YES Termination w.r.t. Q proof of Transformed_CSR_04_Ex5_DLMMU04_iGM.ari

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

0 QTRS
↳1 QTRSRRRProof (⇔, 80 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 7 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 13 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 AND
    ↳9 QDP
        ↳10 UsableRulesProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QDPSizeChangeProof (⇔, 0 ms)
        ↳18 YES
    ↳19 QDP
        ↳20 UsableRulesProof (⇔, 0 ms)
        ↳21 QDP
        ↳22 QDPSizeChangeProof (⇔, 0 ms)
        ↳23 YES
    ↳24 QDP
        ↳25 UsableRulesProof (⇔, 0 ms)
        ↳26 QDP
        ↳27 QDPSizeChangeProof (⇔, 0 ms)
        ↳28 YES
    ↳29 QDP
        ↳30 UsableRulesProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 QDPSizeChangeProof (⇔, 0 ms)
        ↳33 YES
    ↳34 QDP
        ↳35 UsableRulesProof (⇔, 0 ms)
        ↳36 QDP
        ↳37 QDPSizeChangeProof (⇔, 0 ms)
        ↳38 YES
    ↳39 QDP
        ↳40 UsableRulesProof (⇔, 0 ms)
        ↳41 QDP
        ↳42 QDPSizeChangeProof (⇔, 0 ms)
        ↳43 YES
    ↳44 QDP
        ↳45 UsableRulesProof (⇔, 0 ms)
        ↳46 QDP
        ↳47 QDPSizeChangeProof (⇔, 0 ms)
        ↳48 YES
    ↳49 QDP
        ↳50 MRRProof (⇔, 30 ms)
        ↳51 QDP
        ↳52 MRRProof (⇔, 21 ms)
        ↳53 QDP
        ↳54 QDPOrderProof (⇔, 173 ms)
        ↳55 QDP
        ↳56 DependencyGraphProof (⇔, 0 ms)
        ↳57 QDP
        ↳58 QDPOrderProof (⇔, 8 ms)
        ↳59 QDP
        ↳60 QDPOrderProof (⇔, 0 ms)
        ↳61 QDP
        ↳62 QDPOrderProof (⇔, 44 ms)
        ↳63 QDP
        ↳64 QDPOrderProof (⇔, 47 ms)
        ↳65 QDP
        ↳66 DependencyGraphProof (⇔, 0 ms)
        ↳67 QDP
        ↳68 UsableRulesProof (⇔, 0 ms)
        ↳69 QDP
        ↳70 QDPSizeChangeProof (⇔, 0 ms)
        ↳71 YES

(0) Obligation:

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

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(active(x1)) = x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 1   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2 + 2·x1   
POL(s(x1)) = x1   
POL(tail(x1)) = 1 + x1   
POL(take(x1, x2)) = 1 + x1 + x2   
POL(zip(x1, x2)) = 1 + x1 + x2   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

active(zip(nil, XS)) → mark(nil)
active(zip(X, nil)) → mark(nil)
active(tail(cons(X, XS))) → mark(XS)
active(repItems(nil)) → mark(nil)


(2) Obligation:

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

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(active(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2 + 2·x1   
POL(s(x1)) = 2·x1   
POL(tail(x1)) = 2 + x1   
POL(take(x1, x2)) = 1 + x1 + x2   
POL(zip(x1, x2)) = 1 + x1 + x2   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

active(take(0, XS)) → mark(nil)


(4) Obligation:

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

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

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(pairNs) → MARK(cons(0, incr(oddNs)))
ACTIVE(pairNs) → CONS(0, incr(oddNs))
ACTIVE(pairNs) → INCR(oddNs)
ACTIVE(oddNs) → MARK(incr(pairNs))
ACTIVE(oddNs) → INCR(pairNs)
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
ACTIVE(incr(cons(X, XS))) → CONS(s(X), incr(XS))
ACTIVE(incr(cons(X, XS))) → S(X)
ACTIVE(incr(cons(X, XS))) → INCR(XS)
ACTIVE(take(s(N), cons(X, XS))) → MARK(cons(X, take(N, XS)))
ACTIVE(take(s(N), cons(X, XS))) → CONS(X, take(N, XS))
ACTIVE(take(s(N), cons(X, XS))) → TAKE(N, XS)
ACTIVE(zip(cons(X, XS), cons(Y, YS))) → MARK(cons(pair(X, Y), zip(XS, YS)))
ACTIVE(zip(cons(X, XS), cons(Y, YS))) → CONS(pair(X, Y), zip(XS, YS))
ACTIVE(zip(cons(X, XS), cons(Y, YS))) → PAIR(X, Y)
ACTIVE(zip(cons(X, XS), cons(Y, YS))) → ZIP(XS, YS)
ACTIVE(repItems(cons(X, XS))) → MARK(cons(X, cons(X, repItems(XS))))
ACTIVE(repItems(cons(X, XS))) → CONS(X, cons(X, repItems(XS)))
ACTIVE(repItems(cons(X, XS))) → CONS(X, repItems(XS))
ACTIVE(repItems(cons(X, XS))) → REPITEMS(XS)
MARK(pairNs) → ACTIVE(pairNs)
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(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(oddNs) → ACTIVE(oddNs)
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)
MARK(nil) → ACTIVE(nil)
MARK(zip(X1, X2)) → ACTIVE(zip(mark(X1), mark(X2)))
MARK(zip(X1, X2)) → ZIP(mark(X1), mark(X2))
MARK(zip(X1, X2)) → MARK(X1)
MARK(zip(X1, X2)) → MARK(X2)
MARK(pair(X1, X2)) → ACTIVE(pair(mark(X1), mark(X2)))
MARK(pair(X1, X2)) → PAIR(mark(X1), mark(X2))
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(tail(X)) → TAIL(mark(X))
MARK(tail(X)) → MARK(X)
MARK(repItems(X)) → ACTIVE(repItems(mark(X)))
MARK(repItems(X)) → REPITEMS(mark(X))
MARK(repItems(X)) → MARK(X)
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)
INCR(mark(X)) → INCR(X)
INCR(active(X)) → INCR(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)
ZIP(mark(X1), X2) → ZIP(X1, X2)
ZIP(X1, mark(X2)) → ZIP(X1, X2)
ZIP(active(X1), X2) → ZIP(X1, X2)
ZIP(X1, active(X2)) → ZIP(X1, X2)
PAIR(mark(X1), X2) → PAIR(X1, X2)
PAIR(X1, mark(X2)) → PAIR(X1, X2)
PAIR(active(X1), X2) → PAIR(X1, X2)
PAIR(X1, active(X2)) → PAIR(X1, X2)
TAIL(mark(X)) → TAIL(X)
TAIL(active(X)) → TAIL(X)
REPITEMS(mark(X)) → REPITEMS(X)
REPITEMS(active(X)) → REPITEMS(X)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

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 9 SCCs with 24 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

REPITEMS(active(X)) → REPITEMS(X)
REPITEMS(mark(X)) → REPITEMS(X)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

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:

REPITEMS(active(X)) → REPITEMS(X)
REPITEMS(mark(X)) → REPITEMS(X)

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:

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

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

(13) YES

(14) Obligation:

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

TAIL(active(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(X)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

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:

TAIL(active(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(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:

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

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

(18) YES

(19) Obligation:

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

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

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

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:

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

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:

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

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

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

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

(23) YES

(24) Obligation:

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

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

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

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:

ZIP(X1, mark(X2)) → ZIP(X1, X2)
ZIP(mark(X1), X2) → ZIP(X1, X2)
ZIP(active(X1), X2) → ZIP(X1, X2)
ZIP(X1, active(X2)) → ZIP(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:

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

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

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

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

(28) YES

(29) 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(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

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:

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.

(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:

  • 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

(33) YES

(34) 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(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

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

(35) 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.

(36) 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.

(37) 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

(38) YES

(39) Obligation:

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

INCR(active(X)) → INCR(X)
INCR(mark(X)) → INCR(X)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

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

(40) 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.

(41) Obligation:

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

INCR(active(X)) → INCR(X)
INCR(mark(X)) → INCR(X)

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

(42) 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:

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

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

(43) YES

(44) 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(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

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

(45) 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.

(46) 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.

(47) 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

(48) YES

(49) Obligation:

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

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(pairNs) → ACTIVE(pairNs)
ACTIVE(pairNs) → MARK(cons(0, incr(oddNs)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(take(s(N), cons(X, XS))) → MARK(cons(X, take(N, XS)))
MARK(incr(X)) → MARK(X)
MARK(oddNs) → ACTIVE(oddNs)
ACTIVE(oddNs) → MARK(incr(pairNs))
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(zip(cons(X, XS), cons(Y, YS))) → MARK(cons(pair(X, Y), zip(XS, YS)))
MARK(s(X)) → MARK(X)
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
ACTIVE(repItems(cons(X, XS))) → MARK(cons(X, cons(X, repItems(XS))))
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)
MARK(zip(X1, X2)) → ACTIVE(zip(mark(X1), mark(X2)))
MARK(zip(X1, X2)) → MARK(X1)
MARK(zip(X1, X2)) → MARK(X2)
MARK(pair(X1, X2)) → ACTIVE(pair(mark(X1), mark(X2)))
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(tail(X)) → MARK(X)
MARK(repItems(X)) → ACTIVE(repItems(mark(X)))
MARK(repItems(X)) → MARK(X)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

