Termination w.r.t. Q proof of /tmp/tmpYtZMrA/Ex1_2_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 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 PisEmptyProof (⇔)

    ↳18 TRUE

  ↳19 QDP

    ↳20 QDPOrderProof (⇔)

    ↳21 QDP

    ↳22 PisEmptyProof (⇔)

    ↳23 TRUE

  ↳24 QDP

    ↳25 QDPOrderProof (⇔)

    ↳26 QDP

    ↳27 QDPOrderProof (⇔)

    ↳28 QDP

    ↳29 PisEmptyProof (⇔)

    ↳30 TRUE

  ↳31 QDP

    ↳32 QDPOrderProof (⇔)

    ↳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 QDPOrderProof (⇔)

    ↳46 QDP

    ↳47 PisEmptyProof (⇔)

    ↳48 TRUE

  ↳49 QDP

    ↳50 QDPOrderProof (⇔)

    ↳51 QDP

    ↳52 QDPOrderProof (⇔)

    ↳53 QDP

    ↳54 PisEmptyProof (⇔)

    ↳55 TRUE

(0) Obligation:

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

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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(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)
2ND(mark(X)) → 2ND(X)
CONS(mark(X1), X2) → CONS(X1, X2)
FROM(mark(X)) → FROM(X)
S(mark(X)) → S(X)
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)
2ND(ok(X)) → 2ND(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
FROM(ok(X)) → FROM(X)
S(ok(X)) → S(X)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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 13 less nodes.

(4) Complex Obligation (AND)

(5) 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(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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.

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

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

active₁ > 2nd₁ > [ok₁, from₁] > top
active₁ > cons₂ > [ok₁, from₁] > top

Status:

ok₁: [1]
active₁: [1]
2nd₁: [1]
cons₂: [1,2]
from₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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:

S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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.

S(mark(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1) = S(x1)
mark(x1) = mark(x1)
active(x1) = active(x1)
2nd(x1) = x1
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 with status [LPO].
Quasi-Precedence:

active₁ > [cons₂, proper₁] > [S₁, mark₁, s₁]
[ok, top] > [cons₂, proper₁] > [S₁, mark₁, s₁]

Status:

S₁: [1]
mark₁: [1]
active₁: [1]
cons₂: [1,2]
s₁: [1]
proper₁: [1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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:

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

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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.

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

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

active₁ > 2nd₁ > [ok₁, from₁] > top
active₁ > cons₂ > [ok₁, from₁] > top

Status:

ok₁: [1]
active₁: [1]
2nd₁: [1]
cons₂: [1,2]
from₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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:

FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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.

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) = FROM(x1)
mark(x1) = mark(x1)
active(x1) = active(x1)
2nd(x1) = x1
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 with status [LPO].
Quasi-Precedence:

active₁ > [cons₂, proper₁] > [FROM₁, mark₁, s₁]
[ok, top] > [cons₂, proper₁] > [FROM₁, mark₁, s₁]

Status:

FROM₁: [1]
mark₁: [1]
active₁: [1]
cons₂: [1,2]
s₁: [1]
proper₁: [1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

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

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

[proper₁, top] > active₁ > cons₂ > [ok₁, mark₁, from₁]

Status:

ok₁: [1]
mark₁: [1]
active₁: [1]
cons₂: [1,2]
from₁: [1]
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(21) Obligation:

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

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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) PisEmptyProof (EQUIVALENT transformation)

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

(23) TRUE

(24) 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(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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.

(25) 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) = 2nd(x1)
cons(x1, x2) = cons(x1, x2)
from(x1) = from(x1)
s(x1) = x1
proper(x1) = x1
top(x1) = top

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

active₁ > 2nd₁ > [ok₁, from₁] > top
active₁ > cons₂ > [ok₁, from₁] > top

Status:

ok₁: [1]
active₁: [1]
2nd₁: [1]
cons₂: [1,2]
from₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(26) 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(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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.

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) = 2ND(x1)
mark(x1) = mark(x1)
active(x1) = active(x1)
2nd(x1) = x1
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 with status [LPO].
Quasi-Precedence:

active₁ > [cons₂, proper₁] > [2ND₁, mark₁, s₁]
[ok, top] > [cons₂, proper₁] > [2ND₁, mark₁, s₁]

Status:

2ND₁: [1]
mark₁: [1]
active₁: [1]
cons₂: [1,2]
s₁: [1]
proper₁: [1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(28) Obligation:

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

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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) PisEmptyProof (EQUIVALENT transformation)

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

(30) TRUE

(31) 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)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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.

