Termination w.r.t. Q proof of /tmp/tmp5iQT1E/Ex15_Luc98

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 PisEmptyProof (⇔)

    ↳9 TRUE

  ↳10 QDP

    ↳11 QDPOrderProof (⇔)

    ↳12 QDP

    ↳13 PisEmptyProof (⇔)

    ↳14 TRUE

  ↳15 QDP

    ↳16 QDPOrderProof (⇔)

    ↳17 QDP

    ↳18 PisEmptyProof (⇔)

    ↳19 TRUE

  ↳20 QDP

    ↳21 QDPOrderProof (⇔)

    ↳22 QDP

    ↳23 QDPOrderProof (⇔)

    ↳24 QDP

    ↳25 QDPOrderProof (⇔)

    ↳26 QDP

    ↳27 PisEmptyProof (⇔)

    ↳28 TRUE

  ↳29 QDP

    ↳30 QDPOrderProof (⇔)

    ↳31 QDP

    ↳32 QDPOrderProof (⇔)

    ↳33 QDP

    ↳34 PisEmptyProof (⇔)

    ↳35 TRUE

  ↳36 QDP

    ↳37 QDPOrderProof (⇔)

    ↳38 QDP

    ↳39 QDPOrderProof (⇔)

    ↳40 QDP

    ↳41 PisEmptyProof (⇔)

    ↳42 TRUE

  ↳43 QDP

    ↳44 QDPOrderProof (⇔)

    ↳45 QDP

    ↳46 QDPOrderProof (⇔)

    ↳47 QDP

    ↳48 PisEmptyProof (⇔)

    ↳49 TRUE

  ↳50 QDP

    ↳51 QDPOrderProof (⇔)

    ↳52 QDP

    ↳53 QDPOrderProof (⇔)

    ↳54 QDP

    ↳55 QDPOrderProof (⇔)

    ↳56 QDP

    ↳57 PisEmptyProof (⇔)

    ↳58 TRUE

  ↳59 QDP

    ↳60 QDPOrderProof (⇔)

    ↳61 QDP

    ↳62 QDPOrderProof (⇔)

    ↳63 QDP

    ↳64 QDPOrderProof (⇔)

    ↳65 QDP

    ↳66 PisEmptyProof (⇔)

    ↳67 TRUE

  ↳68 QDP

    ↳69 QDPOrderProof (⇔)

    ↳70 QDP

    ↳71 QDPOrderProof (⇔)

    ↳72 QDP

    ↳73 PisEmptyProof (⇔)

    ↳74 TRUE

(0) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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(add(s(X), Y)) → S(add(X, Y))
ACTIVE(add(s(X), Y)) → ADD(X, Y)
ACTIVE(first(s(X), cons(Y, Z))) → CONS(Y, first(X, Z))
ACTIVE(first(s(X), cons(Y, Z))) → FIRST(X, Z)
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(and(X1, X2)) → AND(active(X1), X2)
ACTIVE(and(X1, X2)) → ACTIVE(X1)
ACTIVE(if(X1, X2, X3)) → IF(active(X1), X2, X3)
ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(add(X1, X2)) → ADD(active(X1), X2)
ACTIVE(add(X1, X2)) → ACTIVE(X1)
ACTIVE(first(X1, X2)) → FIRST(active(X1), X2)
ACTIVE(first(X1, X2)) → ACTIVE(X1)
ACTIVE(first(X1, X2)) → FIRST(X1, active(X2))
ACTIVE(first(X1, X2)) → ACTIVE(X2)
AND(mark(X1), X2) → AND(X1, X2)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
ADD(mark(X1), X2) → ADD(X1, X2)
FIRST(mark(X1), X2) → FIRST(X1, X2)
FIRST(X1, mark(X2)) → FIRST(X1, X2)
PROPER(and(X1, X2)) → AND(proper(X1), proper(X2))
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(and(X1, X2)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → IF(proper(X1), proper(X2), proper(X3))
PROPER(if(X1, X2, X3)) → PROPER(X1)
PROPER(if(X1, X2, X3)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → PROPER(X3)
PROPER(add(X1, X2)) → ADD(proper(X1), proper(X2))
PROPER(add(X1, X2)) → PROPER(X1)
PROPER(add(X1, X2)) → PROPER(X2)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(first(X1, X2)) → FIRST(proper(X1), proper(X2))
PROPER(first(X1, X2)) → PROPER(X1)
PROPER(first(X1, X2)) → PROPER(X2)
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)
AND(ok(X1), ok(X2)) → AND(X1, X2)
IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)
ADD(ok(X1), ok(X2)) → ADD(X1, X2)
S(ok(X)) → S(X)
FIRST(ok(X1), ok(X2)) → FIRST(X1, X2)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
FROM(ok(X)) → FROM(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(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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 10 SCCs with 21 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

