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

0 QTRS
↳1 QTRSRRRProof (⇔)
↳2 QTRS
↳3 QTRSRRRProof (⇔)
↳4 QTRS
↳5 DependencyPairsProof (⇔)
↳6 QDP
↳7 DependencyGraphProof (⇔)
↳8 AND
  ↳9 QDP
    ↳10 UsableRulesProof (⇔)
    ↳11 QDP
    ↳12 QDPSizeChangeProof (⇔)
    ↳13 TRUE
  ↳14 QDP
    ↳15 UsableRulesProof (⇔)
    ↳16 QDP
    ↳17 QDPSizeChangeProof (⇔)
    ↳18 TRUE
  ↳19 QDP
    ↳20 UsableRulesProof (⇔)
    ↳21 QDP
    ↳22 QDPSizeChangeProof (⇔)
    ↳23 TRUE
  ↳24 QDP
    ↳25 UsableRulesProof (⇔)
    ↳26 QDP
    ↳27 QDPSizeChangeProof (⇔)
    ↳28 TRUE
  ↳29 QDP
    ↳30 UsableRulesProof (⇔)
    ↳31 QDP
    ↳32 QDPSizeChangeProof (⇔)
    ↳33 TRUE
  ↳34 QDP
    ↳35 MRRProof (⇔)
    ↳36 QDP
    ↳37 MRRProof (⇔)
    ↳38 QDP
    ↳39 QDPOrderProof (⇔)
    ↳40 QDP
    ↳41 QDPOrderProof (⇔)
    ↳42 QDP
    ↳43 QDPOrderProof (⇔)
    ↳44 QDP
    ↳45 QDPOrderProof (⇔)
    ↳46 QDP
    ↳47 DependencyGraphProof (⇔)
    ↳48 QDP
    ↳49 QDPOrderProof (⇔)
    ↳50 QDP
    ↳51 Narrowing (⇔)
    ↳52 QDP
    ↳53 Narrowing (⇔)
    ↳54 QDP
    ↳55 DependencyGraphProof (⇔)
    ↳56 QDP
    ↳57 Narrowing (⇔)
    ↳58 QDP
    ↳59 DependencyGraphProof (⇔)
    ↳60 QDP
    ↳61 Narrowing (⇔)
    ↳62 QDP
    ↳63 DependencyGraphProof (⇔)
    ↳64 QDP
    ↳65 QDPOrderProof (⇔)
    ↳66 QDP
    ↳67 QDPOrderProof (⇔)
    ↳68 QDP
    ↳69 QDPOrderProof (⇔)
    ↳70 QDP
    ↳71 QDPOrderProof (⇔)
    ↳72 QDP
    ↳73 QDPOrderProof (⇔)
    ↳74 QDP
    ↳75 NonTerminationProof (⇔)
    ↳76 FALSE

(0) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(active(x1)) = x1   
POL(and(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = x1 + x2   
POL(length(x1)) = 1 + x1   
POL(mark(x1)) = x1   
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:

active(length(nil)) → mark(0)
active(take(0, IL)) → mark(nil)


(2) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(active(x1)) = x1   
POL(and(x1, x2)) = 1 + x1 + x2   
POL(cons(x1, x2)) = x1 + x2   
POL(length(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(take(x1, x2)) = 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:

active(and(tt, X)) → mark(X)


(4) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

Q is empty.

(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:

ACTIVE(zeros) → MARK(cons(0, zeros))
ACTIVE(zeros) → CONS(0, zeros)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
ACTIVE(length(cons(N, L))) → S(length(L))
ACTIVE(length(cons(N, L))) → LENGTH(L)
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
ACTIVE(take(s(M), cons(N, IL))) → CONS(N, take(M, IL))
ACTIVE(take(s(M), cons(N, IL))) → TAKE(M, IL)
MARK(zeros) → ACTIVE(zeros)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(0) → ACTIVE(0)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
MARK(and(X1, X2)) → AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(tt) → ACTIVE(tt)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(length(X)) → LENGTH(mark(X))
MARK(length(X)) → MARK(X)
MARK(nil) → ACTIVE(nil)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
MARK(take(X1, X2)) → TAKE(mark(X1), mark(X2))
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
AND(X1, mark(X2)) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)
LENGTH(mark(X)) → LENGTH(X)
LENGTH(active(X)) → LENGTH(X)
S(mark(X)) → S(X)
S(active(X)) → S(X)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(active(X1), X2) → TAKE(X1, X2)
TAKE(X1, active(X2)) → TAKE(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 13 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(active(X1), X2) → TAKE(X1, X2)
TAKE(X1, active(X2)) → TAKE(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(10) 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.

