Termination w.r.t. Q proof of /tmp/tmpklCjgN/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 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 QDPOrderProof (⇔)

    ↳18 QDP

    ↳19 QDPOrderProof (⇔)

    ↳20 QDP

    ↳21 PisEmptyProof (⇔)

    ↳22 TRUE

  ↳23 QDP

    ↳24 QDPOrderProof (⇔)

    ↳25 QDP

    ↳26 QDPOrderProof (⇔)

    ↳27 QDP

    ↳28 QDPOrderProof (⇔)

    ↳29 QDP

    ↳30 QDPOrderProof (⇔)

    ↳31 QDP

    ↳32 PisEmptyProof (⇔)

    ↳33 TRUE

  ↳34 QDP

    ↳35 QDPOrderProof (⇔)

    ↳36 QDP

    ↳37 QDPOrderProof (⇔)

    ↳38 QDP

    ↳39 PisEmptyProof (⇔)

    ↳40 TRUE

  ↳41 QDP

    ↳42 QDPOrderProof (⇔)

    ↳43 QDP

    ↳44 QDPOrderProof (⇔)

    ↳45 QDP

    ↳46 QDPOrderProof (⇔)

    ↳47 QDP

    ↳48 QDPOrderProof (⇔)

    ↳49 QDP

    ↳50 PisEmptyProof (⇔)

    ↳51 TRUE

  ↳52 QDP

    ↳53 QDPOrderProof (⇔)

    ↳54 QDP

    ↳55 QDPOrderProof (⇔)

    ↳56 QDP

    ↳57 QDPOrderProof (⇔)

    ↳58 QDP

    ↳59 QDPOrderProof (⇔)

    ↳60 QDP

    ↳61 QDPOrderProof (⇔)

    ↳62 QDP

    ↳63 QDPOrderProof (⇔)

    ↳64 QDP

    ↳65 PisEmptyProof (⇔)

    ↳66 TRUE

  ↳67 QDP

    ↳68 QDPOrderProof (⇔)

    ↳69 QDP

    ↳70 QDPOrderProof (⇔)

    ↳71 QDP

    ↳72 QDPOrderProof (⇔)

    ↳73 QDP

    ↳74 QDPOrderProof (⇔)

    ↳75 QDP

    ↳76 PisEmptyProof (⇔)

    ↳77 TRUE

  ↳78 QDP

    ↳79 QDPOrderProof (⇔)

    ↳80 QDP

    ↳81 PisEmptyProof (⇔)

    ↳82 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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(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(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(add(s(X), Y)) → S(add(X, Y))
ACTIVE(add(s(X), Y)) → ADD(X, Y)
ACTIVE(first(0, X)) → MARK(nil)
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
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)) → MARK(cons(X, from(s(X))))
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
MARK(and(X1, X2)) → AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(true) → ACTIVE(true)
MARK(false) → ACTIVE(false)
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))
MARK(if(X1, X2, X3)) → IF(mark(X1), X2, X3)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), X2))
MARK(add(X1, X2)) → ADD(mark(X1), X2)
MARK(add(X1, X2)) → MARK(X1)
MARK(0) → ACTIVE(0)
MARK(s(X)) → ACTIVE(s(X))
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
MARK(first(X1, X2)) → FIRST(mark(X1), mark(X2))
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
MARK(nil) → ACTIVE(nil)
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
MARK(from(X)) → ACTIVE(from(X))
AND(mark(X1), X2) → AND(X1, X2)
AND(X1, mark(X2)) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
IF(X1, mark(X2), X3) → IF(X1, X2, X3)
IF(X1, X2, mark(X3)) → IF(X1, X2, X3)
IF(active(X1), X2, X3) → IF(X1, X2, X3)
IF(X1, active(X2), X3) → IF(X1, X2, X3)
IF(X1, X2, active(X3)) → IF(X1, X2, X3)
ADD(mark(X1), X2) → ADD(X1, X2)
ADD(X1, mark(X2)) → ADD(X1, X2)
ADD(active(X1), X2) → ADD(X1, X2)
ADD(X1, active(X2)) → ADD(X1, X2)
S(mark(X)) → S(X)
S(active(X)) → S(X)
FIRST(mark(X1), X2) → FIRST(X1, X2)
FIRST(X1, mark(X2)) → FIRST(X1, X2)
FIRST(active(X1), X2) → FIRST(X1, X2)
FIRST(X1, active(X2)) → FIRST(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
FROM(mark(X)) → FROM(X)
FROM(active(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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 8 SCCs with 17 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

FROM(active(X)) → FROM(X)
FROM(mark(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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(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(active(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)
active(x1) = active(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

active₁ > FROM₁

Status:

active₁: [1]
FROM₁: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

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

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FROM(mark(X)) → FROM(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
FROM(x1) = x1
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(9) 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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(active(X1), X2) → CONS(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]
CONS₁: [1]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, active(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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(mark(X1), X2) → CONS(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]
CONS₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(X1, active(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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(X1, active(X2)) → CONS(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]
CONS₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

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

CONS(X1, mark(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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(X1, mark(X2)) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > CONS₂

Status:

CONS₂: [1,2]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(20) 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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE

(23) 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(active(X1), X2) → FIRST(X1, X2)
FIRST(X1, active(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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FIRST(active(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) = FIRST(x1)
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

FIRST₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(25) 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(X1, active(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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(26) 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) = FIRST(x1)
mark(x1) = mark(x1)
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

