Termination w.r.t. Q proof of /tmp/tmpaT6o2H/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 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 PisEmptyProof (⇔)

    ↳18 TRUE

  ↳19 QDP

    ↳20 QDPOrderProof (⇔)

    ↳21 QDP

    ↳22 QDPOrderProof (⇔)

    ↳23 QDP

    ↳24 QDPOrderProof (⇔)

    ↳25 QDP

    ↳26 QDPOrderProof (⇔)

    ↳27 QDP

    ↳28 PisEmptyProof (⇔)

    ↳29 TRUE

  ↳30 QDP

    ↳31 QDPOrderProof (⇔)

    ↳32 QDP

    ↳33 PisEmptyProof (⇔)

    ↳34 TRUE

  ↳35 QDP

    ↳36 QDPOrderProof (⇔)

    ↳37 QDP

    ↳38 QDPOrderProof (⇔)

    ↳39 QDP

    ↳40 QDPOrderProof (⇔)

    ↳41 QDP

    ↳42 QDPOrderProof (⇔)

    ↳43 QDP

    ↳44 PisEmptyProof (⇔)

    ↳45 TRUE

  ↳46 QDP

    ↳47 QDPOrderProof (⇔)

    ↳48 QDP

    ↳49 QDPOrderProof (⇔)

    ↳50 QDP

    ↳51 QDPOrderProof (⇔)

    ↳52 QDP

    ↳53 PisEmptyProof (⇔)

    ↳54 TRUE

  ↳55 QDP

    ↳56 QDPOrderProof (⇔)

    ↳57 QDP

    ↳58 QDPOrderProof (⇔)

    ↳59 QDP

    ↳60 QDPOrderProof (⇔)

    ↳61 QDP

    ↳62 PisEmptyProof (⇔)

    ↳63 TRUE

  ↳64 QDP

    ↳65 QDPOrderProof (⇔)

    ↳66 QDP

    ↳67 DependencyGraphProof (⇔)

    ↳68 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)
FROM(mark(X)) → FROM(X)

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

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

and₁ > [active₁, mark₁, add₁] > false
and₁ > [active₁, mark₁, add₁] > if₂
and₁ > [active₁, mark₁, add₁] > s
and₁ > [active₁, mark₁, add₁] > nil
true > [active₁, mark₁, add₁] > false
true > [active₁, mark₁, add₁] > if₂
true > [active₁, mark₁, add₁] > s
true > [active₁, mark₁, add₁] > nil
0 > [active₁, mark₁, add₁] > false
0 > [active₁, mark₁, add₁] > if₂
0 > [active₁, mark₁, add₁] > s
0 > [active₁, mark₁, add₁] > nil
first > cons > [active₁, mark₁, add₁] > false
first > cons > [active₁, mark₁, add₁] > if₂
first > cons > [active₁, mark₁, add₁] > s
first > cons > [active₁, mark₁, add₁] > nil
from₁ > cons > [active₁, mark₁, add₁] > false
from₁ > cons > [active₁, mark₁, add₁] > if₂
from₁ > cons > [active₁, mark₁, add₁] > s
from₁ > cons > [active₁, mark₁, add₁] > nil

Status:

FROM₁: [1]
active₁: [1]
mark₁: [1]
and₁: [1]
true: []
false: []
if₂: [1,2]
add₁: [1]
0: []
s: []
first: []
nil: []
cons: []
from₁: [1]

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))))
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)

(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))))
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) 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(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.

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

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

true > [mark₁, active₁, and₁] > false
true > [mark₁, active₁, and₁] > if₂
true > [mark₁, active₁, and₁] > 0
true > [mark₁, active₁, and₁] > nil
first > cons > [mark₁, active₁, and₁] > false
first > cons > [mark₁, active₁, and₁] > if₂
first > cons > [mark₁, active₁, and₁] > 0
first > cons > [mark₁, active₁, and₁] > nil
from₁ > cons > [mark₁, active₁, and₁] > false
from₁ > cons > [mark₁, active₁, and₁] > if₂
from₁ > cons > [mark₁, active₁, and₁] > 0
from₁ > cons > [mark₁, active₁, and₁] > nil

Status:

mark₁: [1]
active₁: [1]
and₁: [1]
true: []
false: []
if₂: [1,2]
0: []
first: []
nil: []
cons: []
from₁: [1]

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))))
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)

(12) 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.