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

(50) 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)
MARK(zip(X1, X2)) → MARK(X1)
MARK(zip(X1, X2)) → MARK(X2)
MARK(repItems(X)) → MARK(X)


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = 1 + x1   
POL(MARK(x1)) = 1 + x1   
POL(active(x1)) = x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = 2·x1   
POL(mark(x1)) = x1   
POL(nil) = 1   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 1 + 2·x1   
POL(s(x1)) = x1   
POL(tail(x1)) = 2·x1   
POL(take(x1, x2)) = 1 + x1 + x2   
POL(zip(x1, x2)) = 2 + x1 + 2·x2   

(51) Obligation:

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

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(pairNs) → ACTIVE(pairNs)
ACTIVE(pairNs) → MARK(cons(0, incr(oddNs)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(take(s(N), cons(X, XS))) → MARK(cons(X, take(N, XS)))
MARK(incr(X)) → MARK(X)
MARK(oddNs) → ACTIVE(oddNs)
ACTIVE(oddNs) → MARK(incr(pairNs))
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(zip(cons(X, XS), cons(Y, YS))) → MARK(cons(pair(X, Y), zip(XS, YS)))
MARK(s(X)) → MARK(X)
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
ACTIVE(repItems(cons(X, XS))) → MARK(cons(X, cons(X, repItems(XS))))
MARK(zip(X1, X2)) → ACTIVE(zip(mark(X1), mark(X2)))
MARK(pair(X1, X2)) → ACTIVE(pair(mark(X1), mark(X2)))
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(tail(X)) → MARK(X)
MARK(repItems(X)) → ACTIVE(repItems(mark(X)))

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

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

(52) 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(tail(X)) → MARK(X)


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVE(x1)) = 2 + x1   
POL(MARK(x1)) = 2 + x1   
POL(active(x1)) = x1   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = 2·x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(oddNs) = 0   
POL(pair(x1, x2)) = 2·x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 1 + 2·x1   
POL(s(x1)) = x1   
POL(tail(x1)) = 1 + x1   
POL(take(x1, x2)) = 2 + 2·x1 + x2   
POL(zip(x1, x2)) = 2 + 2·x1 + 2·x2   

(53) Obligation:

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

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(pairNs) → ACTIVE(pairNs)
ACTIVE(pairNs) → MARK(cons(0, incr(oddNs)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(take(s(N), cons(X, XS))) → MARK(cons(X, take(N, XS)))
MARK(incr(X)) → MARK(X)
MARK(oddNs) → ACTIVE(oddNs)
ACTIVE(oddNs) → MARK(incr(pairNs))
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(zip(cons(X, XS), cons(Y, YS))) → MARK(cons(pair(X, Y), zip(XS, YS)))
MARK(s(X)) → MARK(X)
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
ACTIVE(repItems(cons(X, XS))) → MARK(cons(X, cons(X, repItems(XS))))
MARK(zip(X1, X2)) → ACTIVE(zip(mark(X1), mark(X2)))
MARK(pair(X1, X2)) → ACTIVE(pair(mark(X1), mark(X2)))
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(repItems(X)) → ACTIVE(repItems(mark(X)))

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

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

(54) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


ACTIVE(pairNs) → MARK(cons(0, incr(oddNs)))
ACTIVE(oddNs) → MARK(incr(pairNs))
ACTIVE(zip(cons(X, XS), cons(Y, YS))) → MARK(cons(pair(X, Y), zip(XS, YS)))
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  x1
cons(x1, x2)  =  x1
ACTIVE(x1)  =  x1
mark(x1)  =  x1
incr(x1)  =  x1
s(x1)  =  x1
pairNs  =  pairNs
0  =  0
take(x1, x2)  =  x2
oddNs  =  oddNs
zip(x1, x2)  =  zip(x1, x2)
pair(x1, x2)  =  pair(x1, x2)
repItems(x1)  =  x1
tail(x1)  =  tail
active(x1)  =  x1
nil  =  nil