FROM(ok(X)) → FROM(X)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

FROM(ok(X)) → FROM(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FROM(x1) = FROM(x1)
ok(x1) = ok(x1)
active(x1) = active(x1)
and(x1, x2) = x1
true = true
mark(x1) = mark
false = false
if(x1, x2, x3) = x1
add(x1, x2) = x1
0 = 0
s(x1) = x1
first(x1, x2) = first(x1, x2)
nil = nil
cons(x1, x2) = x1
from(x1) = from(x1)
proper(x1) = proper(x1)
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

false > mark > proper₁ > true
false > mark > proper₁ > 0
false > mark > proper₁ > first₂ > ok₁
false > mark > proper₁ > first₂ > nil
false > mark > proper₁ > from₁ > ok₁
top > active₁ > mark > proper₁ > true
top > active₁ > mark > proper₁ > 0
top > active₁ > mark > proper₁ > first₂ > ok₁
top > active₁ > mark > proper₁ > first₂ > nil
top > active₁ > mark > proper₁ > from₁ > ok₁

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(7) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(11) 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) = x1
ok(x1) = ok(x1)
active(x1) = active(x1)
and(x1, x2) = and(x1, x2)
true = true
mark(x1) = mark
false = false
if(x1, x2, x3) = x1
add(x1, x2) = x1
0 = 0
s(x1) = x1
first(x1, x2) = first(x1, x2)
nil = nil
cons(x1, x2) = x1
from(x1) = from(x1)
proper(x1) = proper(x1)
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

active₁ > from₁ > mark > and₂ > ok₁
active₁ > from₁ > mark > first₂ > ok₁
active₁ > from₁ > mark > top > ok₁
false > mark > and₂ > ok₁
false > mark > first₂ > ok₁
false > mark > top > ok₁
nil > ok₁
proper₁ > true > mark > and₂ > ok₁
proper₁ > true > mark > first₂ > ok₁
proper₁ > true > mark > top > ok₁
proper₁ > 0 > ok₁
proper₁ > from₁ > mark > and₂ > ok₁
proper₁ > from₁ > mark > first₂ > ok₁
proper₁ > from₁ > mark > top > ok₁

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(12) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

S(ok(X)) → S(X)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(ok(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1) = S(x1)
ok(x1) = ok(x1)
active(x1) = active(x1)
and(x1, x2) = x1
true = true
mark(x1) = mark
false = false
if(x1, x2, x3) = x1
add(x1, x2) = x1
0 = 0
s(x1) = x1
first(x1, x2) = first(x1, x2)
nil = nil
cons(x1, x2) = x1
from(x1) = from(x1)
proper(x1) = proper(x1)
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

false > mark > proper₁ > true
false > mark > proper₁ > 0
false > mark > proper₁ > first₂ > ok₁
false > mark > proper₁ > first₂ > nil
false > mark > proper₁ > from₁ > ok₁
top > active₁ > mark > proper₁ > true
top > active₁ > mark > proper₁ > 0
top > active₁ > mark > proper₁ > first₂ > ok₁
top > active₁ > mark > proper₁ > first₂ > nil
top > active₁ > mark > proper₁ > from₁ > ok₁

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(17) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) Obligation:

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

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

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FIRST(X1, mark(X2)) → FIRST(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FIRST(x1, x2) = FIRST(x2)
mark(x1) = mark(x1)
ok(x1) = x1
active(x1) = x1
and(x1, x2) = and(x1, x2)
true = true
false = false
if(x1, x2, x3) = if(x1, x2, x3)
add(x1, x2) = add(x1, x2)
0 = 0
s(x1) = s(x1)
first(x1, x2) = first(x1, x2)
nil = nil
cons(x1, x2) = cons
from(x1) = from
proper(x1) = x1
top(x1) = x1

