Termination w.r.t. Q proof of /tmp/tmpVukwbX/Ex6_9_Luc02c

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 QDPOrderProof (⇔)

    ↳11 QDP

    ↳12 PisEmptyProof (⇔)

    ↳13 TRUE

  ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 QDPOrderProof (⇔)

    ↳18 QDP

    ↳19 PisEmptyProof (⇔)

    ↳20 TRUE

  ↳21 QDP

    ↳22 QDPOrderProof (⇔)

    ↳23 QDP

    ↳24 QDPOrderProof (⇔)

    ↳25 QDP

    ↳26 PisEmptyProof (⇔)

    ↳27 TRUE

  ↳28 QDP

    ↳29 QDPOrderProof (⇔)

    ↳30 QDP

    ↳31 QDPOrderProof (⇔)

    ↳32 QDP

    ↳33 PisEmptyProof (⇔)

    ↳34 TRUE

  ↳35 QDP

    ↳36 QDPOrderProof (⇔)

    ↳37 QDP

    ↳38 QDPOrderProof (⇔)

    ↳39 QDP

    ↳40 PisEmptyProof (⇔)

    ↳41 TRUE

  ↳42 QDP

    ↳43 QDPOrderProof (⇔)

    ↳44 QDP

    ↳45 QDPOrderProof (⇔)

    ↳46 QDP

    ↳47 QDPOrderProof (⇔)

    ↳48 QDP

    ↳49 QDPOrderProof (⇔)

    ↳50 QDP

    ↳51 PisEmptyProof (⇔)

    ↳52 TRUE

  ↳53 QDP

    ↳54 QDPOrderProof (⇔)

    ↳55 QDP

    ↳56 QDPOrderProof (⇔)

    ↳57 QDP

    ↳58 QDPOrderProof (⇔)

    ↳59 QDP

    ↳60 QDPOrderProof (⇔)

    ↳61 QDP

    ↳62 QDPOrderProof (⇔)

    ↳63 QDP

    ↳64 PisEmptyProof (⇔)

    ↳65 TRUE

  ↳66 QDP

    ↳67 QDPOrderProof (⇔)

    ↳68 QDP

    ↳69 QDPOrderProof (⇔)

    ↳70 QDP

    ↳71 PisEmptyProof (⇔)

    ↳72 TRUE

(0) Obligation:

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

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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(2nd(cons(X, X1))) → 2ND(cons1(X, X1))
ACTIVE(2nd(cons(X, X1))) → CONS1(X, X1)
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(2nd(X)) → 2ND(active(X))
ACTIVE(2nd(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(from(X)) → FROM(active(X))
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(cons1(X1, X2)) → CONS1(active(X1), X2)
ACTIVE(cons1(X1, X2)) → ACTIVE(X1)
ACTIVE(cons1(X1, X2)) → CONS1(X1, active(X2))
ACTIVE(cons1(X1, X2)) → ACTIVE(X2)
2ND(mark(X)) → 2ND(X)
CONS(mark(X1), X2) → CONS(X1, X2)
FROM(mark(X)) → FROM(X)
S(mark(X)) → S(X)
CONS1(mark(X1), X2) → CONS1(X1, X2)
CONS1(X1, mark(X2)) → CONS1(X1, X2)
PROPER(2nd(X)) → 2ND(proper(X))
PROPER(2nd(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(from(X)) → FROM(proper(X))
PROPER(from(X)) → PROPER(X)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(cons1(X1, X2)) → CONS1(proper(X1), proper(X2))
PROPER(cons1(X1, X2)) → PROPER(X1)
PROPER(cons1(X1, X2)) → PROPER(X2)
2ND(ok(X)) → 2ND(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
FROM(ok(X)) → FROM(X)
S(ok(X)) → S(X)
CONS1(ok(X1), ok(X2)) → CONS1(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(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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 8 SCCs with 18 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

