Termination w.r.t. Q proof of /tmp/tmpbjQymj/Ex1_Luc04b

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 PisEmptyProof (⇔)

    ↳9 TRUE

  ↳10 QDP

    ↳11 QDPOrderProof (⇔)

    ↳12 QDP

    ↳13 PisEmptyProof (⇔)

    ↳14 TRUE

  ↳15 QDP

    ↳16 QDPOrderProof (⇔)

    ↳17 QDP

    ↳18 PisEmptyProof (⇔)

    ↳19 TRUE

  ↳20 QDP

    ↳21 QDPOrderProof (⇔)

    ↳22 QDP

    ↳23 PisEmptyProof (⇔)

    ↳24 TRUE

  ↳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

(0) Obligation:

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

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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(nats) → CONS(0, incr(nats))
ACTIVE(nats) → INCR(nats)
ACTIVE(pairs) → CONS(0, incr(odds))
ACTIVE(pairs) → INCR(odds)
ACTIVE(odds) → INCR(pairs)
ACTIVE(incr(cons(X, XS))) → CONS(s(X), incr(XS))
ACTIVE(incr(cons(X, XS))) → S(X)
ACTIVE(incr(cons(X, XS))) → INCR(XS)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(incr(X)) → INCR(active(X))
ACTIVE(incr(X)) → ACTIVE(X)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(head(X)) → HEAD(active(X))
ACTIVE(head(X)) → ACTIVE(X)
ACTIVE(tail(X)) → TAIL(active(X))
ACTIVE(tail(X)) → ACTIVE(X)
CONS(mark(X1), X2) → CONS(X1, X2)
INCR(mark(X)) → INCR(X)
S(mark(X)) → S(X)
HEAD(mark(X)) → HEAD(X)
TAIL(mark(X)) → TAIL(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(incr(X)) → INCR(proper(X))
PROPER(incr(X)) → PROPER(X)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(head(X)) → HEAD(proper(X))
PROPER(head(X)) → PROPER(X)
PROPER(tail(X)) → TAIL(proper(X))
PROPER(tail(X)) → PROPER(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
INCR(ok(X)) → INCR(X)
S(ok(X)) → S(X)
HEAD(ok(X)) → HEAD(X)
TAIL(ok(X)) → TAIL(X)
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(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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 8 SCCs with 20 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

TAIL(ok(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(X)

The TRS R consists of the following rules:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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.

TAIL(ok(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAIL(x1) = TAIL(x1)
ok(x1) = ok(x1)
mark(x1) = mark(x1)
active(x1) = active(x1)
nats = nats
cons(x1, x2) = cons(x1, x2)
0 = 0
incr(x1) = x1
pairs = pairs
odds = odds
s(x1) = x1
head(x1) = head(x1)
tail(x1) = x1
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

proper₁ > nats
proper₁ > cons₂ > ok₁
proper₁ > cons₂ > mark₁
proper₁ > 0
proper₁ > odds > ok₁
proper₁ > odds > mark₁
proper₁ > odds > pairs
proper₁ > head₁ > ok₁
proper₁ > head₁ > mark₁
top > active₁ > nats
top > active₁ > cons₂ > ok₁
top > active₁ > cons₂ > mark₁
top > active₁ > 0
top > active₁ > odds > ok₁
top > active₁ > odds > mark₁
top > active₁ > odds > pairs
top > active₁ > head₁ > ok₁
top > active₁ > head₁ > mark₁

Status:

head₁: [1]
pairs: []
ok₁: [1]
mark₁: [1]
0: []
active₁: [1]
cons₂: [2,1]
odds: []
nats: []
proper₁: [1]
TAIL₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(7) Obligation:

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

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

HEAD(ok(X)) → HEAD(X)
HEAD(mark(X)) → HEAD(X)

The TRS R consists of the following rules:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