Lexicographic Path Order [LPO].
Precedence:

FIRST₁ > nil
and₂ > mark₁ > nil
and₂ > false > nil
true > mark₁ > nil
if₃ > mark₁ > nil
add₂ > s₁ > first₂ > mark₁ > nil
add₂ > s₁ > first₂ > cons > nil
0 > nil
from > s₁ > first₂ > mark₁ > nil
from > s₁ > first₂ > cons > nil

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(22) Obligation:

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

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

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FIRST(ok(X1), ok(X2)) → FIRST(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FIRST(x1, x2) = FIRST(x2)
mark(x1) = mark
ok(x1) = ok(x1)
active(x1) = active(x1)
and(x1, x2) = and(x1)
true = true
false = false
if(x1, x2, x3) = if(x1)
add(x1, x2) = x1
0 = 0
s(x1) = s(x1)
first(x1, x2) = first(x2)
nil = nil
cons(x1, x2) = cons(x1)
from(x1) = x1
proper(x1) = proper(x1)
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

FIRST₁ > nil
active₁ > and₁ > mark > top > nil
active₁ > and₁ > ok₁ > nil
active₁ > and₁ > false > nil
active₁ > if₁ > mark > top > nil
active₁ > if₁ > ok₁ > nil
active₁ > s₁ > ok₁ > nil
active₁ > cons₁ > first₁ > mark > top > nil
active₁ > cons₁ > first₁ > ok₁ > nil
proper₁ > and₁ > mark > top > nil
proper₁ > and₁ > ok₁ > nil
proper₁ > and₁ > false > nil
proper₁ > true > mark > top > nil
proper₁ > true > ok₁ > nil
proper₁ > if₁ > mark > top > nil
proper₁ > if₁ > ok₁ > nil
proper₁ > 0 > nil
proper₁ > s₁ > ok₁ > nil
proper₁ > cons₁ > first₁ > mark > top > nil
proper₁ > cons₁ > first₁ > ok₁ > nil

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(24) Obligation:

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

FIRST(mark(X1), X2) → FIRST(X1, X2)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

FIRST(mark(X1), X2) → FIRST(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FIRST(x1, x2) = x1
mark(x1) = mark(x1)
active(x1) = x1
and(x1, x2) = and(x1, x2)
true = true
false = false
if(x1, x2, x3) = if(x1, x2, x3)
add(x1, x2) = add(x1, x2)
0 = 0
s(x1) = s
first(x1, x2) = first(x1, x2)
nil = nil
cons(x1, x2) = cons
from(x1) = from
proper(x1) = x1
ok(x1) = x1
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

and₂ > mark₁ > nil
and₂ > false > nil
true > nil
if₃ > mark₁ > nil
0 > nil
s > add₂ > mark₁ > nil
s > first₂ > cons > mark₁ > nil
from > cons > mark₁ > nil
top > nil

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(26) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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) PisEmptyProof (EQUIVALENT transformation)

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

(28) TRUE

(29) Obligation:

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

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

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ADD(ok(X1), ok(X2)) → ADD(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ADD(x1, x2) = x2
ok(x1) = ok(x1)
mark(x1) = x1
active(x1) = x1
and(x1, x2) = and(x2)
true = true
false = false
if(x1, x2, x3) = if(x2, x3)
add(x1, x2) = add(x2)
0 = 0
s(x1) = x1
first(x1, x2) = first(x2)
nil = nil
cons(x1, x2) = x1
from(x1) = x1
proper(x1) = proper(x1)
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

true > ok₁ > false
proper₁ > and₁ > ok₁ > false
proper₁ > if₂ > ok₁ > false
proper₁ > add₁ > ok₁ > false
proper₁ > 0 > ok₁ > false
proper₁ > 0 > nil > false
proper₁ > first₁ > ok₁ > false
proper₁ > first₁ > nil > false
top > false

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(31) Obligation:

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

ADD(mark(X1), X2) → ADD(X1, X2)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

