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

Lexicographic path order with status [LPO].
Precedence:
top > proper > ok1 > active > mark
top > proper > nats
top > proper > zeros > mark
top > proper > 0

Status:
active: []
zeros: []
mark: []
proper: []
ok1: [1]
nats: []
top: []
0: []
S1: [1]

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

Lexicographic path order with status [LPO].
Precedence:
nats > zeros > cons2 > ok1 > CONS1
nats > zeros > 0 > ok1 > CONS1
top > active1 > cons2 > ok1 > CONS1
top > active1 > 0 > ok1 > CONS1
top > proper1 > cons2 > ok1 > CONS1
top > proper1 > 0 > ok1 > CONS1

Status:
active1: [1]
cons2: [1,2]
CONS1: [1]
zeros: []
ok1: [1]
nats: []
proper1: [1]
top: []
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))

(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)  =  TL(x1)
ok(x1)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
nats  =  nats
adx(x1)  =  adx(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

Lexicographic path order with status [LPO].
Precedence:
active1 > adx1 > cons2 > incr1 > mark1 > TL1 > top
active1 > zeros > cons2 > incr1 > mark1 > TL1 > top
active1 > zeros > 0 > top
active1 > s > top
nats > top

Status:
active1: [1]
cons2: [1,2]
adx1: [1]
incr1: [1]
zeros: []
TL1: [1]
mark1: [1]
nats: []
s: []
top: []
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))

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

Lexicographic path order with status [LPO].
Precedence:
top > proper > ok1 > active > mark
top > proper > nats
top > proper > zeros > mark
top > proper > 0

Status:
active: []
zeros: []
mark: []
proper: []
TL1: [1]
ok1: [1]
nats: []
top: []
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))

(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)  =  HD(x1)
ok(x1)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
nats  =  nats
adx(x1)  =  adx(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

Lexicographic path order with status [LPO].
Precedence:
active1 > adx1 > cons2 > incr1 > mark1 > HD1 > top
active1 > zeros > cons2 > incr1 > mark1 > HD1 > top
active1 > zeros > 0 > top
active1 > s > top
nats > top

Status:
active1: [1]
cons2: [1,2]
adx1: [1]
incr1: [1]
zeros: []
mark1: [1]
HD1: [1]
nats: []
s: []
top: []
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))

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

Lexicographic path order with status [LPO].
Precedence:
top > proper > ok1 > active > mark
top > proper > nats
top > proper > zeros > mark
top > proper > 0

Status:
active: []
zeros: []
mark: []
proper: []
ok1: [1]
HD1: [1]
nats: []
top: []
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))

(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)  =  INCR(x1)
ok(x1)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
nats  =  nats
adx(x1)  =  adx(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

Lexicographic path order with status [LPO].
Precedence:
active1 > adx1 > cons2 > incr1 > mark1 > INCR1 > top
active1 > zeros > cons2 > incr1 > mark1 > INCR1 > top
active1 > zeros > 0 > top
active1 > s > top
nats > top

Status:
active1: [1]
cons2: [1,2]
adx1: [1]
incr1: [1]
zeros: []
mark1: [1]
nats: []
s: []
top: []
0: []
INCR1: [1]

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

Lexicographic path order with status [LPO].
Precedence:
top > proper > ok1 > active > mark
top > proper > nats
top > proper > zeros > mark
top > proper > 0

Status:
active: []
zeros: []
mark: []
proper: []
ok1: [1]
nats: []
top: []
0: []
INCR1: [1]

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)  =  ADX(x1)
ok(x1)  =  x1
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
nats  =  nats
adx(x1)  =  adx(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

Lexicographic path order with status [LPO].
Precedence:
active1 > adx1 > cons2 > incr1 > mark1 > ADX1 > top
active1 > zeros > cons2 > incr1 > mark1 > ADX1 > top
active1 > zeros > 0 > top
active1 > s > top
nats > top

Status:
active1: [1]
cons2: [1,2]
ADX1: [1]
adx1: [1]
incr1: [1]
zeros: []
mark1: [1]
nats: []
s: []
top: []
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))

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

Lexicographic path order with status [LPO].
Precedence:
top > proper > ok1 > active > mark
top > proper > nats
top > proper > zeros > mark
top > proper > 0

Status:
active: []
ADX1: [1]
zeros: []
mark: []
proper: []
ok1: [1]
nats: []
top: []
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))

(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)  =  PROPER(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

Lexicographic path order with status [LPO].
Precedence:
top > proper1 > nats > zeros > cons2
top > proper1 > nats > zeros > mark
top > proper1 > nats > zeros > 0
top > proper1 > ok > active > cons2
top > proper1 > ok > active > mark
top > proper1 > ok > active > 0

Status:
active: []
PROPER1: [1]
cons2: [1,2]
zeros: []
mark: []
ok: []
nats: []
proper1: [1]
top: []
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))

