POL(zeros) = 0   
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
take(0, IL) → nil


(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)))
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X

Q is empty.

(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:

LENGTH(cons(N, L)) → LENGTH(activate(L))
LENGTH(cons(N, L)) → ACTIVATE(L)
TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)
ACTIVATE(n__zeros) → ZEROS
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X

Q is empty.
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 with 2 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X

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

(8) 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.

(9) Obligation:

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

ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
take(X1, X2) → n__take(X1, X2)
zeros → cons(0, n__zeros)
zeros → n__zeros

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

(10) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)
The following rules are removed from R:

take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVATE(x1)) = 2·x1   
POL(TAKE(x1, x2)) = x1 + 2·x2   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(n__take(x1, x2)) = 2·x1 + 2·x2   
POL(n__zeros) = 0   
POL(s(x1)) = 2 + 2·x1   
POL(take(x1, x2)) = 2·x1 + 2·x2   
POL(zeros) = 0   

(11) Obligation:

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

ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X
take(X1, X2) → n__take(X1, X2)
zeros → cons(0, n__zeros)
zeros → n__zeros

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

(12) DependencyGraphProof (EQUIVALENT transformation)

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

(13) Obligation:

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

ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X
take(X1, X2) → n__take(X1, X2)
zeros → cons(0, n__zeros)
zeros → n__zeros

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

(14) 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.

(15) Obligation:

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

ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)

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

(16) 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__take(X1, X2)) → ACTIVATE(X2)
    The graph contains the following edges 1 > 1

  • ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
    The graph contains the following edges 1 > 1

(17) TRUE

(18) 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)))
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X

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

(19) 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.

(20) 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
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X
take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
take(X1, X2) → n__take(X1, X2)
zeros → cons(0, n__zeros)
zeros → n__zeros

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

(21) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

take(s(M), cons(N, IL)) → cons(N, n__take(M, activate(IL)))
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LENGTH(x1)) = x1   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = x1 + 2·x2   
POL(n__take(x1, x2)) = x1 + x2   
POL(n__zeros) = 0   
POL(s(x1)) = 2·x1   
POL(take(x1, x2)) = x1 + x2   
POL(zeros) = 0   

(22) 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
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(X) → X
take(X1, X2) → n__take(X1, X2)
zeros → cons(0, n__zeros)
zeros → n__zeros

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

(23) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

activate(X) → X
zeros → n__zeros

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LENGTH(x1)) = x1   
POL(activate(x1)) = 1 + 2·x1   
POL(cons(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(n__take(x1, x2)) = 2 + x1 + 2·x2   
POL(n__zeros) = 0   
POL(take(x1, x2)) = 2 + x1 + 2·x2   
POL(zeros) = 1   

(24) 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
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
take(X1, X2) → n__take(X1, X2)
zeros → cons(0, n__zeros)

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

(25) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(26) 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
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
take(X1, X2) → n__take(X1, X2)
zeros → cons(0, n__zeros)

The set Q consists of the following terms:

activate(n__zeros)
activate(n__take(x0, x1))
take(x0, x1)
zeros

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

(27) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

activate(n__take(X1, X2)) → take(activate(X1), activate(X2))

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LENGTH(x1)) = x1   
POL(activate(x1)) = 2·x1   
POL(cons(x1, x2)) = 2·x1 + 2·x2   
POL(n__take(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(n__zeros) = 0   
POL(take(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(zeros) = 0   

(28) 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
take(X1, X2) → n__take(X1, X2)
zeros → cons(0, n__zeros)

The set Q consists of the following terms:

activate(n__zeros)
activate(n__take(x0, x1))
take(x0, x1)
zeros

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

(29) 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.

(30) 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:

activate(n__zeros)
activate(n__take(x0, x1))
take(x0, x1)
zeros

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

(31) 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].

take(x0, x1)

(32) 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:

activate(n__zeros)
activate(n__take(x0, x1))
zeros

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

(33) 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)

(34) 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:

activate(n__zeros)
activate(n__take(x0, x1))
zeros

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

(35) 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.

(36) 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:

activate(n__zeros)
activate(n__take(x0, x1))
zeros

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

(37) 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)
activate(n__take(x0, x1))

(38) 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.

(39) 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))

(40) 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.

(41) 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.

(42) 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.

(43) 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

(44) 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.

(45) 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))

(46) 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.

(47) 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)).



(48) FALSE