ADD(mark(X1), X2) → ADD(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ADD(x1, x2) = ADD(x1)
mark(x1) = mark(x1)
active(x1) = active(x1)
and(x1, x2) = and(x1, x2)
true = true
false = false
if(x1, x2, x3) = if(x1, x2, x3)
add(x1, x2) = add(x1, x2)
0 = 0
s(x1) = s
first(x1, x2) = first(x1, x2)
nil = nil
cons(x1, x2) = cons(x1)
from(x1) = x1
proper(x1) = x1
ok(x1) = x1
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

ADD₁ > nil
active₁ > and₂ > mark₁ > nil
active₁ > and₂ > false > nil
active₁ > if₃ > mark₁ > nil
active₁ > add₂ > mark₁ > nil
active₁ > s > mark₁ > nil
active₁ > s > cons₁ > nil
active₁ > first₂ > mark₁ > nil
active₁ > first₂ > cons₁ > nil
true > nil
0 > nil
top > nil

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(33) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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) PisEmptyProof (EQUIVALENT transformation)

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

(35) TRUE

(36) Obligation:

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

IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(37) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(mark(X1), X2, X3) → IF(X1, X2, X3)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
IF(x1, x2, x3) = IF(x1, x3)
ok(x1) = x1
mark(x1) = mark(x1)
active(x1) = x1
and(x1, x2) = and(x1, x2)
true = true
false = false
if(x1, x2, x3) = if(x1, x2, x3)
add(x1, x2) = add(x1, x2)
0 = 0
s(x1) = s(x1)
first(x1, x2) = first(x1, x2)
nil = nil
cons(x1, x2) = cons
from(x1) = from
proper(x1) = proper(x1)
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

IF₂ > nil
proper₁ > and₂ > mark₁ > nil
proper₁ > and₂ > false > nil
proper₁ > true > nil
proper₁ > if₃ > mark₁ > nil
proper₁ > add₂ > mark₁ > nil
proper₁ > add₂ > s₁ > nil
proper₁ > 0 > nil
proper₁ > first₂ > mark₁ > nil
proper₁ > first₂ > cons > nil
proper₁ > from > mark₁ > nil
proper₁ > from > s₁ > nil
proper₁ > from > cons > nil
top > nil

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(38) Obligation:

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

IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(39) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(ok(X1), ok(X2), ok(X3)) → IF(X1, X2, X3)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
IF(x1, x2, x3) = x3
ok(x1) = ok(x1)
active(x1) = active(x1)
and(x1, x2) = x1
true = true
mark(x1) = mark
false = false
if(x1, x2, x3) = x1
add(x1, x2) = x2
0 = 0
s(x1) = x1
first(x1, x2) = x1
nil = nil
cons(x1, x2) = x2
from(x1) = x1
proper(x1) = proper
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

top > proper > ok₁ > active₁ > mark
top > proper > true > mark
top > proper > false > mark
top > proper > 0 > mark
top > proper > nil > mark

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(40) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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) PisEmptyProof (EQUIVALENT transformation)

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

(42) TRUE

(43) Obligation:

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

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

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(44) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

AND(ok(X1), ok(X2)) → AND(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
AND(x1, x2) = x2
ok(x1) = ok(x1)
mark(x1) = x1
active(x1) = x1
and(x1, x2) = and(x2)
true = true
false = false
if(x1, x2, x3) = if(x2, x3)
add(x1, x2) = add(x2)
0 = 0
s(x1) = x1
first(x1, x2) = first(x2)
nil = nil
cons(x1, x2) = x1
from(x1) = x1
proper(x1) = proper(x1)
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

true > ok₁ > false
proper₁ > and₁ > ok₁ > false
proper₁ > if₂ > ok₁ > false
proper₁ > add₁ > ok₁ > false
proper₁ > 0 > ok₁ > false
proper₁ > 0 > nil > false
proper₁ > first₁ > ok₁ > false
proper₁ > first₁ > nil > false
top > false

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(45) Obligation:

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

AND(mark(X1), X2) → AND(X1, X2)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(46) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

AND(mark(X1), X2) → AND(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
AND(x1, x2) = AND(x1)
mark(x1) = mark(x1)
active(x1) = active(x1)
and(x1, x2) = and(x1, x2)
true = true
false = false
if(x1, x2, x3) = if(x1, x2, x3)
add(x1, x2) = add(x1, x2)
0 = 0
s(x1) = s
first(x1, x2) = first(x1, x2)
nil = nil
cons(x1, x2) = cons(x1)
from(x1) = x1
proper(x1) = x1
ok(x1) = x1
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

