(0) Obligation:

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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) → ADX(zeros)
ACTIVE(zeros) → CONS(0, zeros)
ACTIVE(incr(cons(X, Y))) → CONS(s(X), incr(Y))
ACTIVE(incr(cons(X, Y))) → S(X)
ACTIVE(incr(cons(X, Y))) → INCR(Y)
ACTIVE(adx(cons(X, Y))) → INCR(cons(X, adx(Y)))
ACTIVE(adx(cons(X, Y))) → CONS(X, adx(Y))
ACTIVE(adx(cons(X, Y))) → ADX(Y)
ACTIVE(adx(X)) → ADX(active(X))
ACTIVE(adx(X)) → ACTIVE(X)
ACTIVE(incr(X)) → INCR(active(X))
ACTIVE(incr(X)) → ACTIVE(X)
ACTIVE(hd(X)) → HD(active(X))
ACTIVE(hd(X)) → ACTIVE(X)
ACTIVE(tl(X)) → TL(active(X))
ACTIVE(tl(X)) → ACTIVE(X)
ADX(mark(X)) → ADX(X)
INCR(mark(X)) → INCR(X)
HD(mark(X)) → HD(X)
TL(mark(X)) → TL(X)
PROPER(adx(X)) → ADX(proper(X))
PROPER(adx(X)) → PROPER(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(incr(X)) → INCR(proper(X))
PROPER(incr(X)) → PROPER(X)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(hd(X)) → HD(proper(X))
PROPER(hd(X)) → PROPER(X)
PROPER(tl(X)) → TL(proper(X))
PROPER(tl(X)) → PROPER(X)
ADX(ok(X)) → ADX(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
INCR(ok(X)) → INCR(X)
S(ok(X)) → S(X)
HD(ok(X)) → HD(X)
TL(ok(X)) → TL(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(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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 9 SCCs with 20 less nodes.

(4) Complex Obligation (AND)

(5) 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(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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.


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)
active(x1)  =  active
nats  =  nats
mark(x1)  =  mark
adx(x1)  =  x1
zeros  =  zeros
cons(x1, x2)  =  x2
0  =  0
incr(x1)  =  x1
s(x1)  =  x1
hd(x1)  =  x1
tl(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
nats > [active, mark] > zeros > [ok1, top]
0 > [ok1, top]
proper1 > zeros > [ok1, top]


The following usable rules [FROCOS05] were oriented:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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:

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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.


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(x1)
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
nats  =  nats
mark(x1)  =  mark
adx(x1)  =  x1
zeros  =  zeros
cons(x1, x2)  =  x2
0  =  0
incr(x1)  =  x1
s(x1)  =  x1
hd(x1)  =  x1
tl(x1)  =  x1
proper(x1)  =  proper
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > mark > top
active1 > zeros
active1 > 0 > ok1 > top
proper > nats > zeros
proper > 0 > ok1 > top


The following usable rules [FROCOS05] were oriented:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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:

TL(ok(X)) → TL(X)
TL(mark(X)) → TL(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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.


TL(mark(X)) → TL(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TL(x1)  =  x1
ok(x1)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
nats  =  nats
adx(x1)  =  x1
zeros  =  zeros
cons(x1, x2)  =  cons(x1, x2)
0  =  0
incr(x1)  =  incr(x1)
s(x1)  =  s
hd(x1)  =  x1
tl(x1)  =  x1
proper(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > zeros > mark1 > top
active1 > zeros > [cons2, s]
active1 > zeros > 0
active1 > incr1 > mark1 > top
active1 > incr1 > [cons2, s]


The following usable rules [FROCOS05] were oriented:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(17) Obligation:

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

TL(ok(X)) → TL(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TL(ok(X)) → TL(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TL(x1)  =  x1
ok(x1)  =  ok(x1)
active(x1)  =  active
nats  =  nats
mark(x1)  =  mark
adx(x1)  =  x1
zeros  =  zeros
cons(x1, x2)  =  x2
0  =  0
incr(x1)  =  x1
s(x1)  =  x1
hd(x1)  =  x1
tl(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
nats > [active, mark] > zeros > [ok1, top]
0 > [ok1, top]
proper1 > zeros > [ok1, top]


