Termination w.r.t. Q proof of /tmp/tmpI0OEYi/Ex3_12_Luc96a

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

(0) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(sel(s(X), cons(Y, Z))) → SEL(X, Z)
ACTIVE(from(X)) → FROM(active(X))
ACTIVE(from(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(sel(X1, X2)) → SEL(active(X1), X2)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → SEL(X1, active(X2))
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
FROM(mark(X)) → FROM(X)
CONS(mark(X1), X2) → CONS(X1, X2)
S(mark(X)) → S(X)
SEL(mark(X1), X2) → SEL(X1, X2)
SEL(X1, mark(X2)) → SEL(X1, X2)
PROPER(from(X)) → FROM(proper(X))
PROPER(from(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(sel(X1, X2)) → SEL(proper(X1), proper(X2))
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X2)
FROM(ok(X)) → FROM(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
S(ok(X)) → S(X)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
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(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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:

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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.

SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(mark(X1), X2) → SEL(X1, X2)

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

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

[active₁, sel₂] > cons₂ > [SEL₂, mark₁, top]
0 > [SEL₂, mark₁, top]

Status:

SEL₂: [2,1]
mark₁: [1]
active₁: [1]
cons₂: [1,2]
sel₂: [2,1]
0: []
top: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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:

SEL(ok(X1), ok(X2)) → SEL(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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.

SEL(ok(X1), ok(X2)) → SEL(X1, X2)

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

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

proper₁ > cons₁ > [ok₁, active₁, mark, 0, top]

Status:

ok₁: [1]
active₁: [1]
mark: []
cons₁: [1]
0: []
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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(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)
mark(x1) = x1
active(x1) = active(x1)
from(x1) = x1
cons(x1, x2) = cons(x1, x2)
s(x1) = x1
sel(x1, x2) = sel(x2)
0 = 0
proper(x1) = proper(x1)
top(x1) = top

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

S₁ > [ok₁, sel₁, top]
active₁ > cons₂ > [ok₁, sel₁, top]
proper₁ > cons₂ > [ok₁, sel₁, top]
proper₁ > 0 > [ok₁, sel₁, top]

Status:

S₁: [1]
ok₁: [1]
active₁: [1]
cons₂: [2,1]
sel₁: [1]
0: []
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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(mark(X)) → S(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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(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
mark(x1) = mark(x1)
active(x1) = active(x1)
from(x1) = x1
cons(x1, x2) = cons(x1, x2)
s(x1) = x1
sel(x1, x2) = sel(x1, x2)
0 = 0
proper(x1) = x1
ok(x1) = ok
top(x1) = top

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

active₁ > [cons₂, sel₂] > [mark₁, ok, top]
0 > [mark₁, ok, top]

Status:

mark₁: [1]
active₁: [1]
cons₂: [1,2]
sel₂: [1,2]
0: []
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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(ok(X1), ok(X2)) → CONS(X1, X2)

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

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

[ok₁, cons₁, proper₁] > [CONS₁, mark, active₁, 0, top]

Status:

CONS₁: [1]
ok₁: [1]
mark: []
active₁: [1]
cons₁: [1]
0: []
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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(mark(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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(mark(X1), X2) → CONS(X1, X2)

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

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

active₁ > [cons₂, sel₂] > [mark₁, ok, top]
0 > [mark₁, ok, top]

Status:

mark₁: [1]
active₁: [1]
cons₂: [1,2]
sel₂: [1,2]
0: []
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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:

FROM(ok(X)) → FROM(X)
FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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.

FROM(ok(X)) → FROM(X)

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

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

FROM₁ > [ok₁, sel₁, top]
active₁ > cons₂ > [ok₁, sel₁, top]
proper₁ > cons₂ > [ok₁, sel₁, top]
proper₁ > 0 > [ok₁, sel₁, top]

Status:

FROM₁: [1]
ok₁: [1]
active₁: [1]
cons₂: [2,1]
sel₁: [1]
0: []
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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:

FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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.

FROM(mark(X)) → FROM(X)

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

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

active₁ > [cons₂, sel₂] > [mark₁, ok, top]
0 > [mark₁, ok, top]

Status:

mark₁: [1]
active₁: [1]
cons₂: [1,2]
sel₂: [1,2]
0: []
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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(from(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(s(X)) → PROPER(X)
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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(from(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X2)

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

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

proper₁ > [cons₂, sel₂] > mark > [ok₁, top]
proper₁ > from₁ > mark > [ok₁, top]
proper₁ > 0 > mark > [ok₁, top]

Status:

cons₂: [2,1]
from₁: [1]
sel₂: [2,1]
mark: []
0: []
proper₁: [1]
ok₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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(s(X)) → PROPER(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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) = PROPER(x1)
s(x1) = s(x1)
active(x1) = x1
from(x1) = from(x1)
mark(x1) = mark
cons(x1, x2) = cons(x1, x2)
sel(x1, x2) = sel(x1, x2)
0 = 0
proper(x1) = x1
ok(x1) = ok
top(x1) = top

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

s₁ > PROPER₁ > [mark, sel₂, ok, top]
from₁ > [mark, sel₂, ok, top]
cons₂ > [mark, sel₂, ok, top]
0 > [mark, sel₂, ok, top]

Status:

PROPER₁: [1]
s₁: [1]
from₁: [1]
mark: []
cons₂: [2,1]
sel₂: [2,1]
0: []
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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:

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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.

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)

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

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

cons₁ > [sel₂, mark, top]
0 > [sel₂, mark, top]

Status:

cons₁: [1]
sel₂: [2,1]
mark: []
0: []
top: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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:

ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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.

ACTIVE(from(X)) → ACTIVE(X)
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)
from(x1) = from(x1)
s(x1) = s(x1)
active(x1) = x1
mark(x1) = mark
cons(x1, x2) = cons(x1, x2)
sel(x1, x2) = sel(x1, x2)
0 = 0
proper(x1) = x1
ok(x1) = ok
top(x1) = top

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

ACTIVE₁ > [s₁, mark, cons₂, sel₂, ok, top]
from₁ > [s₁, mark, cons₂, sel₂, ok, top]
0 > [s₁, mark, cons₂, sel₂, ok, top]

Status:

ACTIVE₁: [1]
from₁: [1]
s₁: [1]
mark: []
cons₂: [2,1]
sel₂: [2,1]
0: []
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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:

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
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.