(11) Obligation:

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

TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(active(X1), X2) → TAKE(X1, X2)
TAKE(X1, active(X2)) → TAKE(X1, X2)

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:

  • TAKE(X1, mark(X2)) → TAKE(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

  • TAKE(mark(X1), X2) → TAKE(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • TAKE(active(X1), X2) → TAKE(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • TAKE(X1, active(X2)) → TAKE(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

(13) TRUE

(14) Obligation:

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(15) 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.

(16) Obligation:

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

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

(17) 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:

  • S(active(X)) → S(X)
    The graph contains the following edges 1 > 1

  • S(mark(X)) → S(X)
    The graph contains the following edges 1 > 1

(18) TRUE

(19) Obligation:

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

LENGTH(active(X)) → LENGTH(X)
LENGTH(mark(X)) → LENGTH(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(20) 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.

(21) Obligation:

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

LENGTH(active(X)) → LENGTH(X)
LENGTH(mark(X)) → LENGTH(X)

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

(22) 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(active(X)) → LENGTH(X)
    The graph contains the following edges 1 > 1

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

(23) TRUE

(24) Obligation:

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

AND(X1, mark(X2)) → AND(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(25) 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.

(26) Obligation:

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

AND(X1, mark(X2)) → AND(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)

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

(27) 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:

  • AND(X1, mark(X2)) → AND(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

  • AND(mark(X1), X2) → AND(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • AND(active(X1), X2) → AND(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • AND(X1, active(X2)) → AND(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

(28) TRUE

(29) Obligation:

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

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(30) 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.

(31) Obligation:

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

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

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

(32) 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:

  • CONS(X1, mark(X2)) → CONS(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

  • CONS(mark(X1), X2) → CONS(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • CONS(active(X1), X2) → CONS(X1, X2)
    The graph contains the following edges 1 > 1, 2 >= 2

  • CONS(X1, active(X2)) → CONS(X1, X2)
    The graph contains the following edges 1 >= 1, 2 > 2

(33) TRUE

(34) Obligation:

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

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
MARK(and(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(length(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(35) 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 dependency pairs:

MARK(and(X1, X2)) → MARK(X1)
MARK(length(X)) → MARK(X)


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(and(x1, x2)) = 1 + x1 + x2   
POL(cons(x1, x2)) = x1 + x2   
POL(length(x1)) = 1 + x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(take(x1, x2)) = x1 + x2   
POL(tt) = 0   
POL(zeros) = 0   

(36) Obligation:

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

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(37) 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 dependency pairs:

MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = x1   
POL(MARK(x1)) = x1   
POL(active(x1)) = x1   
POL(and(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = x1 + x2   
POL(length(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(take(x1, x2)) = 1 + x1 + x2   
POL(tt) = 0   
POL(zeros) = 0   

(38) Obligation:

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

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(39) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
MARK(s(X)) → ACTIVE(s(mark(X)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(MARK(x1)) =
/1\
\1/
 +
/00\
\00/
·x1