Knuth-Bendix order [KBO] with precedence:
oddNs > pairNs > 0

and weight map:

oddNs=1
tail=2
zip_2=2
0=1
pair_2=1
pairNs=1
nil=2

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
mark(incr(X)) → active(incr(mark(X)))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(pairNs) → active(pairNs)
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
mark(0) → active(0)
mark(nil) → active(nil)
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)
s(active(X)) → s(X)
s(mark(X)) → s(X)
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
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)
pair(X1, mark(X2)) → pair(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
repItems(active(X)) → repItems(X)
repItems(mark(X)) → repItems(X)
tail(active(X)) → tail(X)
tail(mark(X)) → tail(X)

(55) Obligation:

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

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(pairNs) → ACTIVE(pairNs)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(take(s(N), cons(X, XS))) → MARK(cons(X, take(N, XS)))
MARK(incr(X)) → MARK(X)
MARK(oddNs) → ACTIVE(oddNs)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
ACTIVE(repItems(cons(X, XS))) → MARK(cons(X, cons(X, repItems(XS))))
MARK(zip(X1, X2)) → ACTIVE(zip(mark(X1), mark(X2)))
MARK(pair(X1, X2)) → ACTIVE(pair(mark(X1), mark(X2)))
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(repItems(X)) → ACTIVE(repItems(mark(X)))

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

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

(56) DependencyGraphProof (EQUIVALENT transformation)

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

(57) Obligation:

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

ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(take(s(N), cons(X, XS))) → MARK(cons(X, take(N, XS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(repItems(cons(X, XS))) → MARK(cons(X, cons(X, repItems(XS))))
MARK(incr(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(zip(X1, X2)) → ACTIVE(zip(mark(X1), mark(X2)))
MARK(pair(X1, X2)) → ACTIVE(pair(mark(X1), mark(X2)))
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(repItems(X)) → ACTIVE(repItems(mark(X)))

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

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

(58) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


ACTIVE(repItems(cons(X, XS))) → MARK(cons(X, cons(X, repItems(XS))))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
incr(x1)  =  x1
cons(x1, x2)  =  x1
MARK(x1)  =  x1
s(x1)  =  x1
mark(x1)  =  x1
take(x1, x2)  =  x2
repItems(x1)  =  repItems(x1)
zip(x1, x2)  =  zip
pair(x1, x2)  =  pair
tail(x1)  =  tail
active(x1)  =  x1
pairNs  =  pairNs
0  =  0
oddNs  =  oddNs
nil  =  nil

Knuth-Bendix order [KBO] with precedence:
oddNs > pairNs > 0

and weight map:

oddNs=1
tail=2
zip=3
0=1
pair=2
pairNs=1
repItems_1=1
nil=2

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
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)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
mark(incr(X)) → active(incr(mark(X)))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(pairNs) → active(pairNs)
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
mark(0) → active(0)
mark(nil) → active(nil)
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)
repItems(active(X)) → repItems(X)
repItems(mark(X)) → repItems(X)
zip(X1, mark(X2)) → zip(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(active(X)) → tail(X)
tail(mark(X)) → tail(X)

(59) Obligation:

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

ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(take(s(N), cons(X, XS))) → MARK(cons(X, take(N, XS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(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(zip(X1, X2)) → ACTIVE(zip(mark(X1), mark(X2)))
MARK(pair(X1, X2)) → ACTIVE(pair(mark(X1), mark(X2)))
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(repItems(X)) → ACTIVE(repItems(mark(X)))

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

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

(60) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


ACTIVE(take(s(N), cons(X, XS))) → MARK(cons(X, take(N, XS)))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
incr(x1)  =  x1
cons(x1, x2)  =  x1
MARK(x1)  =  x1
s(x1)  =  x1
mark(x1)  =  x1
take(x1, x2)  =  take(x2)
zip(x1, x2)  =  zip
pair(x1, x2)  =  pair
tail(x1)  =  tail
repItems(x1)  =  x1
active(x1)  =  x1
pairNs  =  pairNs
0  =  0
oddNs  =  oddNs
nil  =  nil

Knuth-Bendix order [KBO] with precedence:
oddNs > pairNs > 0

and weight map:

oddNs=1
tail=2
zip=3
take_1=1
0=1
pair=2
pairNs=1
nil=2

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
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)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
mark(incr(X)) → active(incr(mark(X)))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(pairNs) → active(pairNs)
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
mark(0) → active(0)
mark(nil) → active(nil)
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)
zip(X1, mark(X2)) → zip(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(active(X)) → tail(X)
tail(mark(X)) → tail(X)
repItems(active(X)) → repItems(X)
repItems(mark(X)) → repItems(X)

(61) Obligation:

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

ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → MARK(X1)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(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(zip(X1, X2)) → ACTIVE(zip(mark(X1), mark(X2)))
MARK(pair(X1, X2)) → ACTIVE(pair(mark(X1), mark(X2)))
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(repItems(X)) → ACTIVE(repItems(mark(X)))

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

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

(62) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
MARK(zip(X1, X2)) → ACTIVE(zip(mark(X1), mark(X2)))
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(repItems(X)) → ACTIVE(repItems(mark(X)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( ACTIVE(x1) ) = x1 + 1

POL( MARK(x1) ) = 2

POL( cons(x1, x2) ) = 0

POL( s(x1) ) = max{0, -2}

POL( active(x1) ) = 2

POL( mark(x1) ) = 2x1 + 1

POL( incr(x1) ) = 1

POL( pair(x1, x2) ) = 1

POL( repItems(x1) ) = 0

POL( tail(x1) ) = 0

POL( take(x1, x2) ) = max{0, -2}

POL( zip(x1, x2) ) = 0

POL( pairNs ) = 0

POL( 0 ) = 0

POL( oddNs ) = 1

POL( nil ) = 0


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
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)
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)
zip(X1, mark(X2)) → zip(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(active(X)) → tail(X)
tail(mark(X)) → tail(X)
repItems(active(X)) → repItems(X)
repItems(mark(X)) → repItems(X)

(63) Obligation:

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

ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(pair(X1, X2)) → ACTIVE(pair(mark(X1), mark(X2)))

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

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

(64) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( MARK(x1) ) = 2x1

POL( cons(x1, x2) ) = 2x1

POL( ACTIVE(x1) ) = max{0, 2x1 - 1}

POL( active(x1) ) = x1

POL( incr(x1) ) = x1 + 1

POL( pair(x1, x2) ) = max{0, -2}

POL( mark(x1) ) = x1

POL( s(x1) ) = x1

POL( pairNs ) = 0

POL( 0 ) = 0

POL( oddNs ) = 2

POL( take(x1, x2) ) = x2

POL( zip(x1, x2) ) = 2x1 + 2

POL( repItems(x1) ) = x1 + 2

POL( tail(x1) ) = 2x1 + 2

POL( nil ) = 1


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
incr(active(X)) → incr(X)
incr(mark(X)) → incr(X)
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)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
mark(incr(X)) → active(incr(mark(X)))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
mark(pairNs) → active(pairNs)
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
mark(0) → active(0)
mark(nil) → active(nil)
pair(X1, mark(X2)) → pair(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
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)
zip(X1, mark(X2)) → zip(X1, X2)
zip(mark(X1), X2) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
repItems(active(X)) → repItems(X)
repItems(mark(X)) → repItems(X)
tail(active(X)) → tail(X)
tail(mark(X)) → tail(X)

(65) Obligation:

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

MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(pair(X1, X2)) → ACTIVE(pair(mark(X1), mark(X2)))

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

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

(66) DependencyGraphProof (EQUIVALENT transformation)

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

(67) Obligation:

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

MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

active(pairNs) → mark(cons(0, incr(oddNs)))
active(oddNs) → mark(incr(pairNs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(zip(cons(X, XS), cons(Y, YS))) → mark(cons(pair(X, Y), zip(XS, YS)))
active(repItems(cons(X, XS))) → mark(cons(X, cons(X, repItems(XS))))
mark(pairNs) → active(pairNs)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(oddNs) → active(oddNs)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(zip(X1, X2)) → active(zip(mark(X1), mark(X2)))
mark(pair(X1, X2)) → active(pair(mark(X1), mark(X2)))
mark(tail(X)) → active(tail(mark(X)))
mark(repItems(X)) → active(repItems(mark(X)))
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)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(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)
zip(mark(X1), X2) → zip(X1, X2)
zip(X1, mark(X2)) → zip(X1, X2)
zip(active(X1), X2) → zip(X1, X2)
zip(X1, active(X2)) → zip(X1, X2)
pair(mark(X1), X2) → pair(X1, X2)
pair(X1, mark(X2)) → pair(X1, X2)
pair(active(X1), X2) → pair(X1, X2)
pair(X1, active(X2)) → pair(X1, X2)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)
repItems(mark(X)) → repItems(X)
repItems(active(X)) → repItems(X)

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

(68) 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.

(69) Obligation:

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

MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

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

(70) 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:

  • MARK(s(X)) → MARK(X)
    The graph contains the following edges 1 > 1

  • MARK(cons(X1, X2)) → MARK(X1)
    The graph contains the following edges 1 > 1

(71) YES