HEAD(ok(X)) → HEAD(X)
HEAD(mark(X)) → HEAD(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
HEAD(x1) = HEAD(x1)
ok(x1) = ok(x1)
mark(x1) = mark(x1)
active(x1) = active(x1)
nats = nats
cons(x1, x2) = cons(x1, x2)
0 = 0
incr(x1) = x1
pairs = pairs
odds = odds
s(x1) = x1
head(x1) = head(x1)
tail(x1) = x1
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

proper₁ > nats
proper₁ > cons₂ > ok₁
proper₁ > cons₂ > mark₁
proper₁ > 0
proper₁ > odds > ok₁
proper₁ > odds > mark₁
proper₁ > odds > pairs
proper₁ > head₁ > ok₁
proper₁ > head₁ > mark₁
top > active₁ > nats
top > active₁ > cons₂ > ok₁
top > active₁ > cons₂ > mark₁
top > active₁ > 0
top > active₁ > odds > ok₁
top > active₁ > odds > mark₁
top > active₁ > odds > pairs
top > active₁ > head₁ > ok₁
top > active₁ > head₁ > mark₁

Status:

head₁: [1]
pairs: []
ok₁: [1]
mark₁: [1]
0: []
HEAD₁: [1]
active₁: [1]
cons₂: [2,1]
odds: []
nats: []
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(12) Obligation:

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

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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.

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) 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(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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.

(16) 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)
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) = S(x1)
ok(x1) = ok(x1)
mark(x1) = mark(x1)
active(x1) = active(x1)
nats = nats
cons(x1, x2) = cons(x1, x2)
0 = 0
incr(x1) = x1
pairs = pairs
odds = odds
s(x1) = x1
head(x1) = head(x1)
tail(x1) = x1
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

proper₁ > nats
proper₁ > cons₂ > ok₁
proper₁ > cons₂ > mark₁
proper₁ > 0
proper₁ > odds > ok₁
proper₁ > odds > mark₁
proper₁ > odds > pairs
proper₁ > head₁ > ok₁
proper₁ > head₁ > mark₁
top > active₁ > nats
top > active₁ > cons₂ > ok₁
top > active₁ > cons₂ > mark₁
top > active₁ > 0
top > active₁ > odds > ok₁
top > active₁ > odds > mark₁
top > active₁ > odds > pairs
top > active₁ > head₁ > ok₁
top > active₁ > head₁ > mark₁

Status:

head₁: [1]
pairs: []
ok₁: [1]
mark₁: [1]
0: []
active₁: [1]
cons₂: [2,1]
odds: []
nats: []
proper₁: [1]
top: []
S₁: [1]

The following usable rules [FROCOS05] were oriented:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(17) Obligation:

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

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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.

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) Obligation:

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

INCR(ok(X)) → INCR(X)
INCR(mark(X)) → INCR(X)

The TRS R consists of the following rules:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

INCR(ok(X)) → INCR(X)
INCR(mark(X)) → INCR(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
INCR(x1) = INCR(x1)
ok(x1) = ok(x1)
mark(x1) = mark(x1)
active(x1) = active(x1)
nats = nats
cons(x1, x2) = cons(x1, x2)
0 = 0
incr(x1) = x1
pairs = pairs
odds = odds
s(x1) = x1
head(x1) = head(x1)
tail(x1) = x1
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

proper₁ > nats
proper₁ > cons₂ > ok₁
proper₁ > cons₂ > mark₁
proper₁ > 0
proper₁ > odds > ok₁
proper₁ > odds > mark₁
proper₁ > odds > pairs
proper₁ > head₁ > ok₁
proper₁ > head₁ > mark₁
top > active₁ > nats
top > active₁ > cons₂ > ok₁
top > active₁ > cons₂ > mark₁
top > active₁ > 0
top > active₁ > odds > ok₁
top > active₁ > odds > mark₁
top > active₁ > odds > pairs
top > active₁ > head₁ > ok₁
top > active₁ > head₁ > mark₁

Status:

head₁: [1]
pairs: []
ok₁: [1]
mark₁: [1]
0: []
active₁: [1]
cons₂: [2,1]
odds: []
nats: []
proper₁: [1]
top: []
INCR₁: [1]

The following usable rules [FROCOS05] were oriented:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(22) Obligation:

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

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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.

(23) PisEmptyProof (EQUIVALENT transformation)

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

(24) TRUE

(25) 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(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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.

