Termination w.r.t. Q proof of /tmp/tmpFtpJVu/ExProp7_Luc06

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 QDPOrderProof (⇔)

    ↳39 QDP

    ↳40 QDPOrderProof (⇔)

    ↳41 QDP

    ↳42 PisEmptyProof (⇔)

    ↳43 TRUE

  ↳44 QDP

    ↳45 QDPOrderProof (⇔)

    ↳46 QDP

    ↳47 QDPOrderProof (⇔)

    ↳48 QDP

    ↳49 QDPOrderProof (⇔)

    ↳50 QDP

    ↳51 QDPOrderProof (⇔)

    ↳52 QDP

    ↳53 PisEmptyProof (⇔)

    ↳54 TRUE

  ↳55 QDP

(0) Obligation:

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

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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(f(0)) → CONS(0, f(s(0)))
ACTIVE(f(0)) → F(s(0))
ACTIVE(f(0)) → S(0)
ACTIVE(f(s(0))) → F(p(s(0)))
ACTIVE(f(s(0))) → P(s(0))
ACTIVE(f(X)) → F(active(X))
ACTIVE(f(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(p(X)) → P(active(X))
ACTIVE(p(X)) → ACTIVE(X)
F(mark(X)) → F(X)
CONS(mark(X1), X2) → CONS(X1, X2)
S(mark(X)) → S(X)
P(mark(X)) → P(X)
PROPER(f(X)) → F(proper(X))
PROPER(f(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(p(X)) → P(proper(X))
PROPER(p(X)) → PROPER(X)
F(ok(X)) → F(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
S(ok(X)) → S(X)
P(ok(X)) → P(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(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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 7 SCCs with 15 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

P(ok(X)) → P(X)
P(mark(X)) → P(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

P(mark(X)) → P(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
P(x1) = x1
ok(x1) = x1
mark(x1) = mark(x1)
active(x1) = active(x1)
f(x1) = f(x1)
0 = 0
cons(x1, x2) = x1
s(x1) = x1
p(x1) = p(x1)
proper(x1) = x1
top(x1) = top

Lexicographic path order with status [LPO].
Quasi-Precedence:

[active₁, 0, p₁, top] > [mark₁, f₁]

Status:

active₁: [1]
f₁: [1]
mark₁: [1]
p₁: [1]
top: []
0: []

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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:

P(ok(X)) → P(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

P(ok(X)) → P(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
P(x1) = P(x1)
ok(x1) = ok(x1)
active(x1) = x1
f(x1) = x1
0 = 0
mark(x1) = x1
cons(x1, x2) = x2
s(x1) = x1
p(x1) = x1
proper(x1) = proper
top(x1) = top

Lexicographic path order with status [LPO].
Quasi-Precedence:

proper > ok₁
proper > 0

Status:

P₁: [1]
proper: []
ok₁: [1]
top: []
0: []

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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:

S(ok(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

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) = x1
ok(x1) = x1
mark(x1) = mark(x1)
active(x1) = active(x1)
f(x1) = f(x1)
0 = 0
cons(x1, x2) = x1
s(x1) = x1
p(x1) = p(x1)
proper(x1) = x1
top(x1) = top

Lexicographic path order with status [LPO].
Quasi-Precedence:

[active₁, 0, p₁, top] > [mark₁, f₁]

Status:

active₁: [1]
f₁: [1]
mark₁: [1]
p₁: [1]
top: []
0: []

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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:

S(ok(X)) → S(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

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) = x1
f(x1) = x1
0 = 0
mark(x1) = x1
cons(x1, x2) = x2
s(x1) = x1
p(x1) = x1
proper(x1) = proper
top(x1) = top

Lexicographic path order with status [LPO].
Quasi-Precedence:

proper > ok₁
proper > 0

Status:

proper: []
ok₁: [1]
top: []
0: []
S₁: [1]

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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:

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

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

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)
f(x1) = f(x1)
0 = 0
cons(x1, x2) = x1
s(x1) = s(x1)
p(x1) = x1
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Quasi-Precedence:

active₁ > s₁ > 0 > [mark₁, f₁] > CONS₂
proper₁ > s₁ > 0 > [mark₁, f₁] > CONS₂

Status:

active₁: [1]
f₁: [1]
CONS₂: [1,2]
mark₁: [1]
s₁: [1]
proper₁: [1]
top: []
0: []

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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:

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

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) = x2
ok(x1) = ok(x1)
active(x1) = active(x1)
f(x1) = f(x1)
0 = 0
mark(x1) = x1
cons(x1, x2) = cons(x1, x2)
s(x1) = s(x1)
p(x1) = p(x1)
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Quasi-Precedence:

[active₁, cons₂, s₁, p₁, proper₁, top] > [ok₁, f₁] > 0

Status:

active₁: [1]
cons₂: [1,2]
f₁: [1]
ok₁: [1]
s₁: [1]
p₁: [1]
proper₁: [1]
top: []
0: []

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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:

F(ok(X)) → F(X)
F(mark(X)) → F(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

F(mark(X)) → F(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1) = x1
ok(x1) = x1
mark(x1) = mark(x1)
active(x1) = active(x1)
f(x1) = f(x1)
0 = 0
cons(x1, x2) = x1
s(x1) = x1
p(x1) = p(x1)
proper(x1) = x1
top(x1) = top

Lexicographic path order with status [LPO].
Quasi-Precedence:

[active₁, 0, p₁, top] > [mark₁, f₁]

Status:

active₁: [1]
f₁: [1]
mark₁: [1]
p₁: [1]
top: []
0: []

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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:

F(ok(X)) → F(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

F(ok(X)) → F(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1) = F(x1)
ok(x1) = ok(x1)
active(x1) = x1
f(x1) = x1
0 = 0
mark(x1) = x1
cons(x1, x2) = x2
s(x1) = x1
p(x1) = x1
proper(x1) = proper
top(x1) = top

Lexicographic path order with status [LPO].
Quasi-Precedence:

proper > ok₁
proper > 0

Status:

proper: []
ok₁: [1]
top: []
0: []
F₁: [1]

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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:

PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(f(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(s(X)) → PROPER(X)
PROPER(p(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1) = PROPER(x1)
cons(x1, x2) = cons(x1, x2)
f(x1) = x1
s(x1) = x1
p(x1) = x1
active(x1) = active(x1)
0 = 0
mark(x1) = mark(x1)
proper(x1) = x1
ok(x1) = x1
top(x1) = top

Lexicographic path order with status [LPO].
Quasi-Precedence:

active₁ > cons₂ > mark₁
active₁ > 0

Status:

PROPER₁: [1]
cons₂: [2,1]
active₁: [1]
mark₁: [1]
top: []
0: []

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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:

PROPER(f(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(p(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

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) = x1
f(x1) = x1
s(x1) = s(x1)
p(x1) = x1
active(x1) = active(x1)
0 = 0
mark(x1) = mark(x1)
cons(x1, x2) = cons(x1)
proper(x1) = proper(x1)
ok(x1) = x1
top(x1) = top

Lexicographic path order with status [LPO].
Quasi-Precedence:

top > active₁ > [s₁, mark₁, cons₁]
top > proper₁ > 0 > [s₁, mark₁, cons₁]

Status:

active₁: [1]
cons₁: [1]
mark₁: [1]
s₁: [1]
proper₁: [1]
top: []
0: []

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(37) Obligation:

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

PROPER(f(X)) → PROPER(X)
PROPER(p(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(p(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)
f(x1) = x1
p(x1) = p(x1)
active(x1) = x1
0 = 0
mark(x1) = mark
cons(x1, x2) = x2
s(x1) = x1
proper(x1) = proper(x1)
ok(x1) = x1
top(x1) = top

Lexicographic path order with status [LPO].
Quasi-Precedence:

PROPER₁ > [p₁, 0, mark, proper₁]
top > [p₁, 0, mark, proper₁]

Status:

PROPER₁: [1]
mark: []
p₁: [1]
proper₁: [1]
top: []
0: []

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(39) Obligation:

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

PROPER(f(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

(40) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(f(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)
f(x1) = f(x1)
active(x1) = active(x1)
0 = 0
mark(x1) = x1
cons(x1, x2) = cons(x1, x2)
s(x1) = s(x1)
p(x1) = p(x1)
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Quasi-Precedence:

PROPER₁ > s₁
active₁ > [f₁, 0, p₁, proper₁, top] > [cons₂, ok] > s₁

Status:

PROPER₁: [1]
active₁: [1]
cons₂: [1,2]
f₁: [1]
s₁: [1]
p₁: [1]
ok: []
proper₁: [1]
top: []
0: []

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(41) Obligation:

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

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.

(42) PisEmptyProof (EQUIVALENT transformation)

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

(43) TRUE

(44) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(p(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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)
f(x1) = x1
s(x1) = x1
p(x1) = x1
active(x1) = active(x1)
0 = 0
mark(x1) = x1
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Quasi-Precedence:

active₁ > [0, proper₁] > ok > [ACTIVE₁, cons₂]
top > [ACTIVE₁, cons₂]

Status:

active₁: [1]
cons₂: [2,1]
ok: []
proper₁: [1]
top: []
0: []
ACTIVE₁: [1]

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(46) Obligation:

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

ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(p(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(47) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(s(X)) → ACTIVE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = x1
f(x1) = x1
s(x1) = s(x1)
p(x1) = x1
active(x1) = active(x1)
0 = 0
mark(x1) = mark(x1)
cons(x1, x2) = cons(x1)
proper(x1) = proper(x1)
ok(x1) = x1
top(x1) = top

Lexicographic path order with status [LPO].
Quasi-Precedence:

top > active₁ > [s₁, mark₁, cons₁]
top > proper₁ > 0 > [s₁, mark₁, cons₁]

Status:

active₁: [1]
cons₁: [1]
mark₁: [1]
s₁: [1]
proper₁: [1]
top: []
0: []

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(48) Obligation:

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

ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(p(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(49) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(p(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)
f(x1) = x1
p(x1) = p(x1)
active(x1) = x1
0 = 0
mark(x1) = mark
cons(x1, x2) = x2
s(x1) = x1
proper(x1) = proper(x1)
ok(x1) = x1
top(x1) = top

Lexicographic path order with status [LPO].
Quasi-Precedence:

ACTIVE₁ > [p₁, 0, mark, proper₁]
top > [p₁, 0, mark, proper₁]

Status:

mark: []
p₁: [1]
proper₁: [1]
top: []
0: []
ACTIVE₁: [1]

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(50) Obligation:

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

ACTIVE(f(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(51) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(f(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)
f(x1) = f(x1)
active(x1) = active(x1)
0 = 0
mark(x1) = x1
cons(x1, x2) = cons(x1, x2)
s(x1) = s(x1)
p(x1) = p(x1)
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Quasi-Precedence:

ACTIVE₁ > s₁
active₁ > [f₁, 0, p₁, proper₁, top] > [cons₂, ok] > s₁

Status:

active₁: [1]
cons₂: [1,2]
f₁: [1]
s₁: [1]
p₁: [1]
ok: []
proper₁: [1]
top: []
0: []
ACTIVE₁: [1]

The following usable rules [FROCOS05] were oriented:

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(52) Obligation:

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

active(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(53) PisEmptyProof (EQUIVALENT transformation)

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

(54) TRUE

(55) 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(f(0)) → mark(cons(0, f(s(0))))
active(f(s(0))) → mark(f(p(s(0))))
active(p(s(X))) → mark(X)
active(f(X)) → f(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(p(X)) → p(active(X))
f(mark(X)) → mark(f(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
p(mark(X)) → mark(p(X))
proper(f(X)) → f(proper(X))
proper(0) → ok(0)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(p(X)) → p(proper(X))
f(ok(X)) → ok(f(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
p(ok(X)) → ok(p(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.