Termination w.r.t. Q proof of /tmp/tmp2SmmkB/ExIntrod_GM01

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

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 PisEmptyProof (⇔)

    ↳18 TRUE

  ↳19 QDP

    ↳20 QDPOrderProof (⇔)

    ↳21 QDP

    ↳22 QDPOrderProof (⇔)

    ↳23 QDP

    ↳24 PisEmptyProof (⇔)

    ↳25 TRUE

  ↳26 QDP

    ↳27 QDPOrderProof (⇔)

    ↳28 QDP

    ↳29 QDPOrderProof (⇔)

    ↳30 QDP

    ↳31 PisEmptyProof (⇔)

    ↳32 TRUE

  ↳33 QDP

    ↳34 QDPOrderProof (⇔)

    ↳35 QDP

    ↳36 QDPOrderProof (⇔)

    ↳37 QDP

    ↳38 PisEmptyProof (⇔)

    ↳39 TRUE

  ↳40 QDP

    ↳41 QDPOrderProof (⇔)

    ↳42 QDP

    ↳43 QDPOrderProof (⇔)

    ↳44 QDP

    ↳45 PisEmptyProof (⇔)

    ↳46 TRUE

  ↳47 QDP

    ↳48 QDPOrderProof (⇔)

    ↳49 QDP

    ↳50 QDPOrderProof (⇔)

    ↳51 QDP

    ↳52 QDPOrderProof (⇔)

    ↳53 QDP

    ↳54 QDPOrderProof (⇔)

    ↳55 QDP

    ↳56 PisEmptyProof (⇔)

    ↳57 TRUE

  ↳58 QDP

    ↳59 QDPOrderProof (⇔)

    ↳60 QDP

    ↳61 QDPOrderProof (⇔)

    ↳62 QDP

    ↳63 QDPOrderProof (⇔)

    ↳64 QDP

    ↳65 QDPOrderProof (⇔)

    ↳66 QDP

    ↳67 QDPOrderProof (⇔)

    ↳68 QDP

    ↳69 PisEmptyProof (⇔)

    ↳70 TRUE

  ↳71 QDP

(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(mark(X)) → TAIL(X)

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

Recursive Path Order [RPO].
Precedence:

TAIL₁ > [mark₁, nats, zeros]
[active₁, nil] > cons₂ > [mark₁, nats, zeros]
[active₁, nil] > 0 > [mark₁, nats, zeros]
top > [mark₁, nats, 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))

(7) Obligation:

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

TAIL(ok(X)) → TAIL(X)

The TRS R consists of the following rules:

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

Recursive Path Order [RPO].
Precedence:

proper > [TAIL₁, ok₁] > active > nil > [mark, top]
proper > [TAIL₁, ok₁] > active > zeros > [mark, top]
proper > [TAIL₁, ok₁] > active > 0
proper > nats > zeros > [mark, top]

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(mark(X)) → HEAD(X)

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

Recursive Path Order [RPO].
Precedence:

HEAD₁ > [mark₁, nats, zeros]
[active₁, nil] > cons₂ > [mark₁, nats, zeros]
[active₁, nil] > 0 > [mark₁, nats, zeros]
top > [mark₁, nats, 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))

(14) Obligation:

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

HEAD(ok(X)) → HEAD(X)

The TRS R consists of the following rules:

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

Recursive Path Order [RPO].
Precedence:

proper > [HEAD₁, ok₁] > active > nil > [mark, top]
proper > [HEAD₁, ok₁] > active > zeros > [mark, top]
proper > [HEAD₁, ok₁] > active > 0
proper > nats > zeros > [mark, top]

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(mark(X)) → ADX(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ADX(x1) = ADX(x1)
ok(x1) = x1
mark(x1) = mark(x1)
active(x1) = active(x1)
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) = x1
tail(x1) = x1
proper(x1) = x1
top(x1) = top

Recursive Path Order [RPO].
Precedence:

ADX₁ > [mark₁, nats, zeros]
[active₁, nil] > cons₂ > [mark₁, nats, zeros]
[active₁, nil] > 0 > [mark₁, nats, zeros]
top > [mark₁, nats, 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))

(21) Obligation:

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

ADX(ok(X)) → ADX(X)

The TRS R consists of the following rules:

active(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(ok(X)) → ADX(X)

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

Recursive Path Order [RPO].
Precedence:

proper > [ADX₁, ok₁] > active > nil > [mark, top]
proper > [ADX₁, ok₁] > active > zeros > [mark, top]
proper > [ADX₁, ok₁] > active > 0
proper > nats > zeros > [mark, top]

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(mark(X)) → S(X)

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

Recursive Path Order [RPO].
Precedence:

S₁ > [mark₁, nats, zeros]
[active₁, nil] > cons₂ > [mark₁, nats, zeros]
[active₁, nil] > 0 > [mark₁, nats, zeros]
top > [mark₁, nats, 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))

(28) Obligation:

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

S(ok(X)) → S(X)

The TRS R consists of the following rules:

active(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(ok(X)) → S(X)

