(0) Obligation:

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

ACTIVE(incr(cons(X, L))) → CONS(s(X), incr(L))
ACTIVE(incr(cons(X, L))) → S(X)
ACTIVE(incr(cons(X, L))) → INCR(L)
ACTIVE(adx(cons(X, L))) → INCR(cons(X, adx(L)))
ACTIVE(adx(cons(X, L))) → CONS(X, adx(L))
ACTIVE(adx(cons(X, L))) → ADX(L)
ACTIVE(nats) → ADX(zeros)
ACTIVE(zeros) → CONS(0, zeros)
ACTIVE(incr(X)) → INCR(active(X))
ACTIVE(incr(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(adx(X)) → ADX(active(X))
ACTIVE(adx(X)) → ACTIVE(X)
ACTIVE(head(X)) → HEAD(active(X))
ACTIVE(head(X)) → ACTIVE(X)
ACTIVE(tail(X)) → TAIL(active(X))
ACTIVE(tail(X)) → ACTIVE(X)
INCR(mark(X)) → INCR(X)
CONS(mark(X1), X2) → CONS(X1, X2)
S(mark(X)) → S(X)
ADX(mark(X)) → ADX(X)
HEAD(mark(X)) → HEAD(X)
TAIL(mark(X)) → TAIL(X)
PROPER(incr(X)) → INCR(proper(X))
PROPER(incr(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(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(adx(X)) → ADX(proper(X))
PROPER(adx(X)) → PROPER(X)
PROPER(head(X)) → HEAD(proper(X))
PROPER(head(X)) → PROPER(X)
PROPER(tail(X)) → TAIL(proper(X))
PROPER(tail(X)) → PROPER(X)
INCR(ok(X)) → INCR(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
S(ok(X)) → S(X)
ADX(ok(X)) → ADX(X)
HEAD(ok(X)) → HEAD(X)
TAIL(ok(X)) → TAIL(X)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 9 SCCs with 22 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
active1 > nil
active1 > cons2 > ok1
active1 > zeros > ok1
active1 > zeros > 0
active1 > head1 > ok1
proper1 > nil
proper1 > cons2 > ok1
proper1 > nats > ok1
proper1 > zeros > ok1
proper1 > zeros > 0
proper1 > head1 > ok1

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(7) Obligation:

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

TAIL(mark(X)) → TAIL(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
TAIL1 > mark1
active1 > nil > mark1
active1 > zeros > cons2 > mark1
active1 > 0 > mark1
active1 > head1 > mark1
active1 > tail1 > mark1
top > proper1 > nats > mark1
top > proper1 > zeros > cons2 > mark1
top > proper1 > head1 > mark1
top > proper1 > tail1 > mark1

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(9) Obligation:

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

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

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
active1 > nil
active1 > cons2 > ok1
active1 > zeros > ok1
active1 > zeros > 0
active1 > head1 > ok1
proper1 > nil
proper1 > cons2 > ok1
proper1 > nats > ok1
proper1 > zeros > ok1
proper1 > zeros > 0
proper1 > head1 > ok1

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(14) Obligation:

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

HEAD(mark(X)) → HEAD(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
HEAD1 > mark1
active1 > nil > mark1
active1 > zeros > cons2 > mark1
active1 > 0 > mark1
active1 > head1 > mark1
active1 > tail1 > mark1
top > proper1 > nats > mark1
top > proper1 > zeros > cons2 > mark1
top > proper1 > head1 > mark1
top > proper1 > tail1 > mark1

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(16) Obligation:

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) Obligation:

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

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

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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)
mark(x1)  =  x1
active(x1)  =  active(x1)
incr(x1)  =  x1
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
adx(x1)  =  x1
nats  =  nats
zeros  =  zeros
0  =  0
head(x1)  =  head(x1)
tail(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > nil
active1 > cons2 > ok1
active1 > zeros > ok1
active1 > zeros > 0
active1 > head1 > ok1
proper1 > nil
proper1 > cons2 > ok1
proper1 > nats > ok1
proper1 > zeros > ok1
proper1 > zeros > 0
proper1 > head1 > ok1