(13) 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: Combined order from the following AFS and order.
CONS(x1, x2) = x2
mark(x1) = mark(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) = s
first(x1, x2) = first
nil = nil
cons(x1, x2) = cons
from(x1) = from(x1)

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

if₂ > [mark₁, and₁, true] > 0 > [false, s, nil, cons]
add₁ > [mark₁, and₁, true] > 0 > [false, s, nil, cons]
first > [mark₁, and₁, true] > 0 > [false, s, nil, cons]
from₁ > [mark₁, and₁, true] > 0 > [false, s, nil, cons]

Status:

mark₁: [1]
and₁: [1]
true: []
false: []
if₂: [1,2]
add₁: [1]
0: []
s: []
first: []
nil: []
cons: []
from₁: [1]

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))))
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)

(14) Obligation:

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

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

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

[active₁, mark₁, if₂, first, from] > CONS₂ > [false, s]
[active₁, mark₁, if₂, first, from] > true > [false, s]
[active₁, mark₁, if₂, first, from] > add₁ > [false, s]
[active₁, mark₁, if₂, first, from] > 0 > nil > [false, s]
[active₁, mark₁, if₂, first, from] > cons > [false, s]

Status:

CONS₂: [2,1]
active₁: [1]
true: []
mark₁: [1]
false: []
if₂: [2,1]
add₁: [1]
0: []
s: []
first: []
nil: []
cons: []
from: []

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))))
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)

(16) 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.

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) Obligation:

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

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.

(20) 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)
active(x1) = x1
and(x1, x2) = and(x1, x2)
true = true
false = false
if(x1, x2, x3) = if(x2, x3)
add(x1, x2) = add(x1, x2)
0 = 0
s(x1) = s
first(x1, x2) = first(x2)
nil = nil
cons(x1, x2) = x1
from(x1) = from(x1)

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

true > [FIRST₁, mark₁, and₂, add₂, first₁, from₁] > 0 > [false, s, nil]
if₂ > [FIRST₁, mark₁, and₂, add₂, first₁, from₁] > 0 > [false, s, nil]

Status:

FIRST₁: [1]
mark₁: [1]
and₂: [2,1]
true: []
false: []
if₂: [2,1]
add₂: [2,1]
0: []
s: []
first₁: [1]
nil: []
from₁: [1]

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))))
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)

(21) Obligation:

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

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.

(22) 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, x2)
mark(x1) = mark(x1)
active(x1) = x1
and(x1, x2) = and(x2)
true = true
false = false
if(x1, x2, x3) = if(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) = from(x1)

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

and₁ > [mark₁, add₂, first₂] > false > [FIRST₂, s, nil]
true > [mark₁, add₂, first₂] > false > [FIRST₂, s, nil]
if₂ > [mark₁, add₂, first₂] > false > [FIRST₂, s, nil]
0 > [mark₁, add₂, first₂] > false > [FIRST₂, s, nil]
from₁ > cons₁ > [mark₁, add₂, first₂] > false > [FIRST₂, s, nil]

Status:

FIRST₂: [1,2]
mark₁: [1]
and₁: [1]
true: []
false: []
if₂: [2,1]
add₂: [2,1]
0: []
s: []
first₂: [2,1]
nil: []
cons₁: [1]
from₁: [1]

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))))
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)

(23) Obligation:

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

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

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

FIRST₁ > [active₁, mark₁, false, if₂]
true > [active₁, mark₁, false, if₂]
0 > [active₁, mark₁, false, if₂]
[first, nil] > cons > [active₁, mark₁, false, if₂]
from₁ > [add₁, s] > [active₁, mark₁, false, if₂]
from₁ > cons > [active₁, mark₁, false, if₂]

Status:

FIRST₁: [1]
active₁: [1]
true: []
mark₁: [1]
false: []
if₂: [2,1]
add₁: [1]
0: []
s: []
first: []
nil: []
cons: []
from₁: [1]

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))))
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)

(25) Obligation:

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

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

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

if₂ > [active₁, mark₁] > true > false
if₂ > [active₁, mark₁] > add₁ > s > false
if₂ > [active₁, mark₁] > 0 > false
first > nil > [active₁, mark₁] > true > false
first > nil > [active₁, mark₁] > add₁ > s > false
first > nil > [active₁, mark₁] > 0 > false
first > cons > [active₁, mark₁] > true > false
first > cons > [active₁, mark₁] > add₁ > s > false
first > cons > [active₁, mark₁] > 0 > false
from > cons > [active₁, mark₁] > true > false
from > cons > [active₁, mark₁] > add₁ > s > false
from > cons > [active₁, mark₁] > 0 > false

Status:

active₁: [1]
true: []
mark₁: [1]
false: []
if₂: [2,1]
add₁: [1]
0: []
s: []
first: []
nil: []
cons: []
from: []

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))))
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)

(27) 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.

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) TRUE

(30) 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.

(31) 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)
S(mark(X)) → S(X)

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

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

and₁ > [active₁, mark₁, add₁] > false
and₁ > [active₁, mark₁, add₁] > if₂
and₁ > [active₁, mark₁, add₁] > s
and₁ > [active₁, mark₁, add₁] > nil
true > [active₁, mark₁, add₁] > false
true > [active₁, mark₁, add₁] > if₂
true > [active₁, mark₁, add₁] > s
true > [active₁, mark₁, add₁] > nil
0 > [active₁, mark₁, add₁] > false
0 > [active₁, mark₁, add₁] > if₂
0 > [active₁, mark₁, add₁] > s
0 > [active₁, mark₁, add₁] > nil
first > cons > [active₁, mark₁, add₁] > false
first > cons > [active₁, mark₁, add₁] > if₂
first > cons > [active₁, mark₁, add₁] > s
first > cons > [active₁, mark₁, add₁] > nil
from₁ > cons > [active₁, mark₁, add₁] > false
from₁ > cons > [active₁, mark₁, add₁] > if₂
from₁ > cons > [active₁, mark₁, add₁] > s
from₁ > cons > [active₁, mark₁, add₁] > nil

Status:

S₁: [1]
active₁: [1]
mark₁: [1]
and₁: [1]
true: []
false: []
if₂: [1,2]
add₁: [1]
0: []
s: []
first: []
nil: []
cons: []
from₁: [1]

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))))
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)

(32) 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.

(33) PisEmptyProof (EQUIVALENT transformation)

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

(34) TRUE

(35) Obligation:

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

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.

(36) 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) = x1
and(x1, x2) = and(x1, x2)
true = true
false = false
if(x1, x2, x3) = if(x2, x3)
add(x1, x2) = add(x2)
0 = 0
s(x1) = s
first(x1, x2) = first(x1, x2)
nil = nil
cons(x1, x2) = x1
from(x1) = from(x1)

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

ADD₁ > [false, nil]
if₂ > [mark₁, and₂, first₂] > true > [false, nil]
if₂ > [mark₁, and₂, first₂] > 0 > [false, nil]
if₂ > [mark₁, and₂, first₂] > s > [false, nil]
add₁ > [mark₁, and₂, first₂] > true > [false, nil]
add₁ > [mark₁, and₂, first₂] > 0 > [false, nil]
add₁ > [mark₁, and₂, first₂] > s > [false, nil]
from₁ > [mark₁, and₂, first₂] > true > [false, nil]
from₁ > [mark₁, and₂, first₂] > 0 > [false, nil]
from₁ > [mark₁, and₂, first₂] > s > [false, nil]

Status:

ADD₁: [1]
mark₁: [1]
and₂: [2,1]
true: []
false: []
if₂: [2,1]
add₁: [1]
0: []
s: []
first₂: [2,1]
nil: []
from₁: [1]

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))))
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)

(37) Obligation:

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

ADD(X1, mark(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.