CONS1(X1, mark(X2)) → CONS1(X1, X2)
CONS1(mark(X1), X2) → CONS1(X1, X2)
CONS1(ok(X1), ok(X2)) → CONS1(X1, X2)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

CONS1(mark(X1), X2) → CONS1(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS1(x1, x2) = CONS1(x1)
mark(x1) = mark(x1)
ok(x1) = x1
active(x1) = active(x1)
2nd(x1) = 2nd(x1)
cons1(x1, x2) = cons1(x1, x2)
cons(x1, x2) = cons(x1, x2)
from(x1) = x1
s(x1) = s(x1)
proper(x1) = x1
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

CONS1₁ > [mark₁, s₁, top]
active₁ > 2nd₁ > [cons1₂, cons₂] > [mark₁, s₁, top]

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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:

CONS1(X1, mark(X2)) → CONS1(X1, X2)
CONS1(ok(X1), ok(X2)) → CONS1(X1, X2)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

CONS1(ok(X1), ok(X2)) → CONS1(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS1(x1, x2) = x1
mark(x1) = mark
ok(x1) = ok(x1)
active(x1) = x1
2nd(x1) = 2nd(x1)
cons1(x1, x2) = cons1(x2)
cons(x1, x2) = cons(x1, x2)
from(x1) = from(x1)
s(x1) = x1
proper(x1) = x1
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

cons₂ > cons1₁ > [ok₁, 2nd₁, from₁] > mark
cons₂ > cons1₁ > [ok₁, 2nd₁, from₁] > top

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(9) Obligation:

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

CONS1(X1, mark(X2)) → CONS1(X1, X2)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS1(X1, mark(X2)) → CONS1(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS1(x1, x2) = CONS1(x2)
mark(x1) = mark(x1)
active(x1) = active(x1)
2nd(x1) = x1
cons1(x1, x2) = cons1(x1, x2)
cons(x1, x2) = cons(x1, x2)
from(x1) = from(x1)
s(x1) = x1
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

proper₁ > [active₁, cons1₂, cons₂, from₁, ok, top] > [CONS1₁, mark₁]

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(11) Obligation:

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

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) TRUE

(14) 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(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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(ok(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1) = x1
ok(x1) = ok(x1)
mark(x1) = x1
active(x1) = active(x1)
2nd(x1) = x1
cons1(x1, x2) = cons1(x1, x2)
cons(x1, x2) = cons(x1, x2)
from(x1) = from(x1)
s(x1) = s(x1)
proper(x1) = proper(x1)
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

top > proper₁ > [active₁, cons1₂, cons₂] > [from₁, s₁] > ok₁

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(16) 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(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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) 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)
2nd(x1) = x1
cons1(x1, x2) = cons1(x1, x2)
cons(x1, x2) = cons(x1, x2)
from(x1) = x1
s(x1) = s(x1)
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

[ok, top] > active₁ > cons1₂ > mark₁
[ok, top] > active₁ > cons₂ > mark₁
[ok, top] > active₁ > s₁ > mark₁
[ok, top] > proper₁ > cons1₂ > mark₁
[ok, top] > proper₁ > cons₂ > mark₁
[ok, top] > proper₁ > s₁ > mark₁

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(18) Obligation:

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

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE

(21) 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(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

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) = x1
ok(x1) = ok(x1)
mark(x1) = x1
active(x1) = active(x1)
2nd(x1) = x1
cons1(x1, x2) = cons1(x1, x2)
cons(x1, x2) = cons(x1, x2)
from(x1) = from(x1)
s(x1) = s(x1)
proper(x1) = proper(x1)
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

top > proper₁ > [active₁, cons1₂, cons₂] > [from₁, s₁] > ok₁

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(23) 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(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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) 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)
2nd(x1) = x1
cons1(x1, x2) = cons1(x1, x2)
cons(x1, x2) = cons(x1, x2)
from(x1) = x1
s(x1) = s(x1)
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