FIRST₁: [1]
active₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(27) Obligation:

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

FIRST(X1, mark(X2)) → FIRST(X1, X2)
FIRST(X1, active(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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(28) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FIRST(X1, active(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(x1, x2)
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]
FIRST₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(29) Obligation:

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

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

FIRST(X1, mark(X2)) → FIRST(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > FIRST₂

Status:

FIRST₂: [1,2]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(31) 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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(32) PisEmptyProof (EQUIVALENT transformation)

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

(33) TRUE

(34) Obligation:

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

S(active(X)) → S(X)
S(mark(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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(active(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)
active(x1) = active(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

active₁ > S₁

Status:

active₁: [1]
S₁: [1]

The following usable rules [FROCOS05] were oriented: none

(36) Obligation:

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

S(mark(X)) → S(X)

The TRS R consists of the following rules:

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

S(mark(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1) = x1
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(38) 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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(39) PisEmptyProof (EQUIVALENT transformation)

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

(40) TRUE

(41) Obligation:

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

ADD(X1, mark(X2)) → ADD(X1, X2)
ADD(mark(X1), X2) → ADD(X1, X2)
ADD(active(X1), X2) → ADD(X1, X2)
ADD(X1, active(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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(42) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ADD(active(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) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]
ADD₁: [1]

The following usable rules [FROCOS05] were oriented: none

(43) Obligation:

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

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

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)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]
mark₁: [1]
ADD₁: [1]

The following usable rules [FROCOS05] were oriented: none

(45) Obligation:

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

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

ADD(X1, active(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, x2)
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]
ADD₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(47) Obligation:

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

ADD(X1, mark(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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(48) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ADD(X1, mark(X2)) → ADD(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > ADD₂

Status:

mark₁: [1]
ADD₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(49) 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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(50) PisEmptyProof (EQUIVALENT transformation)

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

(51) TRUE

(52) Obligation:

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

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

IF(X1, active(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(x2)
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]
IF₁: [1]

The following usable rules [FROCOS05] were oriented: none

(54) Obligation:

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

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

IF(X1, mark(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(x2)
mark(x1) = mark(x1)
active(x1) = active

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active: []
mark₁: [1]
IF₁: [1]

The following usable rules [FROCOS05] were oriented: none

(56) Obligation:

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

IF(mark(X1), X2, X3) → IF(X1, X2, X3)
IF(X1, X2, mark(X3)) → IF(X1, X2, X3)
IF(active(X1), X2, X3) → IF(X1, X2, X3)
IF(X1, X2, active(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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(57) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(X1, X2, mark(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
mark(x1) = mark(x1)
active(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(58) Obligation:

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

IF(mark(X1), X2, X3) → IF(X1, X2, X3)
IF(active(X1), X2, X3) → IF(X1, X2, X3)
IF(X1, X2, active(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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(59) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(X1, X2, active(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
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(60) Obligation:

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

IF(mark(X1), X2, X3) → IF(X1, X2, X3)
IF(active(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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(61) 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, x2)
mark(x1) = mark(x1)
active(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

mark₁ > IF₂

Status:

mark₁: [1]
IF₂: [2,1]

The following usable rules [FROCOS05] were oriented: none

(62) Obligation:

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

IF(active(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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(63) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

IF(active(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) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(64) 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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(65) PisEmptyProof (EQUIVALENT transformation)

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

(66) TRUE

(67) Obligation:

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

AND(X1, mark(X2)) → AND(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, active(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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(68) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

AND(active(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) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]
AND₁: [1]

The following usable rules [FROCOS05] were oriented: none

(69) Obligation:

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

AND(X1, mark(X2)) → AND(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
AND(X1, active(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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(70) 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)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]
AND₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(71) Obligation:

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

AND(X1, mark(X2)) → AND(X1, X2)
AND(X1, active(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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(72) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

AND(X1, active(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, x2)
mark(x1) = x1
active(x1) = active(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

active₁: [1]
AND₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(73) Obligation:

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

AND(X1, mark(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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(74) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

AND(X1, mark(X2)) → AND(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > AND₂

Status:

AND₂: [1,2]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(75) 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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(76) PisEmptyProof (EQUIVALENT transformation)

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

(77) TRUE

(78) Obligation:

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

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

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

(79) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

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

Lexicographic path order with status [LPO].
Precedence:

and₂ > MARK₁ > cons
true > MARK₁ > cons
if₃ > MARK₁ > cons
add₂ > s₁ > MARK₁ > cons
false > MARK₁ > cons
0 > MARK₁ > cons
first₂ > MARK₁ > cons
first₂ > nil > cons
from₁ > MARK₁ > cons

Status:

from₁: [1]
true: []
and₂: [1,2]
0: []
first₂: [2,1]
add₂: [1,2]
MARK₁: [1]
if₃: [3,1,2]
cons: []
false: []
s₁: [1]
nil: []

The following usable rules [FROCOS05] were oriented:

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

(80) 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))))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(add(X1, X2)) → active(add(mark(X1), X2))
mark(0) → active(0)
mark(s(X)) → active(s(X))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(from(X)) → active(from(X))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
add(mark(X1), X2) → add(X1, X2)
add(X1, mark(X2)) → add(X1, X2)
add(active(X1), X2) → add(X1, X2)
add(X1, active(X2)) → add(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)

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

(81) PisEmptyProof (EQUIVALENT transformation)

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