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

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

PROPER₁ > [cons₂, from₁, active₁, mark]
ok > top > [cons₂, from₁, active₁, mark]

Status:

PROPER₁: [1]
cons₂: [2,1]
from₁: [1]
active₁: [1]
mark: []
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(33) Obligation:

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

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

The TRS R consists of the following rules:

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

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

[cons, ok] > active₁ > [2nd₁, mark, top]
[cons, ok] > proper₁ > [2nd₁, mark, top]

Status:

PROPER₁: [1]
2nd₁: [1]
active₁: [1]
cons: []
mark: []
proper₁: [1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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
2nd(x1) = 2nd
cons(x1, x2) = cons
mark(x1) = mark
from(x1) = x1
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

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

cons > [proper₁, top] > [PROPER₁, s₁] > ok > mark
cons > [proper₁, top] > 2nd > ok > mark

Status:

PROPER₁: [1]
s₁: [1]
2nd: []
cons: []
mark: []
proper₁: [1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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(2nd(X)) → ACTIVE(X)
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)

The TRS R consists of the following rules:

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

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

ok > [ACTIVE₁, s₁, proper₁] > from₁ > [mark, top]

Status:

ACTIVE₁: [1]
from₁: [1]
s₁: [1]
mark: []
proper₁: [1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(2nd(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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(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)
2nd(x1) = x1
active(x1) = active(x1)
mark(x1) = mark
from(x1) = x1
s(x1) = s
proper(x1) = proper(x1)
ok(x1) = ok
top(x1) = top

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

active₁ > [s, ok] > [mark, proper₁] > [ACTIVE₁, cons₁]
top > [mark, proper₁] > [ACTIVE₁, cons₁]

Status:

ACTIVE₁: [1]
cons₁: [1]
active₁: [1]
mark: []
s: []
proper₁: [1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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:

ACTIVE(2nd(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

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

proper₁ > 2nd₁ > ACTIVE₁
proper₁ > 2nd₁ > mark
proper₁ > 2nd₁ > [ok, top]
proper₁ > [cons, from] > mark
proper₁ > [cons, from] > [ok, top]

Status:

ACTIVE₁: [1]
2nd₁: [1]
cons: []
mark: []
from: []
proper₁: [1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(46) Obligation:

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

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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) PisEmptyProof (EQUIVALENT transformation)

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

(48) TRUE

(49) 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(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(50) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

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

[ok, active] > mark > [proper, top]

Status:

ok: []
active: []
mark: []
proper: []
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(51) Obligation:

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

TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(52) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

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

[TOP₁, ok₁, 2nd₁, cons₁] > [mark, top]

Status:

TOP₁: [1]
ok₁: [1]
2nd₁: [1]
cons₁: [1]
mark: []
top: []

The following usable rules [FROCOS05] were oriented:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(53) Obligation:

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

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
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))
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))
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))
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))
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) 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) PisEmptyProof (EQUIVALENT transformation)

(11) TRUE

(12) Obligation:

(13) QDPOrderProof (EQUIVALENT transformation)

(14) Obligation:

(15) QDPOrderProof (EQUIVALENT transformation)

(16) Obligation:

(17) PisEmptyProof (EQUIVALENT transformation)

(18) TRUE

(19) Obligation:

(20) QDPOrderProof (EQUIVALENT transformation)

(21) Obligation:

(22) PisEmptyProof (EQUIVALENT transformation)

(23) TRUE

(24) Obligation:

(25) QDPOrderProof (EQUIVALENT transformation)

(26) Obligation:

(27) QDPOrderProof (EQUIVALENT transformation)

(28) Obligation:

(29) PisEmptyProof (EQUIVALENT transformation)

(30) TRUE

(31) Obligation:

(32) QDPOrderProof (EQUIVALENT transformation)

(33) Obligation:

(34) QDPOrderProof (EQUIVALENT transformation)

(35) Obligation:

(36) QDPOrderProof (EQUIVALENT transformation)

(37) Obligation:

(38) PisEmptyProof (EQUIVALENT transformation)

(39) TRUE

(40) Obligation:

(41) QDPOrderProof (EQUIVALENT transformation)

(42) Obligation:

(43) QDPOrderProof (EQUIVALENT transformation)

(44) Obligation:

(45) QDPOrderProof (EQUIVALENT transformation)

(46) Obligation:

(47) PisEmptyProof (EQUIVALENT transformation)

(48) TRUE

(49) Obligation:

(50) QDPOrderProof (EQUIVALENT transformation)

(51) Obligation:

(52) QDPOrderProof (EQUIVALENT transformation)

(53) Obligation:

(54) PisEmptyProof (EQUIVALENT transformation)

(55) TRUE