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) = x1
active(x1) = active(x1)
nats = nats
cons(x1, x2) = cons(x1, x2)
0 = 0
incr(x1) = incr(x1)
pairs = pairs
odds = odds
s(x1) = x1
head(x1) = x1
tail(x1) = x1
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

CONS₁ > top
active₁ > nats > top
active₁ > 0 > ok₁ > top
active₁ > pairs > cons₂ > ok₁ > top
active₁ > pairs > incr₁ > ok₁ > top
active₁ > pairs > odds > ok₁ > top
proper₁ > nats > top
proper₁ > 0 > ok₁ > top
proper₁ > pairs > cons₂ > ok₁ > top
proper₁ > pairs > incr₁ > ok₁ > top
proper₁ > pairs > odds > ok₁ > top

Status:

active₁: [1]
cons₂: [1,2]
odds: []
incr₁: [1]
CONS₁: [1]
pairs: []
ok₁: [1]
nats: []
proper₁: [1]
top: []
0: []

The following usable rules [FROCOS05] were oriented:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(27) 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(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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.

CONS(mark(X1), X2) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = x1
mark(x1) = mark(x1)
active(x1) = active(x1)
nats = nats
cons(x1, x2) = cons(x1, x2)
0 = 0
incr(x1) = x1
pairs = pairs
odds = odds
s(x1) = x1
head(x1) = x1
tail(x1) = x1
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

active₁ > pairs > cons₂ > mark₁ > top
active₁ > pairs > cons₂ > ok > top
active₁ > pairs > 0
active₁ > pairs > odds
nats > cons₂ > mark₁ > top
nats > cons₂ > ok > top
nats > 0
proper₁ > pairs > cons₂ > mark₁ > top
proper₁ > pairs > cons₂ > ok > top
proper₁ > pairs > 0
proper₁ > pairs > odds

Status:

active₁: [1]
cons₂: [2,1]
odds: []
pairs: []
mark₁: [1]
ok: []
nats: []
proper₁: [1]
top: []
0: []

The following usable rules [FROCOS05] were oriented:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(29) Obligation:

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

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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:

PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(incr(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(head(X)) → PROPER(X)
PROPER(tail(X)) → PROPER(X)

The TRS R consists of the following rules:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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.

PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(incr(X)) → PROPER(X)
PROPER(tail(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)
cons(x1, x2) = cons(x1, x2)
incr(x1) = incr(x1)
s(x1) = x1
head(x1) = x1
tail(x1) = tail(x1)
active(x1) = active(x1)
nats = nats
mark(x1) = mark
0 = 0
pairs = pairs
odds = odds
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

PROPER₁ > incr₁
0 > ok > incr₁
proper₁ > tail₁ > mark > cons₂ > incr₁
proper₁ > tail₁ > ok > incr₁
proper₁ > nats > mark > cons₂ > incr₁
proper₁ > nats > ok > incr₁
proper₁ > pairs > mark > cons₂ > incr₁
proper₁ > pairs > ok > incr₁
proper₁ > odds > mark > cons₂ > incr₁
proper₁ > odds > ok > incr₁
top > active₁ > tail₁ > mark > cons₂ > incr₁
top > active₁ > tail₁ > ok > incr₁

Status:

PROPER₁: [1]
tail₁: [1]
pairs: []
0: []
odds: []
active₁: [1]
cons₂: [2,1]
incr₁: [1]
mark: []
ok: []
proper₁: [1]
nats: []
top: []

The following usable rules [FROCOS05] were oriented:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(34) Obligation:

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

PROPER(s(X)) → PROPER(X)
PROPER(head(X)) → PROPER(X)

The TRS R consists of the following rules:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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.