AND₁ > nil
active₁ > and₂ > mark₁ > nil
active₁ > and₂ > false > nil
active₁ > if₃ > mark₁ > nil
active₁ > add₂ > mark₁ > nil
active₁ > s > mark₁ > nil
active₁ > s > cons₁ > nil
active₁ > first₂ > mark₁ > nil
active₁ > first₂ > cons₁ > nil
true > nil
0 > nil
top > nil

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(47) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(48) PisEmptyProof (EQUIVALENT transformation)

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

(49) TRUE

(50) Obligation:

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

PROPER(and(X1, X2)) → PROPER(X2)
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(if(X1, X2, X3)) → PROPER(X1)
PROPER(if(X1, X2, X3)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → PROPER(X3)
PROPER(add(X1, X2)) → PROPER(X1)
PROPER(add(X1, X2)) → PROPER(X2)
PROPER(s(X)) → PROPER(X)
PROPER(first(X1, X2)) → PROPER(X1)
PROPER(first(X1, X2)) → PROPER(X2)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(from(X)) → PROPER(X)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(51) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(and(X1, X2)) → PROPER(X2)
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(if(X1, X2, X3)) → PROPER(X1)
PROPER(if(X1, X2, X3)) → PROPER(X2)
PROPER(if(X1, X2, X3)) → PROPER(X3)
PROPER(add(X1, X2)) → PROPER(X1)
PROPER(add(X1, X2)) → PROPER(X2)
PROPER(first(X1, X2)) → PROPER(X1)
PROPER(first(X1, X2)) → PROPER(X2)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1) = x1
and(x1, x2) = and(x1, x2)
if(x1, x2, x3) = if(x1, x2, x3)
add(x1, x2) = add(x1, x2)
s(x1) = x1
first(x1, x2) = first(x1, x2)
cons(x1, x2) = cons(x1, x2)
from(x1) = x1
active(x1) = x1
true = true
mark(x1) = mark
false = false
0 = 0
nil = nil
proper(x1) = x1
ok(x1) = ok
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

add₂ > ok > and₂ > mark
add₂ > ok > if₃ > mark
add₂ > ok > cons₂ > mark
add₂ > ok > top > mark
first₂ > nil > ok > and₂ > mark
first₂ > nil > ok > if₃ > mark
first₂ > nil > ok > cons₂ > mark
first₂ > nil > ok > top > mark
true > ok > and₂ > mark
true > ok > if₃ > mark
true > ok > cons₂ > mark
true > ok > top > mark
false > ok > and₂ > mark
false > ok > if₃ > mark
false > ok > cons₂ > mark
false > ok > top > mark
0 > nil > ok > and₂ > mark
0 > nil > ok > if₃ > mark
0 > nil > ok > cons₂ > mark
0 > nil > ok > top > mark

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(52) Obligation:

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

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

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(53) 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)
from(x1) = x1
active(x1) = active(x1)
and(x1, x2) = x1
true = true
mark(x1) = mark
false = false
if(x1, x2, x3) = x1
add(x1, x2) = x1
0 = 0
first(x1, x2) = first(x2)
nil = nil
cons(x1, x2) = x2
proper(x1) = proper(x1)
ok(x1) = ok(x1)
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

PROPER₁ > nil
proper₁ > s₁ > first₁ > mark > nil
proper₁ > s₁ > first₁ > ok₁ > nil
proper₁ > true > mark > nil
proper₁ > true > ok₁ > nil
proper₁ > false > mark > nil
proper₁ > false > ok₁ > nil
proper₁ > 0 > nil
top > active₁ > s₁ > first₁ > mark > nil
top > active₁ > s₁ > first₁ > ok₁ > nil
top > active₁ > false > mark > nil
top > active₁ > false > ok₁ > nil

