Termination w.r.t. Q proof of /tmp/tmpdPVRIL/Ex4_7_37_Bor03

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 PisEmptyProof (⇔)

    ↳13 TRUE

  ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 QDPOrderProof (⇔)

    ↳18 QDP

    ↳19 QDPOrderProof (⇔)

    ↳20 QDP

    ↳21 PisEmptyProof (⇔)

    ↳22 TRUE

  ↳23 QDP

    ↳24 QDPOrderProof (⇔)

    ↳25 QDP

    ↳26 QDPOrderProof (⇔)

    ↳27 QDP

    ↳28 QDPOrderProof (⇔)

    ↳29 QDP

    ↳30 PisEmptyProof (⇔)

    ↳31 TRUE

  ↳32 QDP

    ↳33 QDPOrderProof (⇔)

    ↳34 QDP

    ↳35 QDPOrderProof (⇔)

    ↳36 QDP

    ↳37 QDPOrderProof (⇔)

    ↳38 QDP

    ↳39 PisEmptyProof (⇔)

    ↳40 TRUE

  ↳41 QDP

    ↳42 QDPOrderProof (⇔)

    ↳43 QDP

    ↳44 QDPOrderProof (⇔)

    ↳45 QDP

    ↳46 PisEmptyProof (⇔)

    ↳47 TRUE

  ↳48 QDP

    ↳49 QDPOrderProof (⇔)

    ↳50 QDP

    ↳51 QDPOrderProof (⇔)

    ↳52 QDP

    ↳53 PisEmptyProof (⇔)

    ↳54 TRUE

  ↳55 QDP

    ↳56 QDPOrderProof (⇔)

    ↳57 QDP

    ↳58 QDPOrderProof (⇔)

    ↳59 QDP

    ↳60 PisEmptyProof (⇔)

    ↳61 TRUE

  ↳62 QDP

    ↳63 QDPOrderProof (⇔)

    ↳64 QDP

    ↳65 QDPOrderProof (⇔)

    ↳66 QDP

    ↳67 QDPOrderProof (⇔)

    ↳68 QDP

    ↳69 PisEmptyProof (⇔)

    ↳70 TRUE

  ↳71 QDP

    ↳72 QDPOrderProof (⇔)

    ↳73 QDP

    ↳74 QDPOrderProof (⇔)

    ↳75 QDP

    ↳76 QDPOrderProof (⇔)

    ↳77 QDP

    ↳78 QDPOrderProof (⇔)

    ↳79 QDP

    ↳80 PisEmptyProof (⇔)

    ↳81 TRUE

  ↳82 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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

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 10 SCCs with 31 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(ok(X1), ok(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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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, 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) = ZWQUOT(x2)
mark(x1) = mark(x1)
ok(x1) = x1

Lexicographic Path Order [LPO].
Precedence:

mark₁ > ZWQUOT₁

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

ZWQUOT(mark(X1), X2) → ZWQUOT(X1, X2)
ZWQUOT(ok(X1), ok(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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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(ok(X1), ok(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) = ZWQUOT(x2)
mark(x1) = mark
ok(x1) = ok(x1)

Lexicographic Path Order [LPO].
Precedence:

ok₁ > [ZWQUOT₁, mark]

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)

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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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(mark(X1), X2) → ZWQUOT(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(11) 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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) TRUE

(14) 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(ok(X1), ok(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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(15) 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) = QUOT(x2)
mark(x1) = mark(x1)
ok(x1) = x1

Lexicographic Path Order [LPO].
Precedence:

mark₁ > QUOT₁

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

QUOT(mark(X1), X2) → QUOT(X1, X2)
QUOT(ok(X1), ok(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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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(ok(X1), ok(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) = QUOT(x2)
mark(x1) = mark
ok(x1) = ok(x1)

Lexicographic Path Order [LPO].
Precedence:

ok₁ > [QUOT₁, mark]

The following usable rules [FROCOS05] were oriented: none

(18) 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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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(mark(X1), X2) → QUOT(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(20) 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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE

(23) 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(ok(X1), ok(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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(24) 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) = MINUS(x2)
mark(x1) = mark(x1)
ok(x1) = x1