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(21) Obligation:

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

ADX(mark(X)) → ADX(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
incr(x1)  =  x1
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
adx(x1)  =  x1
nats  =  nats
zeros  =  zeros
0  =  0
head(x1)  =  head(x1)
tail(x1)  =  tail(x1)
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
ADX1 > mark1
active1 > nil > mark1
active1 > zeros > cons2 > mark1
active1 > 0 > mark1
active1 > head1 > mark1
active1 > tail1 > mark1
top > proper1 > nats > mark1
top > proper1 > zeros > cons2 > mark1
top > proper1 > head1 > mark1
top > proper1 > tail1 > mark1

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(23) Obligation:

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE

(26) Obligation:

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

S(ok(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(27) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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)
mark(x1)  =  x1
active(x1)  =  active(x1)
incr(x1)  =  x1
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
adx(x1)  =  x1
nats  =  nats
zeros  =  zeros
0  =  0
head(x1)  =  head(x1)
tail(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > nil
active1 > cons2 > ok1
active1 > zeros > ok1
active1 > zeros > 0
active1 > head1 > ok1
proper1 > nil
proper1 > cons2 > ok1
proper1 > nats > ok1
proper1 > zeros > ok1
proper1 > zeros > 0
proper1 > head1 > ok1

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(28) Obligation:

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

S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
S1 > mark1
active1 > nil > mark1
active1 > zeros > cons2 > mark1
active1 > 0 > mark1
active1 > head1 > mark1
active1 > tail1 > mark1
top > proper1 > nats > mark1
top > proper1 > zeros > cons2 > mark1
top > proper1 > head1 > mark1
top > proper1 > tail1 > mark1

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(30) Obligation:

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(31) PisEmptyProof (EQUIVALENT transformation)

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

(32) TRUE

(33) Obligation:

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

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

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(34) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
proper1 > nil > mark > CONS1 > ok1
proper1 > nil > mark > top > ok1
proper1 > cons1 > s1 > mark > CONS1 > ok1
proper1 > cons1 > s1 > mark > top > ok1
proper1 > nats > mark > CONS1 > ok1
proper1 > nats > mark > top > ok1
proper1 > zeros > mark > CONS1 > ok1
proper1 > zeros > mark > top > ok1
proper1 > zeros > 0 > ok1
proper1 > head1 > ok1

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(35) Obligation:

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

CONS(mark(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(36) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
active1 > nil > mark1
active1 > adx1 > cons2 > mark1
nats > adx1 > cons2 > mark1
nats > zeros > cons2 > mark1
nats > zeros > 0 > mark1
top > proper1 > nil > mark1
top > proper1 > adx1 > cons2 > mark1

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(37) Obligation:

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(38) PisEmptyProof (EQUIVALENT transformation)

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

(39) TRUE

(40) Obligation:

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

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

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(41) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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)
mark(x1)  =  x1
active(x1)  =  active(x1)
incr(x1)  =  x1
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
adx(x1)  =  x1
nats  =  nats
zeros  =  zeros
0  =  0
head(x1)  =  head(x1)
tail(x1)  =  x1
proper(x1)  =  proper(x1)
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > nil
active1 > cons2 > ok1
active1 > zeros > ok1
active1 > zeros > 0
active1 > head1 > ok1
proper1 > nil
proper1 > cons2 > ok1
proper1 > nats > ok1
proper1 > zeros > ok1
proper1 > zeros > 0
proper1 > head1 > ok1

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(42) Obligation:

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

INCR(mark(X)) → INCR(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(43) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
incr(x1)  =  x1
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
adx(x1)  =  x1
nats  =  nats
zeros  =  zeros
0  =  0
head(x1)  =  head(x1)
tail(x1)  =  tail(x1)
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
INCR1 > mark1
active1 > nil > mark1
active1 > zeros > cons2 > mark1
active1 > 0 > mark1
active1 > head1 > mark1
active1 > tail1 > mark1
top > proper1 > nats > mark1
top > proper1 > zeros > cons2 > mark1
top > proper1 > head1 > mark1
top > proper1 > tail1 > mark1