The following usable rules [FROCOS05] were oriented:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(19) Obligation:

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) Obligation:

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

HD(ok(X)) → HD(X)
HD(mark(X)) → HD(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


HD(mark(X)) → HD(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
HD(x1)  =  x1
ok(x1)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
nats  =  nats
adx(x1)  =  x1
zeros  =  zeros
cons(x1, x2)  =  cons(x1, x2)
0  =  0
incr(x1)  =  incr(x1)
s(x1)  =  s
hd(x1)  =  x1
tl(x1)  =  x1
proper(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > zeros > mark1 > top
active1 > zeros > [cons2, s]
active1 > zeros > 0
active1 > incr1 > mark1 > top
active1 > incr1 > [cons2, s]


The following usable rules [FROCOS05] were oriented:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(24) Obligation:

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

HD(ok(X)) → HD(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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.

(25) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


HD(ok(X)) → HD(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
HD(x1)  =  x1
ok(x1)  =  ok(x1)
active(x1)  =  active
nats  =  nats
mark(x1)  =  mark
adx(x1)  =  x1
zeros  =  zeros
cons(x1, x2)  =  x2
0  =  0
incr(x1)  =  x1
s(x1)  =  x1
hd(x1)  =  x1
tl(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
nats > [active, mark] > zeros > [ok1, top]
0 > [ok1, top]
proper1 > zeros > [ok1, top]


The following usable rules [FROCOS05] were oriented:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(26) Obligation:

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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) PisEmptyProof (EQUIVALENT transformation)

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

(28) TRUE

(29) 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(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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) 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)  =  x1
ok(x1)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
nats  =  nats
adx(x1)  =  x1
zeros  =  zeros
cons(x1, x2)  =  cons(x1, x2)
0  =  0
incr(x1)  =  incr(x1)
s(x1)  =  s
hd(x1)  =  x1
tl(x1)  =  x1
proper(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > zeros > mark1 > top
active1 > zeros > [cons2, s]
active1 > zeros > 0
active1 > incr1 > mark1 > top
active1 > incr1 > [cons2, s]


The following usable rules [FROCOS05] were oriented:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(31) 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(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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.

(32) 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)
active(x1)  =  active
nats  =  nats
mark(x1)  =  mark
adx(x1)  =  x1
zeros  =  zeros
cons(x1, x2)  =  x2
0  =  0
incr(x1)  =  x1
s(x1)  =  x1
hd(x1)  =  x1
tl(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
nats > [active, mark] > zeros > [ok1, top]
0 > [ok1, top]
proper1 > zeros > [ok1, top]


The following usable rules [FROCOS05] were oriented:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(33) Obligation:

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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) PisEmptyProof (EQUIVALENT transformation)

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

(35) TRUE

(36) Obligation:

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

ADX(ok(X)) → ADX(X)
ADX(mark(X)) → ADX(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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.


ADX(mark(X)) → ADX(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ADX(x1)  =  x1
ok(x1)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
nats  =  nats
adx(x1)  =  x1
zeros  =  zeros
cons(x1, x2)  =  cons(x1, x2)
0  =  0
incr(x1)  =  incr(x1)
s(x1)  =  s
hd(x1)  =  x1
tl(x1)  =  x1
proper(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > zeros > mark1 > top
active1 > zeros > [cons2, s]
active1 > zeros > 0
active1 > incr1 > mark1 > top
active1 > incr1 > [cons2, s]


The following usable rules [FROCOS05] were oriented:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(38) Obligation:

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

ADX(ok(X)) → ADX(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ADX(ok(X)) → ADX(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ADX(x1)  =  x1
ok(x1)  =  ok(x1)
active(x1)  =  active
nats  =  nats
mark(x1)  =  mark
adx(x1)  =  x1
zeros  =  zeros
cons(x1, x2)  =  x2
0  =  0
incr(x1)  =  x1
s(x1)  =  x1
hd(x1)  =  x1
tl(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
nats > [active, mark] > zeros > [ok1, top]
0 > [ok1, top]
proper1 > zeros > [ok1, top]


The following usable rules [FROCOS05] were oriented:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(40) Obligation:

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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) PisEmptyProof (EQUIVALENT transformation)

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

(42) TRUE

(43) Obligation:

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

PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(adx(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(incr(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(hd(X)) → PROPER(X)
PROPER(tl(X)) → PROPER(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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.


PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
adx(x1)  =  x1
incr(x1)  =  x1
s(x1)  =  x1
hd(x1)  =  x1
tl(x1)  =  x1
active(x1)  =  active
nats  =  nats
mark(x1)  =  mark
zeros  =  zeros
0  =  0
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active > [zeros, 0] > [proper1, ok, top] > cons2 > mark
nats > [zeros, 0] > [proper1, ok, top] > cons2 > mark


The following usable rules [FROCOS05] were oriented:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(45) Obligation:

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

PROPER(adx(X)) → PROPER(X)
PROPER(incr(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(hd(X)) → PROPER(X)
PROPER(tl(X)) → PROPER(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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) 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)
adx(x1)  =  x1
incr(x1)  =  x1
s(x1)  =  s(x1)
hd(x1)  =  x1
tl(x1)  =  x1
active(x1)  =  active
nats  =  nats
mark(x1)  =  mark
zeros  =  zeros
cons(x1, x2)  =  x1
0  =  0
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
nats > [zeros, 0, ok] > [PROPER1, s1, proper1] > [active, mark]
top > [PROPER1, s1, proper1] > [active, mark]


The following usable rules [FROCOS05] were oriented:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(47) Obligation:

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

PROPER(adx(X)) → PROPER(X)
PROPER(incr(X)) → PROPER(X)
PROPER(hd(X)) → PROPER(X)
PROPER(tl(X)) → PROPER(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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.

(48) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(adx(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)
adx(x1)  =  adx(x1)
incr(x1)  =  x1
hd(x1)  =  x1
tl(x1)  =  x1
active(x1)  =  x1
nats  =  nats
mark(x1)  =  mark
zeros  =  zeros
cons(x1, x2)  =  cons
0  =  0
s(x1)  =  x1
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
[adx1, nats] > PROPER1
[adx1, nats] > [mark, zeros, cons, top]


The following usable rules [FROCOS05] were oriented:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(49) Obligation:

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

PROPER(incr(X)) → PROPER(X)
PROPER(hd(X)) → PROPER(X)
PROPER(tl(X)) → PROPER(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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.

(50) 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)  =  PROPER(x1)
incr(x1)  =  incr(x1)
hd(x1)  =  x1
tl(x1)  =  x1
active(x1)  =  active(x1)
nats  =  nats
mark(x1)  =  mark
adx(x1)  =  x1
zeros  =  zeros
cons(x1, x2)  =  x1
0  =  0
s(x1)  =  x1
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
[active1, 0, proper1] > [PROPER1, incr1] > [mark, top]
[active1, 0, proper1] > [PROPER1, incr1] > ok
[active1, 0, proper1] > nats > zeros


The following usable rules [FROCOS05] were oriented:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(51) Obligation:

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

PROPER(hd(X)) → PROPER(X)
PROPER(tl(X)) → PROPER(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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.

(52) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(hd(X)) → PROPER(X)
PROPER(tl(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
hd(x1)  =  hd(x1)
tl(x1)  =  tl(x1)
active(x1)  =  active(x1)
nats  =  nats
mark(x1)  =  mark
adx(x1)  =  x1
zeros  =  zeros
cons(x1, x2)  =  cons(x2)
0  =  0
incr(x1)  =  x1
s(x1)  =  x1
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
[hd1, active1] > [zeros, 0] > cons1 > [tl1, mark, top]
nats > [zeros, 0] > cons1 > [tl1, mark, top]


The following usable rules [FROCOS05] were oriented:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(53) Obligation:

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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) PisEmptyProof (EQUIVALENT transformation)

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

(55) TRUE

(56) Obligation:

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

ACTIVE(incr(X)) → ACTIVE(X)
ACTIVE(adx(X)) → ACTIVE(X)
ACTIVE(hd(X)) → ACTIVE(X)
ACTIVE(tl(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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.

