Termination w.r.t. Q of the given QTRS could not be shown:

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

(0) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(sel(0, cons(X, XS))) → MARK(X)
ACTIVE(sel(s(N), cons(X, XS))) → MARK(sel(N, XS))
ACTIVE(sel(s(N), cons(X, XS))) → SEL(N, XS)
ACTIVE(minus(X, 0)) → MARK(0)
ACTIVE(minus(s(X), s(Y))) → MARK(minus(X, Y))
ACTIVE(minus(s(X), s(Y))) → MINUS(X, Y)
ACTIVE(quot(0, s(Y))) → MARK(0)
ACTIVE(quot(s(X), s(Y))) → MARK(s(quot(minus(X, Y), s(Y))))
ACTIVE(quot(s(X), s(Y))) → S(quot(minus(X, Y), s(Y)))
ACTIVE(quot(s(X), s(Y))) → QUOT(minus(X, Y), s(Y))
ACTIVE(quot(s(X), s(Y))) → MINUS(X, Y)
ACTIVE(zWquot(XS, nil)) → MARK(nil)
ACTIVE(zWquot(nil, XS)) → MARK(nil)
ACTIVE(zWquot(cons(X, XS), cons(Y, YS))) → MARK(cons(quot(X, Y), zWquot(XS, YS)))
ACTIVE(zWquot(cons(X, XS), cons(Y, YS))) → CONS(quot(X, Y), zWquot(XS, YS))
ACTIVE(zWquot(cons(X, XS), cons(Y, YS))) → QUOT(X, Y)
ACTIVE(zWquot(cons(X, XS), cons(Y, YS))) → ZWQUOT(XS, YS)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(from(X)) → FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2)))
MARK(sel(X1, X2)) → SEL(mark(X1), mark(X2))
MARK(sel(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → MARK(X2)
MARK(0) → ACTIVE(0)
MARK(minus(X1, X2)) → ACTIVE(minus(mark(X1), mark(X2)))
MARK(minus(X1, X2)) → MINUS(mark(X1), mark(X2))
MARK(minus(X1, X2)) → MARK(X1)
MARK(minus(X1, X2)) → MARK(X2)
MARK(quot(X1, X2)) → ACTIVE(quot(mark(X1), mark(X2)))
MARK(quot(X1, X2)) → QUOT(mark(X1), mark(X2))
MARK(quot(X1, X2)) → MARK(X1)
MARK(quot(X1, X2)) → MARK(X2)
MARK(zWquot(X1, X2)) → ACTIVE(zWquot(mark(X1), mark(X2)))
MARK(zWquot(X1, X2)) → ZWQUOT(mark(X1), mark(X2))
MARK(zWquot(X1, X2)) → MARK(X1)
MARK(zWquot(X1, X2)) → MARK(X2)
MARK(nil) → ACTIVE(nil)
FROM(mark(X)) → FROM(X)
FROM(active(X)) → FROM(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)
S(mark(X)) → S(X)
S(active(X)) → S(X)
SEL(mark(X1), X2) → SEL(X1, X2)
SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(active(X1), X2) → SEL(X1, X2)
SEL(X1, active(X2)) → SEL(X1, X2)
MINUS(mark(X1), X2) → MINUS(X1, X2)
MINUS(X1, mark(X2)) → MINUS(X1, X2)
MINUS(active(X1), X2) → MINUS(X1, X2)
MINUS(X1, active(X2)) → MINUS(X1, X2)
QUOT(mark(X1), X2) → QUOT(X1, X2)
QUOT(X1, mark(X2)) → QUOT(X1, X2)
QUOT(active(X1), X2) → QUOT(X1, X2)
QUOT(X1, active(X2)) → QUOT(X1, X2)
ZWQUOT(mark(X1), X2) → ZWQUOT(X1, X2)
ZWQUOT(X1, mark(X2)) → ZWQUOT(X1, X2)
ZWQUOT(active(X1), X2) → ZWQUOT(X1, X2)
ZWQUOT(X1, active(X2)) → ZWQUOT(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 8 SCCs with 24 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ZWQUOT(X1, active(X2)) → ZWQUOT(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ZWQUOT(x1, x2)  =  x2
mark(x1)  =  x1
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ZWQUOT(X1, mark(X2)) → ZWQUOT(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ZWQUOT(x1, x2)  =  x2
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ZWQUOT(active(X1), X2) → ZWQUOT(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ZWQUOT(x1, x2)  =  x1
mark(x1)  =  x1
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(11) Obligation:

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

ZWQUOT(mark(X1), X2) → ZWQUOT(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ZWQUOT(mark(X1), X2) → ZWQUOT(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ZWQUOT(x1, x2)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(13) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(14) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(15) TRUE

(16) Obligation:

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

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


QUOT(X1, active(X2)) → QUOT(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2)  =  x2
mark(x1)  =  x1
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

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

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


QUOT(X1, mark(X2)) → QUOT(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2)  =  x2
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(20) Obligation:

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

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


QUOT(active(X1), X2) → QUOT(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2)  =  x1
mark(x1)  =  x1
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(22) Obligation:

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

QUOT(mark(X1), X2) → QUOT(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


QUOT(mark(X1), X2) → QUOT(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(24) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(25) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(26) TRUE

(27) Obligation:

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

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(28) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MINUS(X1, active(X2)) → MINUS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  x2
mark(x1)  =  x1
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(29) Obligation:

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

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MINUS(X1, mark(X2)) → MINUS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  x2
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(31) Obligation:

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

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(32) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MINUS(active(X1), X2) → MINUS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  x1
mark(x1)  =  x1
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(33) Obligation:

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

MINUS(mark(X1), X2) → MINUS(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(34) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MINUS(mark(X1), X2) → MINUS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(35) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(36) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(37) TRUE

(38) Obligation:

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

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(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.