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(54) 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(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(55) 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) = PROPER(x1)
from(x1) = from(x1)
active(x1) = x1
and(x1, x2) = and(x2)
true = true
mark(x1) = x1
false = false
if(x1, x2, x3) = if(x2, x3)
add(x1, x2) = add(x1, x2)
0 = 0
s(x1) = s(x1)
first(x1, x2) = first(x2)
nil = nil
cons(x1, x2) = x1
proper(x1) = x1
ok(x1) = x1
top(x1) = top(x1)

Lexicographic Path Order [LPO].
Precedence:

PROPER₁ > nil
from₁ > nil
and₁ > false > nil
true > nil
if₂ > nil
add₂ > s₁ > first₁ > nil
0 > nil
top₁ > nil

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(56) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(57) PisEmptyProof (EQUIVALENT transformation)

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

(58) TRUE

(59) Obligation:

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

ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(and(X1, X2)) → ACTIVE(X1)
ACTIVE(add(X1, X2)) → ACTIVE(X1)
ACTIVE(first(X1, X2)) → ACTIVE(X1)
ACTIVE(first(X1, X2)) → ACTIVE(X2)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(60) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(if(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(first(X1, X2)) → ACTIVE(X1)
ACTIVE(first(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
if(x1, x2, x3) = if(x1, x2, x3)
and(x1, x2) = x1
add(x1, x2) = x1
first(x1, x2) = first(x1, x2)
active(x1) = x1
true = true
mark(x1) = mark
false = false
0 = 0
s(x1) = s
nil = nil
cons(x1, x2) = cons
from(x1) = from
proper(x1) = proper(x1)
ok(x1) = x1
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

false > mark
proper₁ > if₃ > mark
proper₁ > first₂ > nil
proper₁ > first₂ > cons > mark
proper₁ > true > mark
proper₁ > 0 > mark
proper₁ > s > cons > mark
proper₁ > from > cons > mark

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
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(and(X1, X2)) → ACTIVE(X1)
ACTIVE(add(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(62) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(and(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) = x1
and(x1, x2) = and(x1, x2)
add(x1, x2) = x1
active(x1) = active(x1)
true = true
mark(x1) = mark
false = false
if(x1, x2, x3) = if(x1, x2, x3)
0 = 0
s(x1) = s
first(x1, x2) = first(x2)
nil = nil
cons(x1, x2) = cons
from(x1) = from
proper(x1) = x1
ok(x1) = x1
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

active₁ > false
active₁ > if₃ > mark > and₂
active₁ > s > first₁ > mark > and₂
active₁ > s > first₁ > nil
active₁ > s > first₁ > cons
active₁ > from > mark > and₂
active₁ > from > cons

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(63) Obligation:

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

ACTIVE(add(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(64) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(add(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) = x1
add(x1, x2) = add(x1, x2)
active(x1) = x1
and(x1, x2) = and(x1, x2)
true = true
mark(x1) = mark(x1)
false = false
if(x1, x2, x3) = if(x1, x2, x3)
0 = 0
s(x1) = s
first(x1, x2) = first(x1, x2)
nil = nil
cons(x1, x2) = cons
from(x1) = from
proper(x1) = x1
ok(x1) = x1
top(x1) = top(x1)

Lexicographic Path Order [LPO].
Precedence:

and₂ > false > mark₁ > nil
true > mark₁ > nil
if₃ > mark₁ > nil
0 > nil
s > add₂ > mark₁ > nil
s > first₂ > mark₁ > nil
s > first₂ > cons > nil
from > mark₁ > nil
from > cons > nil
top₁ > nil

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(65) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(66) PisEmptyProof (EQUIVALENT transformation)

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

(67) TRUE

(68) 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(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(69) 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)
ok(x1) = x1
active(x1) = x1
mark(x1) = mark(x1)
proper(x1) = x1
and(x1, x2) = and(x1, x2)
true = true
false = false
if(x1, x2, x3) = if(x1, x2, x3)
add(x1, x2) = add(x1, x2)
0 = 0
s(x1) = s
first(x1, x2) = first(x1, x2)
nil = nil
cons(x1, x2) = x1
from(x1) = from(x1)
top(x1) = top(x1)

Lexicographic Path Order [LPO].
Precedence:

and₂ > mark₁ > top₁
and₂ > false
if₃ > mark₁ > top₁
s > add₂ > mark₁ > top₁
s > first₂ > mark₁ > top₁
s > first₂ > nil
from₁ > mark₁ > top₁