(57) 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)
adx(x1)  =  x1
hd(x1)  =  x1
tl(x1)  =  x1
active(x1)  =  x1
nats  =  nats
mark(x1)  =  mark
zeros  =  zeros
cons(x1, x2)  =  cons(x1)
0  =  0
s(x1)  =  s
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
cons1 > [nats, mark, zeros] > incr1 > [s, ok]
cons1 > [nats, mark, zeros] > top > [s, ok]
0 > [s, ok]


The following usable rules [FROCOS05] were oriented:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(58) Obligation:

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

ACTIVE(adx(X)) → ACTIVE(X)
ACTIVE(hd(X)) → ACTIVE(X)
ACTIVE(tl(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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.

(59) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(tl(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
adx(x1)  =  x1
hd(x1)  =  x1
tl(x1)  =  tl(x1)
active(x1)  =  active(x1)
nats  =  nats
mark(x1)  =  mark
zeros  =  zeros
cons(x1, x2)  =  cons
0  =  0
incr(x1)  =  x1
s(x1)  =  x1
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > zeros > [mark, top] > [cons, 0, proper1, ok] > tl1
nats > [mark, top] > [cons, 0, proper1, ok] > tl1


The following usable rules [FROCOS05] were oriented:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(60) Obligation:

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

ACTIVE(adx(X)) → ACTIVE(X)
ACTIVE(hd(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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.

(61) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(adx(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
adx(x1)  =  adx(x1)
hd(x1)  =  x1
active(x1)  =  active(x1)
nats  =  nats
mark(x1)  =  mark
zeros  =  zeros
cons(x1, x2)  =  x1
0  =  0
incr(x1)  =  incr(x1)
s(x1)  =  s(x1)
tl(x1)  =  x1
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > [mark, proper1] > nats > adx1 > top
active1 > [mark, proper1] > nats > [zeros, 0] > top
active1 > [mark, proper1] > [incr1, s1] > top


The following usable rules [FROCOS05] were oriented:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(62) Obligation:

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

ACTIVE(hd(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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.

(63) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(hd(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
hd(x1)  =  hd(x1)
active(x1)  =  active(x1)
nats  =  nats
mark(x1)  =  mark
adx(x1)  =  adx
zeros  =  zeros
cons(x1, x2)  =  x2
0  =  0
incr(x1)  =  incr
s(x1)  =  s(x1)
tl(x1)  =  tl(x1)
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
[active1, tl1] > hd1 > mark
[active1, tl1] > [nats, zeros] > mark
[active1, tl1] > adx > [incr, s1] > mark
[active1, tl1] > 0


The following usable rules [FROCOS05] were oriented:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(64) Obligation:

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

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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.

(65) PisEmptyProof (EQUIVALENT transformation)

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

(66) TRUE

(67) Obligation:

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

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

The TRS R consists of the following rules:

active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(incr(cons(X, Y))) → mark(cons(s(X), incr(Y)))
active(adx(cons(X, Y))) → mark(incr(cons(X, adx(Y))))
active(hd(cons(X, Y))) → mark(X)
active(tl(cons(X, Y))) → mark(Y)
active(adx(X)) → adx(active(X))
active(incr(X)) → incr(active(X))
active(hd(X)) → hd(active(X))
active(tl(X)) → tl(active(X))
adx(mark(X)) → mark(adx(X))
incr(mark(X)) → mark(incr(X))
hd(mark(X)) → mark(hd(X))
tl(mark(X)) → mark(tl(X))
proper(nats) → ok(nats)
proper(adx(X)) → adx(proper(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(s(X)) → s(proper(X))
proper(hd(X)) → hd(proper(X))
proper(tl(X)) → tl(proper(X))
adx(ok(X)) → ok(adx(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
hd(ok(X)) → ok(hd(X))
tl(ok(X)) → ok(tl(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.