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(44) Obligation:

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(45) PisEmptyProof (EQUIVALENT transformation)

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

(46) TRUE

(47) Obligation:

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

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

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(48) 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)
incr(x1)  =  x1
s(x1)  =  x1
adx(x1)  =  x1
head(x1)  =  x1
tail(x1)  =  x1
active(x1)  =  x1
nil  =  nil
mark(x1)  =  mark
nats  =  nats
zeros  =  zeros
0  =  0
proper(x1)  =  x1
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
cons2 > mark > PROPER1
nil > mark > PROPER1
nats > mark > PROPER1
zeros > mark > PROPER1
0 > PROPER1
top > PROPER1

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(49) Obligation:

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

PROPER(incr(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(adx(X)) → PROPER(X)
PROPER(head(X)) → PROPER(X)
PROPER(tail(X)) → PROPER(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(50) 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)  =  x1
incr(x1)  =  x1
s(x1)  =  x1
adx(x1)  =  adx(x1)
head(x1)  =  x1
tail(x1)  =  x1
active(x1)  =  active(x1)
nil  =  nil
mark(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
nats  =  nats
zeros  =  zeros
0  =  0
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > nil > adx1
active1 > zeros > cons2 > adx1
active1 > 0 > adx1
proper1 > nil > adx1
proper1 > nats > adx1
proper1 > zeros > cons2 > adx1
proper1 > 0 > adx1
top > adx1

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(51) Obligation:

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

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

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(52) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
active1 > tail1
active1 > nil
active1 > cons2
active1 > zeros > 0
nats > zeros > 0

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(53) Obligation:

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

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

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(54) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
nats > zeros > 0 > cons2
nats > ok > head1 > PROPER1 > cons2
top > active1 > nil > ok > head1 > PROPER1 > cons2
top > active1 > zeros > 0 > cons2
top > proper1 > zeros > 0 > cons2
top > proper1 > ok > head1 > PROPER1 > cons2

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(55) Obligation:

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

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

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(56) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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)
s(x1)  =  x1
active(x1)  =  active(x1)
nil  =  nil
mark(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)
adx(x1)  =  x1
nats  =  nats
zeros  =  zeros
0  =  0
head(x1)  =  x1
tail(x1)  =  x1
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > incr1 > PROPER1
active1 > cons2 > PROPER1
active1 > 0 > PROPER1
nil > ok > PROPER1
nats > zeros > cons2 > PROPER1
nats > zeros > 0 > PROPER1
nats > zeros > ok > PROPER1
proper1 > incr1 > PROPER1
proper1 > cons2 > PROPER1
proper1 > 0 > PROPER1
proper1 > ok > PROPER1
top > PROPER1

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(57) Obligation:

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

PROPER(s(X)) → PROPER(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(58) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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)
s(x1)  =  s(x1)
active(x1)  =  active(x1)
incr(x1)  =  x1
nil  =  nil
mark(x1)  =  mark
cons(x1, x2)  =  x2
adx(x1)  =  adx
nats  =  nats
zeros  =  zeros
0  =  0
head(x1)  =  head
tail(x1)  =  tail
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
0 > mark
tail > active1 > s1 > PROPER1 > mark
tail > active1 > nil > mark
tail > active1 > adx > mark
tail > active1 > zeros > mark
tail > active1 > head > mark
tail > proper1 > s1 > PROPER1 > mark
tail > proper1 > nil > mark
tail > proper1 > adx > mark
tail > proper1 > nats > mark
tail > proper1 > zeros > mark
tail > proper1 > head > mark
top > active1 > s1 > PROPER1 > mark
top > active1 > nil > mark
top > active1 > adx > mark
top > active1 > zeros > mark
top > active1 > head > mark
top > proper1 > s1 > PROPER1 > mark
top > proper1 > nil > mark
top > proper1 > adx > mark
top > proper1 > nats > mark
top > proper1 > zeros > mark
top > proper1 > head > mark

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(59) Obligation:

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(60) PisEmptyProof (EQUIVALENT transformation)

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

(61) TRUE

(62) Obligation:

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

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

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(63) 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)
cons(x1, x2)  =  x1
incr(x1)  =  incr(x1)
s(x1)  =  x1
adx(x1)  =  x1
head(x1)  =  x1
tail(x1)  =  x1
active(x1)  =  active(x1)
nil  =  nil
mark(x1)  =  mark
nats  =  nats
zeros  =  zeros
0  =  0
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
proper1 > incr1 > ACTIVE1
proper1 > incr1 > nil
proper1 > incr1 > mark
proper1 > nats > mark
proper1 > nats > zeros
proper1 > 0
top > active1 > incr1 > ACTIVE1
top > active1 > incr1 > nil
top > active1 > incr1 > mark
top > active1 > zeros

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(64) Obligation:

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

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

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(65) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(cons(X1, X2)) → ACTIVE(X1)
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)  =  ACTIVE(x1)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  x1
adx(x1)  =  adx(x1)
head(x1)  =  x1
tail(x1)  =  x1
active(x1)  =  active(x1)
incr(x1)  =  incr(x1)
nil  =  nil
mark(x1)  =  x1
nats  =  nats
zeros  =  zeros
0  =  0
proper(x1)  =  proper(x1)
ok(x1)  =  ok
top(x1)  =  top

Recursive Path Order [RPO].
Precedence:
active1 > adx1 > nil > incr1
active1 > adx1 > ok > cons2 > ACTIVE1 > incr1
active1 > zeros > cons2 > ACTIVE1 > incr1
active1 > zeros > 0 > incr1
nats > adx1 > nil > incr1
nats > adx1 > ok > cons2 > ACTIVE1 > incr1
nats > zeros > cons2 > ACTIVE1 > incr1
nats > zeros > 0 > incr1
proper1 > adx1 > nil > incr1
proper1 > adx1 > ok > cons2 > ACTIVE1 > incr1
proper1 > zeros > cons2 > ACTIVE1 > incr1
proper1 > zeros > 0 > incr1
top > incr1

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(66) Obligation:

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

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

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(67) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
top > active1 > s1 > ACTIVE1
top > active1 > tail1 > ACTIVE1
top > active1 > nil
top > active1 > cons2
top > active1 > adx1
top > active1 > zeros
top > active1 > 0
top > proper1 > s1 > ACTIVE1
top > proper1 > tail1 > ACTIVE1
top > proper1 > nil
top > proper1 > cons2
top > proper1 > adx1
top > proper1 > nats
top > proper1 > zeros
top > proper1 > 0

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(68) Obligation:

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

ACTIVE(head(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(69) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
proper1 > head1 > mark
proper1 > head1 > ok > s1
proper1 > nil > mark
proper1 > nil > ok > s1
proper1 > nats > mark
proper1 > zeros > cons2 > mark
proper1 > zeros > cons2 > ok > s1
proper1 > zeros > 0
proper1 > tail > mark
proper1 > tail > ok > s1

The following usable rules [FROCOS05] were oriented:

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(70) Obligation:

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

active(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(71) PisEmptyProof (EQUIVALENT transformation)

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

(72) TRUE

(73) 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(incr(nil)) → mark(nil)
active(incr(cons(X, L))) → mark(cons(s(X), incr(L)))
active(adx(nil)) → mark(nil)
active(adx(cons(X, L))) → mark(incr(cons(X, adx(L))))
active(nats) → mark(adx(zeros))
active(zeros) → mark(cons(0, zeros))
active(head(cons(X, L))) → mark(X)
active(tail(cons(X, L))) → mark(L)
active(incr(X)) → incr(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(adx(X)) → adx(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
incr(mark(X)) → mark(incr(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
adx(mark(X)) → mark(adx(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(incr(X)) → incr(proper(X))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(adx(X)) → adx(proper(X))
proper(nats) → ok(nats)
proper(zeros) → ok(zeros)
proper(0) → ok(0)
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
incr(ok(X)) → ok(incr(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
adx(ok(X)) → ok(adx(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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