(38) 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: Combined order from the following AFS and order.
ADD(x1, x2) = ADD(x2)
mark(x1) = mark(x1)
active(x1) = x1
and(x1, x2) = and(x2)
true = true
false = false
if(x1, x2, x3) = if(x2, x3)
add(x1, x2) = add(x1, x2)
0 = 0
s(x1) = s
first(x1, x2) = first(x2)
nil = nil
cons(x1, x2) = cons
from(x1) = from(x1)

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

ADD₁ > [s, nil]
and₁ > false > [mark₁, add₂, first₁] > 0 > [s, nil]
true > [mark₁, add₂, first₁] > 0 > [s, nil]
if₂ > [mark₁, add₂, first₁] > 0 > [s, nil]
from₁ > cons > [mark₁, add₂, first₁] > 0 > [s, nil]

Status:

ADD₁: [1]
mark₁: [1]
and₁: [1]
true: []
false: []
if₂: [1,2]
add₂: [2,1]
0: []
s: []
first₁: [1]
nil: []
cons: []
from₁: [1]

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))))
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)

(39) Obligation:

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

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.

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

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

ADD₁ > [active₁, mark₁, false, if₂]
true > [active₁, mark₁, false, if₂]
add₁ > s > [active₁, mark₁, false, if₂]
0 > nil > [active₁, mark₁, false, if₂]
from₁ > s > [active₁, mark₁, false, if₂]
from₁ > [first, cons] > nil > [active₁, mark₁, false, if₂]

Status:

ADD₁: [1]
active₁: [1]
true: []
mark₁: [1]
false: []
if₂: [2,1]
add₁: [1]
0: []
s: []
first: []
nil: []
cons: []
from₁: [1]

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))))
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)

(41) Obligation:

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

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

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

and₁ > [active₁, mark₁] > true
and₁ > [active₁, mark₁] > false
and₁ > [active₁, mark₁] > add₁ > s
and₁ > [active₁, mark₁] > [0, nil]
if₂ > [active₁, mark₁] > true
if₂ > [active₁, mark₁] > false
if₂ > [active₁, mark₁] > add₁ > s
if₂ > [active₁, mark₁] > [0, nil]
first > cons > [active₁, mark₁] > true
first > cons > [active₁, mark₁] > false
first > cons > [active₁, mark₁] > add₁ > s
first > cons > [active₁, mark₁] > [0, nil]
from > cons > [active₁, mark₁] > true
from > cons > [active₁, mark₁] > false
from > cons > [active₁, mark₁] > add₁ > s
from > cons > [active₁, mark₁] > [0, nil]

Status:

active₁: [1]
and₁: [1]
true: []
mark₁: [1]
false: []
if₂: [2,1]
add₁: [1]
0: []
s: []
first: []
nil: []
cons: []
from: []

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))))
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)

(43) 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.

(44) PisEmptyProof (EQUIVALENT transformation)

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

(45) TRUE

(46) 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.

(47) 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)
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) = IF(x1, x3)
mark(x1) = mark(x1)
active(x1) = x1
and(x1, x2) = and(x1, x2)
true = true
false = false
if(x1, x2, x3) = if(x2, x3)
add(x1, x2) = add(x1, x2)
0 = 0
s(x1) = s
first(x1, x2) = first
nil = nil
cons(x1, x2) = cons
from(x1) = from(x1)

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

true > [IF₂, mark₁, and₂, add₂] > 0 > [false, s]
true > [IF₂, mark₁, and₂, add₂] > nil > [false, s]
if₂ > [IF₂, mark₁, and₂, add₂] > 0 > [false, s]
if₂ > [IF₂, mark₁, and₂, add₂] > nil > [false, s]
first > cons > [IF₂, mark₁, and₂, add₂] > 0 > [false, s]
first > cons > [IF₂, mark₁, and₂, add₂] > nil > [false, s]
from₁ > cons > [IF₂, mark₁, and₂, add₂] > 0 > [false, s]
from₁ > cons > [IF₂, mark₁, and₂, add₂] > nil > [false, s]

Status:

IF₂: [1,2]
mark₁: [1]
and₂: [2,1]
true: []
false: []
if₂: [1,2]
add₂: [2,1]
0: []
s: []
first: []
nil: []
cons: []
from₁: [1]

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))))
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)

(48) Obligation:

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

IF(X1, mark(X2), 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.