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(70) 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(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(71) 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
and(x1, x2) = x2
true = true
mark(x1) = mark
false = false
if(x1, x2, x3) = x1
add(x1, x2) = x2
0 = 0
s(x1) = s(x1)
first(x1, x2) = first(x2)
nil = nil
cons(x1, x2) = x2
from(x1) = x1
proper(x1) = proper(x1)
top(x1) = top

Lexicographic Path Order [LPO].
Precedence:

TOP₁ > mark
proper₁ > true > ok₁ > mark
proper₁ > false > mark
proper₁ > 0 > mark
proper₁ > s₁ > first₁ > ok₁ > mark
proper₁ > s₁ > first₁ > nil > mark
top > mark

The following usable rules [FROCOS05] were oriented:

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(72) Obligation:

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

active(and(true, X)) → mark(X)
active(and(false, Y)) → mark(false)
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(and(X1, X2)) → and(active(X1), X2)
active(if(X1, X2, X3)) → if(active(X1), X2, X3)
active(add(X1, X2)) → add(active(X1), X2)
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
and(mark(X1), X2) → mark(and(X1, X2))
if(mark(X1), X2, X3) → mark(if(X1, X2, X3))
add(mark(X1), X2) → mark(add(X1, X2))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(true) → ok(true)
proper(false) → ok(false)
proper(if(X1, X2, X3)) → if(proper(X1), proper(X2), proper(X3))
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
if(ok(X1), ok(X2), ok(X3)) → ok(if(X1, X2, X3))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
s(ok(X)) → ok(s(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(73) 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) PisEmptyProof (EQUIVALENT transformation)

(9) TRUE

(10) Obligation:

(11) QDPOrderProof (EQUIVALENT transformation)

(12) Obligation:

(13) PisEmptyProof (EQUIVALENT transformation)

(14) TRUE

(15) Obligation:

(16) QDPOrderProof (EQUIVALENT transformation)

(17) Obligation:

(18) PisEmptyProof (EQUIVALENT transformation)

(19) TRUE

(20) Obligation:

(21) QDPOrderProof (EQUIVALENT transformation)

(22) Obligation:

(23) QDPOrderProof (EQUIVALENT transformation)

(24) Obligation:

(25) QDPOrderProof (EQUIVALENT transformation)

(26) Obligation:

(27) PisEmptyProof (EQUIVALENT transformation)

(28) TRUE

(29) Obligation:

(30) QDPOrderProof (EQUIVALENT transformation)

(31) Obligation:

(32) QDPOrderProof (EQUIVALENT transformation)

(33) Obligation:

(34) PisEmptyProof (EQUIVALENT transformation)

(35) TRUE

(36) Obligation:

(37) QDPOrderProof (EQUIVALENT transformation)

(38) Obligation:

(39) QDPOrderProof (EQUIVALENT transformation)

(40) Obligation:

(41) PisEmptyProof (EQUIVALENT transformation)

(42) TRUE

(43) Obligation:

(44) QDPOrderProof (EQUIVALENT transformation)

(45) Obligation:

(46) QDPOrderProof (EQUIVALENT transformation)

(47) Obligation:

(48) PisEmptyProof (EQUIVALENT transformation)

(49) TRUE

(50) Obligation:

(51) QDPOrderProof (EQUIVALENT transformation)

(52) Obligation:

(53) QDPOrderProof (EQUIVALENT transformation)

(54) Obligation:

(55) QDPOrderProof (EQUIVALENT transformation)

(56) Obligation:

(57) PisEmptyProof (EQUIVALENT transformation)

(58) TRUE

(59) Obligation:

(60) QDPOrderProof (EQUIVALENT transformation)

(61) Obligation:

(62) QDPOrderProof (EQUIVALENT transformation)

(63) Obligation:

(64) QDPOrderProof (EQUIVALENT transformation)

(65) Obligation:

(66) PisEmptyProof (EQUIVALENT transformation)

(67) TRUE

(68) Obligation:

(69) QDPOrderProof (EQUIVALENT transformation)

(70) Obligation:

(71) QDPOrderProof (EQUIVALENT transformation)

(72) Obligation:

(73) PisEmptyProof (EQUIVALENT transformation)

(74) TRUE