PROPER(head(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)
s(x1) = x1
head(x1) = head(x1)
active(x1) = x1
nats = nats
mark(x1) = mark
cons(x1, x2) = cons
0 = 0
incr(x1) = x1
pairs = pairs
odds = odds
tail(x1) = x1
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

nats > mark > proper₁ > head₁ > PROPER₁
nats > mark > top > PROPER₁
nats > ok > head₁ > PROPER₁
nats > ok > top > PROPER₁
0 > ok > head₁ > PROPER₁
0 > ok > top > PROPER₁
odds > pairs > cons > mark > proper₁ > head₁ > PROPER₁
odds > pairs > cons > mark > top > PROPER₁
odds > pairs > cons > ok > head₁ > PROPER₁
odds > pairs > cons > ok > top > PROPER₁

Status:

PROPER₁: [1]
odds: []
cons: []
head₁: [1]
mark: []
pairs: []
ok: []
nats: []
proper₁: [1]
top: []
0: []

The following usable rules [FROCOS05] were oriented:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(36) 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(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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.

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)
active(x1) = active(x1)
nats = nats
mark(x1) = mark
cons(x1, x2) = cons(x1)
0 = 0
incr(x1) = incr
pairs = pairs
odds = odds
head(x1) = x1
tail(x1) = x1
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

nats > 0 > ok > s₁ > mark > cons₁
nats > 0 > ok > top > cons₁
incr > active₁ > s₁ > mark > cons₁
incr > proper₁ > s₁ > mark > cons₁
incr > ok > s₁ > mark > cons₁
incr > ok > top > cons₁
pairs > ok > s₁ > mark > cons₁
pairs > ok > top > cons₁
odds > ok > s₁ > mark > cons₁
odds > ok > top > cons₁

Status:

cons₁: [1]
incr: []
pairs: []
0: []
active₁: [1]
odds: []
mark: []
s₁: [1]
ok: []
nats: []
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(38) Obligation:

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

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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:

ACTIVE(incr(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(head(X)) → ACTIVE(X)
ACTIVE(tail(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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.

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)
incr(x1) = x1
cons(x1, x2) = cons(x1)
s(x1) = x1
head(x1) = x1
tail(x1) = x1
active(x1) = x1
nats = nats
mark(x1) = mark
0 = 0
pairs = pairs
odds = odds
proper(x1) = x1
ok(x1) = x1
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

cons₁ > ACTIVE₁ > top
cons₁ > mark > top
nats > mark > top
0 > top
pairs > mark > top
odds > mark > top

Status:

odds: []
cons₁: [1]
pairs: []
mark: []
nats: []
top: []
0: []
ACTIVE₁: [1]

The following usable rules [FROCOS05] were oriented:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(43) Obligation:

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

ACTIVE(incr(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(head(X)) → ACTIVE(X)
ACTIVE(tail(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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.

ACTIVE(incr(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(head(X)) → ACTIVE(X)
ACTIVE(tail(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)
incr(x1) = incr(x1)
s(x1) = s(x1)
head(x1) = head(x1)
tail(x1) = tail(x1)
active(x1) = x1
nats = nats
mark(x1) = mark
cons(x1, x2) = cons(x1, x2)
0 = 0
pairs = pairs
odds = odds
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

nats > cons₂ > incr₁ > s₁
nats > cons₂ > mark > head₁ > s₁
nats > ok > incr₁ > s₁
nats > ok > head₁ > s₁
0 > ok > incr₁ > s₁
0 > ok > head₁ > s₁
top > proper₁ > tail₁ > ACTIVE₁ > s₁
top > proper₁ > tail₁ > mark > head₁ > s₁
top > proper₁ > cons₂ > incr₁ > s₁
top > proper₁ > cons₂ > mark > head₁ > s₁
top > proper₁ > pairs > incr₁ > s₁
top > proper₁ > pairs > mark > head₁ > s₁
top > proper₁ > odds > incr₁ > s₁
top > proper₁ > odds > mark > head₁ > s₁
top > proper₁ > ok > incr₁ > s₁
top > proper₁ > ok > head₁ > s₁

Status:

tail₁: [1]
head₁: [1]
pairs: []
0: []
ACTIVE₁: [1]
odds: []
cons₂: [2,1]
incr₁: [1]
mark: []
ok: []
s₁: [1]
proper₁: [1]
nats: []
top: []

The following usable rules [FROCOS05] were oriented:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(45) Obligation:

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

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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:

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
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.