[ok, top] > active₁ > cons1₂ > mark₁
[ok, top] > active₁ > cons₂ > mark₁
[ok, top] > active₁ > s₁ > mark₁
[ok, top] > proper₁ > cons1₂ > mark₁
[ok, top] > proper₁ > cons₂ > mark₁
[ok, top] > proper₁ > s₁ > mark₁

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(25) Obligation:

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

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(26) PisEmptyProof (EQUIVALENT transformation)

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

(27) TRUE

(28) 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(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

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)
2nd(x1) = 2nd(x1)
cons1(x1, x2) = cons1(x1, x2)
cons(x1, x2) = cons(x1, x2)
from(x1) = x1
s(x1) = x1
proper(x1) = x1
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

CONS₂ > top
[active₁, 2nd₁, cons1₂] > cons₂ > mark₁ > top

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(30) 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(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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) 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, x2)
ok(x1) = ok(x1)
active(x1) = x1
2nd(x1) = x1
cons1(x1, x2) = x2
cons(x1, x2) = x2
mark(x1) = mark
from(x1) = x1
s(x1) = x1
proper(x1) = x1
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

ok₁ > CONS₂ > mark
ok₁ > top > mark

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(32) Obligation:

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

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(33) PisEmptyProof (EQUIVALENT transformation)

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

(34) TRUE

(35) Obligation:

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

2ND(ok(X)) → 2ND(X)
2ND(mark(X)) → 2ND(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

2ND(ok(X)) → 2ND(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
2ND(x1) = x1
ok(x1) = ok(x1)
mark(x1) = x1
active(x1) = active(x1)
2nd(x1) = x1
cons1(x1, x2) = cons1(x1, x2)
cons(x1, x2) = cons(x1, x2)
from(x1) = from(x1)
s(x1) = s(x1)
proper(x1) = proper(x1)
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

top > proper₁ > [active₁, cons1₂, cons₂] > [from₁, s₁] > ok₁

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
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:

2ND(mark(X)) → 2ND(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

2ND(mark(X)) → 2ND(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
2ND(x1) = x1
mark(x1) = mark(x1)
active(x1) = active(x1)
2nd(x1) = x1
cons1(x1, x2) = cons1(x1, x2)
cons(x1, x2) = cons(x1, x2)
from(x1) = x1
s(x1) = s(x1)
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

[ok, top] > active₁ > cons1₂ > mark₁
[ok, top] > active₁ > cons₂ > mark₁
[ok, top] > active₁ > s₁ > mark₁
[ok, top] > proper₁ > cons1₂ > mark₁
[ok, top] > proper₁ > cons₂ > mark₁
[ok, top] > proper₁ > s₁ > mark₁

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(39) Obligation:

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

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(40) PisEmptyProof (EQUIVALENT transformation)

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

(41) TRUE

(42) Obligation:

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

PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(2nd(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(from(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(cons1(X1, X2)) → PROPER(X1)
PROPER(cons1(X1, X2)) → PROPER(X2)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(cons1(X1, X2)) → PROPER(X1)
PROPER(cons1(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)
2nd(x1) = x1
from(x1) = x1
s(x1) = x1
cons1(x1, x2) = cons1(x1, x2)
active(x1) = active(x1)
mark(x1) = x1
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

PROPER₁ > [ok, top]
active₁ > cons₂ > [ok, top]
active₁ > cons1₂ > [ok, top]
proper₁ > cons₂ > [ok, top]
proper₁ > cons1₂ > [ok, top]

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(44) Obligation:

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

PROPER(2nd(X)) → PROPER(X)
PROPER(from(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(2nd(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)
2nd(x1) = 2nd(x1)
from(x1) = x1
s(x1) = x1
active(x1) = active(x1)
cons1(x1, x2) = cons1
cons(x1, x2) = x1
mark(x1) = mark
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

proper₁ > [2nd₁, active₁, cons1, ok, top] > PROPER₁
proper₁ > [2nd₁, active₁, cons1, ok, top] > mark

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
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:

PROPER(from(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(47) 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
from(x1) = x1
s(x1) = s(x1)
active(x1) = active(x1)
2nd(x1) = 2nd(x1)
cons1(x1, x2) = x2
cons(x1, x2) = cons(x1, x2)
mark(x1) = x1
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

[ok, top] > active₁ > s₁ > 2nd₁
[ok, top] > active₁ > cons₂ > 2nd₁
[ok, top] > proper₁ > s₁ > 2nd₁
[ok, top] > proper₁ > cons₂ > 2nd₁

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
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:

PROPER(from(X)) → PROPER(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(49) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(from(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
from(x1) = from(x1)
active(x1) = x1
2nd(x1) = 2nd
cons1(x1, x2) = x2
cons(x1, x2) = cons(x1, x2)
mark(x1) = mark
s(x1) = x1
proper(x1) = x1
ok(x1) = x1
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

from₁ > cons₂ > 2nd > [mark, top]

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(50) Obligation:

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

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(51) PisEmptyProof (EQUIVALENT transformation)

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

(52) TRUE

(53) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(2nd(X)) → ACTIVE(X)
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(cons1(X1, X2)) → ACTIVE(X1)
ACTIVE(cons1(X1, X2)) → ACTIVE(X2)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(54) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(cons1(X1, X2)) → ACTIVE(X1)
ACTIVE(cons1(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) = x1
2nd(x1) = x1
from(x1) = x1
s(x1) = x1
cons1(x1, x2) = cons1(x1, x2)
active(x1) = active(x1)
mark(x1) = mark
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

active₁ > [mark, ok, top] > proper₁ > cons1₂

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(55) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(2nd(X)) → ACTIVE(X)
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(56) 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)
2nd(x1) = x1
from(x1) = x1
s(x1) = x1
active(x1) = x1
cons1(x1, x2) = cons1
mark(x1) = mark
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

ACTIVE₁ > mark
proper₁ > [cons₂, cons1, ok, top] > mark

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(57) Obligation:

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

ACTIVE(2nd(X)) → ACTIVE(X)
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(58) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(2nd(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)
2nd(x1) = 2nd(x1)
from(x1) = x1
s(x1) = x1
active(x1) = active(x1)
cons1(x1, x2) = cons1
cons(x1, x2) = x1
mark(x1) = mark
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

proper₁ > [2nd₁, active₁, cons1, ok, top] > ACTIVE₁
proper₁ > [2nd₁, active₁, cons1, ok, top] > mark

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(59) 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(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(60) 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
from(x1) = x1
s(x1) = s(x1)
active(x1) = active(x1)
2nd(x1) = 2nd(x1)
cons1(x1, x2) = x2
cons(x1, x2) = cons(x1, x2)
mark(x1) = x1
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

[ok, top] > active₁ > s₁ > 2nd₁
[ok, top] > active₁ > cons₂ > 2nd₁
[ok, top] > proper₁ > s₁ > 2nd₁
[ok, top] > proper₁ > cons₂ > 2nd₁

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(61) Obligation:

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

ACTIVE(from(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(62) 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)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = x1
from(x1) = from(x1)
active(x1) = x1
2nd(x1) = 2nd
cons1(x1, x2) = x2
cons(x1, x2) = cons(x1, x2)
mark(x1) = mark
s(x1) = x1
proper(x1) = x1
ok(x1) = x1
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

from₁ > cons₂ > 2nd > [mark, top]

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(63) Obligation:

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

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(64) PisEmptyProof (EQUIVALENT transformation)

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

(65) TRUE

(66) 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(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(67) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TOP(ok(X)) → TOP(active(X))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TOP(x1) = x1
ok(x1) = ok(x1)
active(x1) = x1
mark(x1) = mark
proper(x1) = proper
2nd(x1) = x1
cons1(x1, x2) = x1
cons(x1, x2) = x2
from(x1) = x1
s(x1) = x1
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

ok₁ > [mark, proper]
top > [mark, proper]