(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(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)  =  PROPER(x1)
adx(x1)  =  x1
incr(x1)  =  x1
s(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
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
active1 > mark > proper1 > nats
active1 > mark > proper1 > ok > tl1 > PROPER1
active1 > mark > proper1 > ok > top
active1 > zeros > ok > tl1 > PROPER1
active1 > zeros > ok > top
active1 > cons > ok > tl1 > PROPER1
active1 > cons > ok > top
active1 > 0 > ok > tl1 > PROPER1
active1 > 0 > ok > top

Status:
PROPER1: [1]
active1: [1]
cons: []
tl1: [1]
zeros: []
mark: []
ok: []
nats: []
proper1: [1]
top: []
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))

(47) 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)

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

Lexicographic path order with status [LPO].
Precedence:
adx1 > PROPER1 > mark
adx1 > ok > mark
incr1 > PROPER1 > mark
incr1 > ok > mark
s1 > PROPER1 > mark
s1 > ok > mark
hd1 > PROPER1 > mark
nats > ok > mark
zeros > ok > mark
0 > ok > mark
top > mark

Status:
PROPER1: [1]
adx1: [1]
incr1: [1]
zeros: []
mark: []
s1: [1]
ok: []
nats: []
top: []
0: []
hd1: [1]

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:
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.

(50) PisEmptyProof (EQUIVALENT transformation)

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

(51) TRUE

(52) 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.

(53) 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)
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)  =  ACTIVE(x1)
incr(x1)  =  x1
adx(x1)  =  adx(x1)
hd(x1)  =  hd(x1)
tl(x1)  =  x1
active(x1)  =  active(x1)
nats  =  nats
mark(x1)  =  mark
zeros  =  zeros
cons(x1, x2)  =  cons(x1)
0  =  0
s(x1)  =  s(x1)
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
active1 > hd1 > ACTIVE1 > top
active1 > mark > adx1 > ACTIVE1 > top
active1 > mark > adx1 > cons1 > top
active1 > s1 > top
zeros > mark > adx1 > ACTIVE1 > top
zeros > mark > adx1 > cons1 > top
zeros > ok > adx1 > ACTIVE1 > top
zeros > ok > adx1 > cons1 > top
zeros > ok > hd1 > ACTIVE1 > top
zeros > ok > s1 > top
0 > ok > adx1 > ACTIVE1 > top
0 > ok > adx1 > cons1 > top
0 > ok > hd1 > ACTIVE1 > top
0 > ok > s1 > top
proper1 > nats > top
proper1 > ok > adx1 > ACTIVE1 > top
proper1 > ok > adx1 > cons1 > top
proper1 > ok > hd1 > ACTIVE1 > top
proper1 > ok > s1 > top

Status:
cons1: [1]
0: []
hd1: [1]
ACTIVE1: [1]
active1: [1]
adx1: [1]
zeros: []
mark: []
ok: []
s1: [1]
proper1: [1]
nats: []
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))

(54) Obligation:

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

ACTIVE(incr(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.

(55) 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
incr(x1)  =  x1
tl(x1)  =  tl(x1)
active(x1)  =  active(x1)
nats  =  nats
mark(x1)  =  x1
adx(x1)  =  x1
zeros  =  zeros
cons(x1, x2)  =  cons(x1, x2)
0  =  0
s(x1)  =  s
hd(x1)  =  x1
proper(x1)  =  x1
ok(x1)  =  ok
top(x1)  =  top

Lexicographic path order with status [LPO].
Precedence:
nats > zeros > cons2 > ok
nats > zeros > 0 > ok
top > active1 > tl1 > ok
top > active1 > cons2 > ok
top > active1 > 0 > ok
top > active1 > s > ok

Status:
active1: [1]
cons2: [1,2]
tl1: [1]
zeros: []
ok: []
nats: []
s: []
top: []
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))

(56) Obligation:

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

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

Lexicographic path order with status [LPO].
Precedence:
active1 > zeros > ok > incr1
active1 > zeros > ok > cons1 > adx1
active1 > 0 > ok > incr1
active1 > 0 > ok > cons1 > adx1
active1 > s > ok > incr1
active1 > s > ok > cons1 > adx1
active1 > hd1 > mark > incr1
active1 > hd1 > mark > adx1
active1 > hd1 > ok > incr1
active1 > hd1 > ok > cons1 > adx1
top > proper1 > nats > ok > incr1
top > proper1 > nats > ok > cons1 > adx1
top > proper1 > zeros > ok > incr1
top > proper1 > zeros > ok > cons1 > adx1
top > proper1 > 0 > ok > incr1
top > proper1 > 0 > ok > cons1 > adx1
top > proper1 > s > ok > incr1
top > proper1 > s > ok > cons1 > adx1
top > proper1 > hd1 > mark > incr1
top > proper1 > hd1 > mark > adx1
top > proper1 > hd1 > ok > incr1
top > proper1 > hd1 > ok > cons1 > adx1

Status:
cons1: [1]
s: []
0: []
hd1: [1]
ACTIVE1: [1]
active1: [1]
adx1: [1]
incr1: [1]
zeros: []
mark: []
ok: []
proper1: [1]
nats: []
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))

(58) 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.

(59) PisEmptyProof (EQUIVALENT transformation)

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

(60) TRUE

(61) 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.