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

(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(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)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:

mark₁ > TAIL₁

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

TAIL(ok(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.

(8) 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)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

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)
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.

(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(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.

(13) 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: Combined order from the following AFS and order.
HEAD(x1)  =  HEAD(x1)
ok(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:

mark₁ > HEAD₁

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

HEAD(ok(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.

(15) 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)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

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)
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.

(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(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.

(20) 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)  =  S(x1)
ok(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:

mark₁ > S₁

The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

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

S(ok(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.

(22) 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)  =  x1
ok(x1)  =  ok(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

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)
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.

(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(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.

(27) 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: Combined order from the following AFS and order.
INCR(x1)  =  INCR(x1)
ok(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:

mark₁ > INCR₁

The following usable rules [FROCOS05] were oriented: none

(28) Obligation:

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

INCR(ok(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.

(29) 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)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

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)
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.

(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(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.

(34) 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(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(35) 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.

(36) 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)  =  CONS(x1)
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(37) 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.

(38) PisEmptyProof (EQUIVALENT transformation)

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

(39) TRUE

(40) 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.

(41) 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)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(42) Obligation:

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

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.

(43) 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)  =  x1
incr(x1)  =  x1
s(x1)  =  x1
head(x1)  =  head(x1)
tail(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(44) Obligation:

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

PROPER(incr(X)) → PROPER(X)
PROPER(s(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.

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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)
incr(x1)  =  x1
s(x1)  =  x1
tail(x1)  =  tail(x1)

Lexicographic Path Order [LPO].
Precedence:

tail₁ > PROPER₁

The following usable rules [FROCOS05] were oriented: none

(46) Obligation:

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

PROPER(incr(X)) → PROPER(X)
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.

(47) 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)  =  PROPER(x1)
incr(x1)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:

s₁ > PROPER₁

The following usable rules [FROCOS05] were oriented: none

(48) Obligation:

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

PROPER(incr(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.

(49) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(incr(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
incr(x1)  =  incr(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(50) 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.

(51) PisEmptyProof (EQUIVALENT transformation)

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

(52) TRUE

(53) 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.

(54) 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)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(55) Obligation:

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

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.

(56) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(head(X)) → ACTIVE(X)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(57) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(s(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.

(58) 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, x2)
s(x1)  =  x1
tail(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:

cons₂ > ACTIVE₁

The following usable rules [FROCOS05] were oriented: none

(59) Obligation:

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

ACTIVE(s(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.

(60) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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)
s(x1)  =  x1
tail(x1)  =  tail(x1)

Lexicographic Path Order [LPO].
Precedence:

tail₁ > ACTIVE₁

The following usable rules [FROCOS05] were oriented: none

(61) Obligation:

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

ACTIVE(s(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.

(62) 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)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(63) 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.

(64) PisEmptyProof (EQUIVALENT transformation)

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

(65) TRUE

(66) 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.