SEL(X1, active(X2)) → SEL(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  x2
mark(x1)  =  x1
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(40) Obligation:

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

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(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.


SEL(X1, mark(X2)) → SEL(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  x2
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(42) Obligation:

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

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(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.


SEL(active(X1), X2) → SEL(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  x1
mark(x1)  =  x1
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(44) Obligation:

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

SEL(mark(X1), X2) → SEL(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(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.


SEL(mark(X1), X2) → SEL(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(46) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(47) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(48) TRUE

(49) 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(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(50) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


S(mark(X)) → S(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1)  =  x1
active(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(51) Obligation:

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

S(active(X)) → S(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(52) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


S(active(X)) → S(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1)  =  x1
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(53) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(54) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(55) TRUE

(56) 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(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(57) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


CONS(X1, active(X2)) → CONS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2)  =  x2
mark(x1)  =  x1
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(58) 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)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(59) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


CONS(X1, mark(X2)) → CONS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2)  =  x2
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(60) Obligation:

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

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(61) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


CONS(active(X1), X2) → CONS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2)  =  x1
mark(x1)  =  x1
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(62) Obligation:

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

CONS(mark(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(63) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


CONS(mark(X1), X2) → CONS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(64) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(65) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(66) TRUE

(67) Obligation:

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

FROM(active(X)) → FROM(X)
FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(68) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FROM(mark(X)) → FROM(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FROM(x1)  =  x1
active(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(69) Obligation:

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

FROM(active(X)) → FROM(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(70) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FROM(active(X)) → FROM(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FROM(x1)  =  x1
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(71) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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

(72) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(73) TRUE

(74) Obligation:

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

MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(from(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(sel(0, cons(X, XS))) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(sel(s(N), cons(X, XS))) → MARK(sel(N, XS))
MARK(s(X)) → MARK(X)
MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2)))
ACTIVE(minus(s(X), s(Y))) → MARK(minus(X, Y))
MARK(sel(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → MARK(X2)
MARK(minus(X1, X2)) → ACTIVE(minus(mark(X1), mark(X2)))
ACTIVE(quot(s(X), s(Y))) → MARK(s(quot(minus(X, Y), s(Y))))
MARK(minus(X1, X2)) → MARK(X1)
MARK(minus(X1, X2)) → MARK(X2)
MARK(quot(X1, X2)) → ACTIVE(quot(mark(X1), mark(X2)))
ACTIVE(zWquot(cons(X, XS), cons(Y, YS))) → MARK(cons(quot(X, Y), zWquot(XS, YS)))
MARK(quot(X1, X2)) → MARK(X1)
MARK(quot(X1, X2)) → MARK(X2)
MARK(zWquot(X1, X2)) → ACTIVE(zWquot(mark(X1), mark(X2)))
MARK(zWquot(X1, X2)) → MARK(X1)
MARK(zWquot(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(minus(X, 0)) → mark(0)
active(minus(s(X), s(Y))) → mark(minus(X, Y))
active(quot(0, s(Y))) → mark(0)
active(quot(s(X), s(Y))) → mark(s(quot(minus(X, Y), s(Y))))
active(zWquot(XS, nil)) → mark(nil)
active(zWquot(nil, XS)) → mark(nil)
active(zWquot(cons(X, XS), cons(Y, YS))) → mark(cons(quot(X, Y), zWquot(XS, YS)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
mark(0) → active(0)
mark(minus(X1, X2)) → active(minus(mark(X1), mark(X2)))
mark(quot(X1, X2)) → active(quot(mark(X1), mark(X2)))
mark(zWquot(X1, X2)) → active(zWquot(mark(X1), mark(X2)))
mark(nil) → active(nil)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)
minus(mark(X1), X2) → minus(X1, X2)
minus(X1, mark(X2)) → minus(X1, X2)
minus(active(X1), X2) → minus(X1, X2)
minus(X1, active(X2)) → minus(X1, X2)
quot(mark(X1), X2) → quot(X1, X2)
quot(X1, mark(X2)) → quot(X1, X2)
quot(active(X1), X2) → quot(X1, X2)
quot(X1, active(X2)) → quot(X1, X2)
zWquot(mark(X1), X2) → zWquot(X1, X2)
zWquot(X1, mark(X2)) → zWquot(X1, X2)
zWquot(active(X1), X2) → zWquot(X1, X2)
zWquot(X1, active(X2)) → zWquot(X1, X2)

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