Lexicographic Path Order [LPO].
Precedence:

mark₁ > MINUS₁

The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

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

MINUS(mark(X1), X2) → MINUS(X1, X2)
MINUS(ok(X1), ok(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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MINUS(ok(X1), ok(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) = MINUS(x2)
mark(x1) = mark
ok(x1) = ok(x1)

Lexicographic Path Order [LPO].
Precedence:

ok₁ > [MINUS₁, mark]

The following usable rules [FROCOS05] were oriented: none

(27) 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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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(mark(X1), X2) → MINUS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(29) 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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(30) PisEmptyProof (EQUIVALENT transformation)

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

(31) TRUE

(32) 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(ok(X1), ok(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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(33) 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) = SEL(x2)
mark(x1) = mark(x1)
ok(x1) = x1

Lexicographic Path Order [LPO].
Precedence:

mark₁ > SEL₁

The following usable rules [FROCOS05] were oriented: none

(34) Obligation:

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

SEL(mark(X1), X2) → SEL(X1, X2)
SEL(ok(X1), ok(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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SEL(ok(X1), ok(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) = SEL(x2)
mark(x1) = mark
ok(x1) = ok(x1)

Lexicographic Path Order [LPO].
Precedence:

ok₁ > [SEL₁, mark]

The following usable rules [FROCOS05] were oriented: none

(36) 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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(37) 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: Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(38) 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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(39) PisEmptyProof (EQUIVALENT transformation)

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

(40) TRUE

(41) Obligation:

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

S(ok(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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(42) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(ok(X)) → S(X)

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

Lexicographic Path Order [LPO].
Precedence:

[S₁, ok₁]

The following usable rules [FROCOS05] were oriented: none

(43) Obligation:

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

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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(44) 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
mark(x1) = mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(45) 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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(46) PisEmptyProof (EQUIVALENT transformation)

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

(47) TRUE

(48) Obligation:

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)
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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(49) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(ok(X1), ok(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) = CONS(x2)
ok(x1) = ok(x1)
mark(x1) = mark

Lexicographic Path Order [LPO].
Precedence:

ok₁ > CONS₁
mark > CONS₁

The following usable rules [FROCOS05] were oriented: none

(50) 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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(51) 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: Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(52) 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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(53) PisEmptyProof (EQUIVALENT transformation)

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

(54) TRUE

(55) Obligation:

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

FROM(ok(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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(56) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FROM(ok(X)) → FROM(X)

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

Lexicographic Path Order [LPO].
Precedence:

[FROM₁, ok₁]

The following usable rules [FROCOS05] were oriented: none

(57) Obligation:

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

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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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].

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
mark(x1) = mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(59) 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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(60) PisEmptyProof (EQUIVALENT transformation)

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

(61) TRUE

(62) Obligation:

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

PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(from(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(s(X)) → PROPER(X)
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X2)
PROPER(minus(X1, X2)) → PROPER(X1)
PROPER(minus(X1, X2)) → PROPER(X2)
PROPER(quot(X1, X2)) → PROPER(X1)
PROPER(quot(X1, X2)) → PROPER(X2)
PROPER(zWquot(X1, X2)) → PROPER(X1)
PROPER(zWquot(X1, X2)) → PROPER(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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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.

PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X2)
PROPER(minus(X1, X2)) → PROPER(X1)
PROPER(minus(X1, X2)) → PROPER(X2)
PROPER(quot(X1, X2)) → PROPER(X1)
PROPER(quot(X1, X2)) → PROPER(X2)
PROPER(zWquot(X1, X2)) → PROPER(X1)
PROPER(zWquot(X1, X2)) → PROPER(X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1) = PROPER(x1)
cons(x1, x2) = cons(x1, x2)
from(x1) = x1
s(x1) = x1
sel(x1, x2) = sel(x1, x2)
minus(x1, x2) = minus(x1, x2)
quot(x1, x2) = quot(x1, x2)
zWquot(x1, x2) = zWquot(x1, x2)