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

Recursive Path Order [RPO].
Precedence:

proper > [S₁, ok₁] > active > nil > [mark, top]
proper > [S₁, ok₁] > active > zeros > [mark, top]
proper > [S₁, ok₁] > active > 0
proper > nats > zeros > [mark, top]

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(mark(X1), X2) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = CONS(x1, x2)
ok(x1) = 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) = x1
tail(x1) = x1
proper(x1) = x1
top(x1) = top

Recursive Path Order [RPO].
Precedence:

active₁ > nil > mark₁ > CONS₂
active₁ > nil > mark₁ > top
active₁ > cons₂ > mark₁ > CONS₂
active₁ > cons₂ > mark₁ > top
active₁ > zeros > mark₁ > CONS₂
active₁ > zeros > mark₁ > top
active₁ > 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))

(35) Obligation:

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

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

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

Recursive Path Order [RPO].
Precedence:

top > active₁ > [CONS₁, ok₁, nil, zeros] > 0 > mark
top > proper > nats > [CONS₁, ok₁, nil, zeros] > 0 > 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))

(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(mark(X)) → INCR(X)

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

Recursive Path Order [RPO].
Precedence:

INCR₁ > [mark₁, nats, zeros]
[active₁, nil] > cons₂ > [mark₁, nats, zeros]
[active₁, nil] > 0 > [mark₁, nats, zeros]
top > [mark₁, nats, 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))

(42) Obligation:

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

INCR(ok(X)) → INCR(X)

The TRS R consists of the following rules:

active(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(ok(X)) → INCR(X)

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

Recursive Path Order [RPO].
Precedence:

proper > [INCR₁, ok₁] > active > nil > [mark, top]
proper > [INCR₁, ok₁] > active > zeros > [mark, top]
proper > [INCR₁, ok₁] > active > 0
proper > nats > zeros > [mark, top]

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)
PROPER(tail(X)) → PROPER(X)

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

Recursive Path Order [RPO].
Precedence:

[active₁, zeros] > cons₂ > [nil, mark₁]
[active₁, zeros] > tail₁ > [nil, mark₁]
[active₁, 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))

(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)

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

Recursive Path Order [RPO].
Precedence:

[head₁, active₁] > cons₂ > [PROPER₁, adx₁] > ok
[head₁, active₁] > 0 > ok
nil > ok
nats > zeros > cons₂ > [PROPER₁, adx₁] > ok
nats > zeros > 0 > ok
top > ok

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)

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

Recursive Path Order [RPO].
Precedence:

PROPER₁ > [mark, cons₂, head₁]
[adx, nats] > [incr₁, active₁, nil, zeros] > [mark, cons₂, head₁]
0 > [mark, cons₂, head₁]
top > [mark, cons₂, head₁]

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(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.

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

Recursive Path Order [RPO].
Precedence:

[s₁, active₁, zeros] > [nil, mark, cons₁, head] > incr₁ > PROPER₁
[s₁, active₁, zeros] > [nil, mark, cons₁, head] > top > PROPER₁
[s₁, active₁, zeros] > 0 > PROPER₁
nats > 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))

(55) 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.

(56) PisEmptyProof (EQUIVALENT transformation)

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

(57) TRUE

(58) 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.

(59) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(cons(X1, X2)) → ACTIVE(X1)

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

Recursive Path Order [RPO].
Precedence:

active₁ > nil
active₁ > zeros > cons₂ > ACTIVE₁
active₁ > 0
nats > zeros > cons₂ > ACTIVE₁

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))

(60) Obligation:

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

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.

(61) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(tail(X)) → ACTIVE(X)

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

Recursive Path Order [RPO].
Precedence:

proper₁ > [ACTIVE₁, tail₁] > [mark, ok, top]
proper₁ > nil > [mark, ok, top]
proper₁ > cons > [mark, ok, top]
proper₁ > [nats, zeros] > [mark, ok, top]
proper₁ > [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))

(62) Obligation:

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

ACTIVE(incr(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(adx(X)) → ACTIVE(X)
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.

(63) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

Recursive Path Order [RPO].
Precedence:

[head₁, active₁] > cons₂ > [ACTIVE₁, adx₁] > ok
[head₁, active₁] > 0 > ok
nil > ok
nats > zeros > cons₂ > [ACTIVE₁, adx₁] > ok
nats > zeros > 0 > ok
top > ok

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(incr(X)) → ACTIVE(X)
ACTIVE(s(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(incr(X)) → ACTIVE(X)

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

Recursive Path Order [RPO].
Precedence:

ACTIVE₁ > [mark, cons₂, head₁]
[adx, nats] > [incr₁, active₁, nil, zeros] > [mark, cons₂, head₁]
0 > [mark, cons₂, head₁]
top > [mark, cons₂, head₁]

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)

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)

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

Recursive Path Order [RPO].
Precedence:

[s₁, active₁, zeros] > [nil, mark, cons₁, head] > incr₁ > ACTIVE₁
[s₁, active₁, zeros] > [nil, mark, cons₁, head] > top > ACTIVE₁
[s₁, active₁, zeros] > 0 > ACTIVE₁
nats > ACTIVE₁

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:
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.

(69) PisEmptyProof (EQUIVALENT transformation)

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

(70) TRUE

(71) 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.