POL(cons(x1, x2)) =
/1\
\0/
 +
/00\
\00/
·x1 +
/01\
\00/
·x2

POL(ACTIVE(x1)) =
/0\
\1/
 +
/01\
\00/
·x1

POL(mark(x1)) =
/0\
\0/
 +
/00\
\01/
·x1

POL(length(x1)) =
/0\
\1/
 +
/00\
\00/
·x1

POL(s(x1)) =
/1\
\0/
 +
/00\
\00/
·x1

POL(zeros) =
/0\
\1/

POL(0) =
/1\
\1/

POL(and(x1, x2)) =
/1\
\0/
 +
/00\
\00/
·x1 +
/01\
\00/
·x2

POL(take(x1, x2)) =
/1\
\1/
 +
/00\
\00/
·x1 +
/00\
\00/
·x2

POL(active(x1)) =
/0\
\1/
 +
/11\
\01/
·x1

POL(nil) =
/1\
\0/

POL(tt) =
/0\
\1/

The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
take(X1, mark(X2)) → take(X1, X2)
take(mark(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)

(40) Obligation:

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

ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(s(X)) → MARK(X)
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(41) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(ACTIVE(x1)) =
/0\
\0/
 +
/01\
\10/
·x1

POL(length(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(cons(x1, x2)) =
/0\
\0/
 +
/11\
\00/
·x1 +
/00\
\00/
·x2

POL(MARK(x1)) =
/0\
\0/
 +
/10\
\10/
·x1

POL(s(x1)) =
/0\
\0/
 +
/11\
\00/
·x1

POL(zeros) =
/0\
\0/

POL(0) =
/0\
\0/

POL(take(x1, x2)) =
/1\
\0/
 +
/11\
\00/
·x1 +
/11\
\11/
·x2

POL(mark(x1)) =
/0\
\0/
 +
/01\
\10/
·x1

POL(active(x1)) =
/0\
\0/
 +
/01\
\10/
·x1

POL(and(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/11\
\00/
·x2

POL(nil) =
/1\
\0/

POL(tt) =
/1\
\0/

The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
take(X1, mark(X2)) → take(X1, X2)
take(mark(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
mark(nil) → active(nil)
active(zeros) → mark(cons(0, zeros))
mark(length(X)) → active(length(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(s(X)) → active(s(mark(X)))
active(length(cons(N, L))) → mark(s(length(L)))
mark(0) → active(0)
mark(tt) → active(tt)

(42) Obligation:

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

ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(43) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(ACTIVE(x1)) =
/0\
\0/
 +
/01\
\00/
·x1

POL(length(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(cons(x1, x2)) =
/0\
\1/
 +
/00\
\00/
·x1 +
/00\
\00/
·x2

POL(MARK(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(s(x1)) =
/0\
\1/
 +
/10\
\11/
·x1

POL(zeros) =
/0\
\0/

POL(0) =
/0\
\1/

POL(take(x1, x2)) =
/0\
\1/
 +
/00\
\00/
·x1 +
/00\
\00/
·x2

POL(mark(x1)) =
/1\
\1/
 +
/00\
\10/
·x1

POL(active(x1)) =
/1\
\0/
 +
/11\
\11/
·x1

POL(and(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/01\
\10/
·x2

POL(nil) =
/0\
\0/

POL(tt) =
/0\
\1/

The following usable rules [FROCOS05] were oriented:

length(active(X)) → length(X)
length(mark(X)) → length(X)

(44) Obligation:

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

ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(zeros) → ACTIVE(zeros)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(ACTIVE(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(length(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(cons(x1, x2)) =
/0\
\0/
 +
/01\
\01/
·x1 +
/00\
\00/
·x2

POL(MARK(x1)) =
/0\
\0/
 +
/01\
\00/
·x1

POL(s(x1)) =
/0\
\0/
 +
/00\
\11/
·x1

POL(zeros) =
/0\
\1/

POL(0) =
/0\
\0/

POL(mark(x1)) =
/0\
\0/
 +
/11\
\00/
·x1

POL(active(x1)) =
/0\
\0/
 +
/11\
\00/
·x1

POL(nil) =
/0\
\0/

POL(take(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/01\
\01/
·x2

POL(and(x1, x2)) =
/0\
\0/
 +
/00\
\01/
·x1 +
/01\
\10/
·x2

POL(tt) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)

(46) Obligation:

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

ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(47) DependencyGraphProof (EQUIVALENT transformation)

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

(48) Obligation:

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

MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(49) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(cons(X1, X2)) → MARK(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(MARK(x1)) =
/0\
\0/
 +
/11\
\10/
·x1

POL(cons(x1, x2)) =
/0\
\1/
 +
/10\
\11/
·x1 +
/00\
\10/
·x2

POL(length(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(ACTIVE(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(mark(x1)) =
/0\
\0/
 +
/10\
\01/
·x1

POL(s(x1)) =
/0\
\0/
 +
/11\
\00/
·x1

POL(active(x1)) =
/0\
\1/
 +
/01\
\10/
·x1

POL(nil) =
/0\
\0/

POL(zeros) =
/1\
\0/

POL(0) =
/0\
\1/

POL(take(x1, x2)) =
/1\
\0/
 +
/01\
\00/
·x1 +
/01\
\00/
·x2

POL(and(x1, x2)) =
/0\
\0/
 +
/01\
\10/
·x1 +
/00\
\00/
·x2

POL(tt) =
/0\
\1/

The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)

(50) Obligation:

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

MARK(length(X)) → ACTIVE(length(mark(X)))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(51) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule MARK(length(X)) → ACTIVE(length(mark(X))) at position [0] we obtained the following new rules [LPAR04]:

MARK(length(x0)) → ACTIVE(length(x0))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(0)) → ACTIVE(length(active(0)))
MARK(length(and(x0, x1))) → ACTIVE(length(active(and(mark(x0), x1))))
MARK(length(tt)) → ACTIVE(length(active(tt)))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(nil)) → ACTIVE(length(active(nil)))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))

(52) Obligation:

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

ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(length(x0)) → ACTIVE(length(x0))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(0)) → ACTIVE(length(active(0)))
MARK(length(and(x0, x1))) → ACTIVE(length(active(and(mark(x0), x1))))
MARK(length(tt)) → ACTIVE(length(active(tt)))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(nil)) → ACTIVE(length(active(nil)))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(53) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule MARK(length(0)) → ACTIVE(length(active(0))) at position [0] we obtained the following new rules [LPAR04]:

MARK(length(0)) → ACTIVE(length(0))

(54) Obligation:

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

ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(length(x0)) → ACTIVE(length(x0))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(and(x0, x1))) → ACTIVE(length(active(and(mark(x0), x1))))
MARK(length(tt)) → ACTIVE(length(active(tt)))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(nil)) → ACTIVE(length(active(nil)))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))
MARK(length(0)) → ACTIVE(length(0))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(55) DependencyGraphProof (EQUIVALENT transformation)

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

(56) Obligation:

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

MARK(s(X)) → MARK(X)
MARK(length(x0)) → ACTIVE(length(x0))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(and(x0, x1))) → ACTIVE(length(active(and(mark(x0), x1))))
MARK(length(tt)) → ACTIVE(length(active(tt)))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(nil)) → ACTIVE(length(active(nil)))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(57) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule MARK(length(tt)) → ACTIVE(length(active(tt))) at position [0] we obtained the following new rules [LPAR04]:

MARK(length(tt)) → ACTIVE(length(tt))

(58) Obligation:

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

MARK(s(X)) → MARK(X)
MARK(length(x0)) → ACTIVE(length(x0))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(and(x0, x1))) → ACTIVE(length(active(and(mark(x0), x1))))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(nil)) → ACTIVE(length(active(nil)))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))
MARK(length(tt)) → ACTIVE(length(tt))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(59) DependencyGraphProof (EQUIVALENT transformation)

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

(60) Obligation:

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

MARK(s(X)) → MARK(X)
MARK(length(x0)) → ACTIVE(length(x0))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(and(x0, x1))) → ACTIVE(length(active(and(mark(x0), x1))))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(nil)) → ACTIVE(length(active(nil)))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(61) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule MARK(length(nil)) → ACTIVE(length(active(nil))) at position [0] we obtained the following new rules [LPAR04]:

MARK(length(nil)) → ACTIVE(length(nil))

(62) Obligation:

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

MARK(s(X)) → MARK(X)
MARK(length(x0)) → ACTIVE(length(x0))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(and(x0, x1))) → ACTIVE(length(active(and(mark(x0), x1))))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))
MARK(length(nil)) → ACTIVE(length(nil))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(63) DependencyGraphProof (EQUIVALENT transformation)

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

(64) Obligation:

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

MARK(s(X)) → MARK(X)
MARK(length(x0)) → ACTIVE(length(x0))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(and(x0, x1))) → ACTIVE(length(active(and(mark(x0), x1))))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(65) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(length(and(x0, x1))) → ACTIVE(length(active(and(mark(x0), x1))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(MARK(x1)) =
/0\
\0/
 +
/11\
\00/
·x1

POL(s(x1)) =
/0\
\0/
 +
/00\
\11/
·x1

POL(length(x1)) =
/0\
\0/
 +
/01\
\10/
·x1

POL(ACTIVE(x1)) =
/0\
\0/
 +
/01\
\00/
·x1

POL(cons(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/11\
\00/
·x2

POL(zeros) =
/0\
\0/

POL(active(x1)) =
/0\
\0/
 +
/10\
\01/
·x1

POL(mark(x1)) =
/0\
\0/
 +
/10\
\01/
·x1

POL(and(x1, x2)) =
/0\
\1/
 +
/00\
\00/
·x1 +
/10\
\10/
·x2

POL(take(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/00\
\00/
·x2

POL(nil) =
/0\
\1/

POL(0) =
/0\
\0/

POL(tt) =
/0\
\1/

The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
take(X1, mark(X2)) → take(X1, X2)
take(mark(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
mark(nil) → active(nil)
active(zeros) → mark(cons(0, zeros))
mark(length(X)) → active(length(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(s(X)) → active(s(mark(X)))
active(length(cons(N, L))) → mark(s(length(L)))
mark(0) → active(0)
mark(tt) → active(tt)

(66) Obligation:

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

MARK(s(X)) → MARK(X)
MARK(length(x0)) → ACTIVE(length(x0))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(67) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(MARK(x1)) =
/0\
\0/
 +
/10\
\00/
·x1

POL(s(x1)) =
/0\
\0/
 +
/10\
\01/
·x1

POL(length(x1)) =
/1\
\0/
 +
/11\
\01/
·x1

POL(ACTIVE(x1)) =
/1\
\0/
 +
/01\
\00/
·x1

POL(cons(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/01\
\11/
·x2

POL(zeros) =
/0\
\0/

POL(active(x1)) =
/0\
\0/
 +
/10\
\01/
·x1

POL(mark(x1)) =
/0\
\0/
 +
/11\
\01/
·x1

POL(take(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/00\
\00/
·x2

POL(and(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/10\
\00/
·x2

POL(nil) =
/1\
\1/

POL(0) =
/0\
\0/

POL(tt) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
take(X1, mark(X2)) → take(X1, X2)
take(mark(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
mark(nil) → active(nil)
active(zeros) → mark(cons(0, zeros))
mark(length(X)) → active(length(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(s(X)) → active(s(mark(X)))
active(length(cons(N, L))) → mark(s(length(L)))
mark(0) → active(0)
mark(tt) → active(tt)

(68) Obligation:

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

MARK(s(X)) → MARK(X)
MARK(length(x0)) → ACTIVE(length(x0))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(69) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(MARK(x1)) =
/1\
\0/
 +
/10\
\10/
·x1

POL(s(x1)) =
/0\
\1/
 +
/10\
\00/
·x1

POL(length(x1)) =
/0\
\1/
 +
/11\
\10/
·x1

POL(ACTIVE(x1)) =
/0\
\0/
 +
/01\
\10/
·x1

POL(cons(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/11\
\01/
·x2

POL(zeros) =
/0\
\0/

POL(active(x1)) =
/0\
\0/
 +
/10\
\01/
·x1

POL(mark(x1)) =
/0\
\0/
 +
/10\
\01/
·x1

POL(take(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/00\
\00/
·x2

POL(and(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/00\
\00/
·x2

POL(nil) =
/0\
\1/

POL(0) =
/0\
\1/

POL(tt) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
take(X1, mark(X2)) → take(X1, X2)
take(mark(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
mark(nil) → active(nil)
active(zeros) → mark(cons(0, zeros))
mark(length(X)) → active(length(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(s(X)) → active(s(mark(X)))
active(length(cons(N, L))) → mark(s(length(L)))
mark(0) → active(0)
mark(tt) → active(tt)