Lexicographic Path Order [LPO].
Precedence:

cons₂ > PROPER₁
sel₂ > PROPER₁
minus₂ > PROPER₁
zWquot₂ > PROPER₁

The following usable rules [FROCOS05] were oriented: none

(64) Obligation:

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

PROPER(from(X)) → PROPER(X)
PROPER(s(X)) → PROPER(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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(65) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(from(X)) → PROPER(X)

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

Lexicographic Path Order [LPO].
Precedence:

[PROPER₁, from₁]

The following usable rules [FROCOS05] were oriented: none

(66) Obligation:

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

PROPER(s(X)) → PROPER(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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(67) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(s(X)) → PROPER(X)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(68) 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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(69) PisEmptyProof (EQUIVALENT transformation)

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

(70) TRUE

(71) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(minus(X1, X2)) → ACTIVE(X1)
ACTIVE(minus(X1, X2)) → ACTIVE(X2)
ACTIVE(quot(X1, X2)) → ACTIVE(X1)
ACTIVE(quot(X1, X2)) → ACTIVE(X2)
ACTIVE(zWquot(X1, X2)) → ACTIVE(X1)
ACTIVE(zWquot(X1, X2)) → ACTIVE(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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(72) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(minus(X1, X2)) → ACTIVE(X1)
ACTIVE(minus(X1, X2)) → ACTIVE(X2)
ACTIVE(quot(X1, X2)) → ACTIVE(X1)
ACTIVE(quot(X1, X2)) → ACTIVE(X2)
ACTIVE(zWquot(X1, X2)) → ACTIVE(X1)
ACTIVE(zWquot(X1, X2)) → ACTIVE(X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = ACTIVE(x1)
cons(x1, x2) = x1
from(x1) = x1
s(x1) = x1
sel(x1, x2) = sel(x1, x2)
minus(x1, x2) = minus(x1, x2)
quot(x1, x2) = quot(x1, x2)
zWquot(x1, x2) = zWquot(x1, x2)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(73) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(74) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(s(X)) → ACTIVE(X)

trivial

The following usable rules [FROCOS05] were oriented: none

(75) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(from(X)) → ACTIVE(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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(76) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(from(X)) → ACTIVE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = ACTIVE(x1)
cons(x1, x2) = x1
from(x1) = from(x1)

Lexicographic Path Order [LPO].
Precedence:

from₁ > ACTIVE₁

The following usable rules [FROCOS05] were oriented: none

(77) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)

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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(78) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(cons(X1, X2)) → ACTIVE(X1)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = ACTIVE(x1)
cons(x1, x2) = cons(x1)

Lexicographic Path Order [LPO].
Precedence:

[ACTIVE₁, cons₁]

The following usable rules [FROCOS05] were oriented: none

(79) 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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(80) PisEmptyProof (EQUIVALENT transformation)

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

(81) TRUE

(82) Obligation:

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

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(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)))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
active(minus(X1, X2)) → minus(active(X1), X2)
active(minus(X1, X2)) → minus(X1, active(X2))
active(quot(X1, X2)) → quot(active(X1), X2)
active(quot(X1, X2)) → quot(X1, active(X2))
active(zWquot(X1, X2)) → zWquot(active(X1), X2)
active(zWquot(X1, X2)) → zWquot(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
minus(mark(X1), X2) → mark(minus(X1, X2))
minus(X1, mark(X2)) → mark(minus(X1, X2))
quot(mark(X1), X2) → mark(quot(X1, X2))
quot(X1, mark(X2)) → mark(quot(X1, X2))
zWquot(mark(X1), X2) → mark(zWquot(X1, X2))
zWquot(X1, mark(X2)) → mark(zWquot(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(minus(X1, X2)) → minus(proper(X1), proper(X2))
proper(quot(X1, X2)) → quot(proper(X1), proper(X2))
proper(zWquot(X1, X2)) → zWquot(proper(X1), proper(X2))
proper(nil) → ok(nil)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
minus(ok(X1), ok(X2)) → ok(minus(X1, X2))
quot(ok(X1), ok(X2)) → ok(quot(X1, X2))
zWquot(ok(X1), ok(X2)) → ok(zWquot(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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