(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)
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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) → MARK(cons(0, incr(nats)))
ACTIVE(nats) → CONS(0, incr(nats))
ACTIVE(nats) → INCR(nats)
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
ACTIVE(pairs) → CONS(0, incr(odds))
ACTIVE(pairs) → INCR(odds)
ACTIVE(odds) → MARK(incr(pairs))
ACTIVE(odds) → INCR(pairs)
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
ACTIVE(incr(cons(X, XS))) → CONS(s(X), incr(XS))
ACTIVE(incr(cons(X, XS))) → S(X)
ACTIVE(incr(cons(X, XS))) → INCR(XS)
ACTIVE(head(cons(X, XS))) → MARK(X)
ACTIVE(tail(cons(X, XS))) → MARK(XS)
MARK(nats) → ACTIVE(nats)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(0) → ACTIVE(0)
MARK(incr(X)) → ACTIVE(incr(mark(X)))
MARK(incr(X)) → INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(pairs) → ACTIVE(pairs)
MARK(odds) → ACTIVE(odds)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(head(X)) → HEAD(mark(X))
MARK(head(X)) → MARK(X)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(tail(X)) → TAIL(mark(X))
MARK(tail(X)) → MARK(X)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
INCR(mark(X)) → INCR(X)
INCR(active(X)) → INCR(X)
S(mark(X)) → S(X)
S(active(X)) → S(X)
HEAD(mark(X)) → HEAD(X)
HEAD(active(X)) → HEAD(X)
TAIL(mark(X)) → TAIL(X)
TAIL(active(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)
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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 6 SCCs with 14 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

active(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)
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(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(active(X)) → TAIL(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAIL(x1)  =  x1
active(x1)  =  active(x1)
mark(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
active1: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

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


TAIL(mark(X)) → TAIL(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
mark1 > TAIL1

Status:
TAIL1: [1]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(9) 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)
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

HEAD(active(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)
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
active1: [1]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

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


HEAD(mark(X)) → HEAD(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
mark1 > HEAD1

Status:
HEAD1: [1]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(16) 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)
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) Obligation:

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

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

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

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
active1: [1]

The following usable rules [FROCOS05] were oriented: none

(21) 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(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)
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(22) 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: Lexicographic path order with status [LPO].
Precedence:
mark1 > S1

Status:
S1: [1]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(23) 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)
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE

(26) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(27) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
active1: [1]

The following usable rules [FROCOS05] were oriented: none

(28) Obligation:

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

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

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

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


INCR(mark(X)) → INCR(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
mark1 > INCR1

Status:
INCR1: [1]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(30) 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)
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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

(31) PisEmptyProof (EQUIVALENT transformation)

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

(32) TRUE

(33) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(34) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
mark1 > CONS2
active1 > CONS2

Status:
CONS2: [1,2]
mark1: [1]
active1: [1]

The following usable rules [FROCOS05] were oriented: none

(35) Obligation:

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

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

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

(36) PisEmptyProof (EQUIVALENT transformation)

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

(37) TRUE

(38) Obligation:

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

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(incr(cons(X, XS))) → MARK(cons(s(X), incr(XS)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(nats) → ACTIVE(nats)
ACTIVE(nats) → MARK(cons(0, incr(nats)))
MARK(incr(X)) → ACTIVE(incr(mark(X)))
ACTIVE(head(cons(X, XS))) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(pairs) → ACTIVE(pairs)
ACTIVE(pairs) → MARK(cons(0, incr(odds)))
MARK(odds) → ACTIVE(odds)
ACTIVE(odds) → MARK(incr(pairs))
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(tail(cons(X, XS))) → MARK(XS)
MARK(s(X)) → MARK(X)
MARK(head(X)) → ACTIVE(head(mark(X)))
MARK(head(X)) → MARK(X)
MARK(tail(X)) → ACTIVE(tail(mark(X)))
MARK(tail(X)) → MARK(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)
mark(nats) → active(nats)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(incr(X)) → active(incr(mark(X)))
mark(pairs) → active(pairs)
mark(odds) → active(odds)
mark(s(X)) → active(s(mark(X)))
mark(head(X)) → active(head(mark(X)))
mark(tail(X)) → active(tail(mark(X)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
incr(mark(X)) → incr(X)
incr(active(X)) → incr(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
head(mark(X)) → head(X)
head(active(X)) → head(X)
tail(mark(X)) → tail(X)
tail(active(X)) → tail(X)

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