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(68) Obligation:

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

TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(69) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TOP(mark(X)) → TOP(proper(X))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TOP(x1) = TOP(x1)
mark(x1) = mark(x1)
proper(x1) = x1
active(x1) = active(x1)
2nd(x1) = 2nd(x1)
cons1(x1, x2) = cons1(x1, x2)
cons(x1, x2) = cons(x1, x2)
from(x1) = from(x1)
s(x1) = s(x1)
ok(x1) = ok
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

[active₁, cons1₂, from₁] > cons₂ > [mark₁, 2nd₁, s₁] > [ok, top]

The following usable rules [FROCOS05] were oriented:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(70) Obligation:

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

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
active(cons1(X1, X2)) → cons1(active(X1), X2)
active(cons1(X1, X2)) → cons1(X1, active(X2))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
cons1(mark(X1), X2) → mark(cons1(X1, X2))
cons1(X1, mark(X2)) → mark(cons1(X1, X2))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
proper(cons1(X1, X2)) → cons1(proper(X1), proper(X2))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
cons1(ok(X1), ok(X2)) → ok(cons1(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.

(71) PisEmptyProof (EQUIVALENT transformation)

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

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Complex Obligation (AND)

(5) Obligation:

(6) QDPOrderProof (EQUIVALENT transformation)

(7) Obligation:

(8) QDPOrderProof (EQUIVALENT transformation)

(9) Obligation:

(10) QDPOrderProof (EQUIVALENT transformation)

(11) Obligation:

(12) PisEmptyProof (EQUIVALENT transformation)

(13) TRUE

(14) Obligation:

(15) QDPOrderProof (EQUIVALENT transformation)

(16) Obligation:

(17) QDPOrderProof (EQUIVALENT transformation)

(18) Obligation:

(19) PisEmptyProof (EQUIVALENT transformation)

(20) TRUE

(21) Obligation:

(22) QDPOrderProof (EQUIVALENT transformation)

(23) Obligation:

(24) QDPOrderProof (EQUIVALENT transformation)

(25) Obligation:

(26) PisEmptyProof (EQUIVALENT transformation)

(27) TRUE

(28) Obligation:

(29) QDPOrderProof (EQUIVALENT transformation)

(30) Obligation:

(31) QDPOrderProof (EQUIVALENT transformation)

(32) Obligation:

(33) PisEmptyProof (EQUIVALENT transformation)

(34) TRUE

(35) Obligation:

(36) QDPOrderProof (EQUIVALENT transformation)

(37) Obligation:

(38) QDPOrderProof (EQUIVALENT transformation)

(39) Obligation:

(40) PisEmptyProof (EQUIVALENT transformation)

(41) TRUE

(42) Obligation:

(43) QDPOrderProof (EQUIVALENT transformation)

(44) Obligation:

(45) QDPOrderProof (EQUIVALENT transformation)

(46) Obligation:

(47) QDPOrderProof (EQUIVALENT transformation)

(48) Obligation:

(49) QDPOrderProof (EQUIVALENT transformation)

(50) Obligation:

(51) PisEmptyProof (EQUIVALENT transformation)

(52) TRUE

(53) Obligation:

(54) QDPOrderProof (EQUIVALENT transformation)

(55) Obligation:

(56) QDPOrderProof (EQUIVALENT transformation)

(57) Obligation:

(58) QDPOrderProof (EQUIVALENT transformation)

(59) Obligation:

(60) QDPOrderProof (EQUIVALENT transformation)

(61) Obligation:

(62) QDPOrderProof (EQUIVALENT transformation)

(63) Obligation:

(64) PisEmptyProof (EQUIVALENT transformation)

(65) TRUE

(66) Obligation:

(67) QDPOrderProof (EQUIVALENT transformation)

(68) Obligation:

(69) QDPOrderProof (EQUIVALENT transformation)

(70) Obligation:

(71) PisEmptyProof (EQUIVALENT transformation)

(72) TRUE