(70) Obligation:

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

MARK(s(X)) → MARK(X)
MARK(length(x0)) → ACTIVE(length(x0))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(71) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(MARK(x1)) = -I + 1A·x1

POL(s(x1)) = -I + 0A·x1

POL(length(x1)) = 0A + 0A·x1

POL(ACTIVE(x1)) = 1A + 0A·x1

POL(cons(x1, x2)) = 1A + 0A·x1 + 1A·x2

POL(zeros) = 0A

POL(active(x1)) = 1A + 0A·x1

POL(mark(x1)) = 1A + 0A·x1

POL(take(x1, x2)) = -I + -I·x1 + 0A·x2

POL(nil) = 0A

POL(0) = 0A

POL(and(x1, x2)) = -I + 0A·x1 + 0A·x2

POL(tt) = 0A

The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
mark(nil) → active(nil)
length(active(X)) → length(X)
length(mark(X)) → length(X)
active(zeros) → mark(cons(0, zeros))
mark(length(X)) → active(length(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(s(X)) → active(s(mark(X)))
active(length(cons(N, L))) → mark(s(length(L)))
mark(0) → active(0)
take(X1, mark(X2)) → take(X1, X2)
take(mark(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
mark(tt) → active(tt)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)

(72) Obligation:

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

MARK(s(X)) → MARK(X)
MARK(length(x0)) → ACTIVE(length(x0))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(73) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(length(x0)) → ACTIVE(length(x0))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(MARK(x1)) = -I + 4A·x1

POL(s(x1)) = -I + 0A·x1

POL(length(x1)) = 2A + 0A·x1

POL(ACTIVE(x1)) = -I + 1A·x1

POL(cons(x1, x2)) = 5A + -I·x1 + 3A·x2

POL(zeros) = 0A

POL(active(x1)) = 5A + 0A·x1

POL(take(x1, x2)) = 0A + -I·x1 + -I·x2

POL(mark(x1)) = 5A + 0A·x1

POL(nil) = 0A

POL(0) = 0A

POL(and(x1, x2)) = 0A + -I·x1 + 0A·x2

POL(tt) = 0A

The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
mark(nil) → active(nil)
length(active(X)) → length(X)
length(mark(X)) → length(X)
active(zeros) → mark(cons(0, zeros))
mark(length(X)) → active(length(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(s(X)) → active(s(mark(X)))
active(length(cons(N, L))) → mark(s(length(L)))
mark(0) → active(0)
take(X1, mark(X2)) → take(X1, X2)
take(mark(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
mark(tt) → active(tt)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)

(74) Obligation:

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

MARK(s(X)) → MARK(X)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)

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

(75) 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 = MARK(length(zeros)) evaluates to t =MARK(length(zeros))

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



Rewriting sequence

MARK(length(zeros)) → ACTIVE(length(active(zeros)))
with rule MARK(length(zeros)) → ACTIVE(length(active(zeros))) at position [] and matcher [ ]

ACTIVE(length(active(zeros))) → ACTIVE(length(mark(cons(0, zeros))))
with rule active(zeros) → mark(cons(0, zeros)) at position [0,0] and matcher [ ]

ACTIVE(length(mark(cons(0, zeros)))) → ACTIVE(length(cons(0, zeros)))
with rule length(mark(X)) → length(X) at position [0] and matcher [X / cons(0, zeros)]

ACTIVE(length(cons(0, zeros))) → MARK(s(length(zeros)))
with rule ACTIVE(length(cons(N, L))) → MARK(s(length(L))) at position [] and matcher [N / 0, L / zeros]

MARK(s(length(zeros))) → MARK(length(zeros))
with rule MARK(s(X)) → MARK(X)

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.



(76) FALSE