(49) 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)
IF(active(X1), X2, X3) → IF(X1, X2, X3)
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(x1, x2)
mark(x1) = mark(x1)
active(x1) = active(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) = s
first(x1, x2) = first(x1, x2)
nil = nil
cons(x1, x2) = x1
from(x1) = from(x1)

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

and₁ > [mark₁, 0, first₂, from₁] > IF₂ > nil
and₁ > [mark₁, 0, first₂, from₁] > [active₁, s] > false > nil
and₁ > [mark₁, 0, first₂, from₁] > true > nil
if₂ > [mark₁, 0, first₂, from₁] > IF₂ > nil
if₂ > [mark₁, 0, first₂, from₁] > [active₁, s] > false > nil
if₂ > [mark₁, 0, first₂, from₁] > true > nil
add₁ > [mark₁, 0, first₂, from₁] > IF₂ > nil
add₁ > [mark₁, 0, first₂, from₁] > [active₁, s] > false > nil
add₁ > [mark₁, 0, first₂, from₁] > true > nil

Status:

IF₂: [1,2]
mark₁: [1]
active₁: [1]
and₁: [1]
true: []
false: []
if₂: [2,1]
add₁: [1]
0: []
s: []
first₂: [2,1]
nil: []
from₁: [1]

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))))
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)

(50) Obligation:

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

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.

(51) 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) = IF(x3)
active(x1) = active(x1)
and(x1, x2) = and(x2)
true = true
mark(x1) = mark(x1)
false = false
if(x1, x2, x3) = if(x2, x3)
add(x1, x2) = add(x2)
0 = 0
s(x1) = s
first(x1, x2) = first(x1, x2)
nil = nil
cons(x1, x2) = cons
from(x1) = from

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

IF₁ > s
[and₁, false] > [mark₁, first₂] > cons > active₁ > nil > s
true > [mark₁, first₂] > cons > active₁ > nil > s
if₂ > [mark₁, first₂] > cons > active₁ > nil > s
add₁ > [mark₁, first₂] > cons > active₁ > nil > s
0 > [mark₁, first₂] > cons > active₁ > nil > s
from > [mark₁, first₂] > cons > active₁ > nil > s

Status:

IF₁: [1]
active₁: [1]
and₁: [1]
true: []
mark₁: [1]
false: []
if₂: [2,1]
add₁: [1]
0: []
s: []
first₂: [2,1]
nil: []
cons: []
from: []

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))))
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)

(52) 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.

(53) PisEmptyProof (EQUIVALENT transformation)

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

(54) TRUE

(55) 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.

(56) 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)
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) = x1
mark(x1) = mark(x1)
active(x1) = active(x1)
and(x1, x2) = and(x2)
true = true
false = false
if(x1, x2, x3) = if(x2, x3)
add(x1, x2) = x2
0 = 0
s(x1) = x1
first(x1, x2) = first
nil = nil
cons(x1, x2) = cons
from(x1) = from(x1)

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

true > [mark₁, active₁, and₁] > false
true > [mark₁, active₁, and₁] > if₂
true > [mark₁, active₁, and₁] > 0
true > [mark₁, active₁, and₁] > nil
first > cons > [mark₁, active₁, and₁] > false
first > cons > [mark₁, active₁, and₁] > if₂
first > cons > [mark₁, active₁, and₁] > 0
first > cons > [mark₁, active₁, and₁] > nil
from₁ > cons > [mark₁, active₁, and₁] > false
from₁ > cons > [mark₁, active₁, and₁] > if₂
from₁ > cons > [mark₁, active₁, and₁] > 0
from₁ > cons > [mark₁, active₁, and₁] > nil

Status:

mark₁: [1]
active₁: [1]
and₁: [1]
true: []
false: []
if₂: [1,2]
0: []
first: []
nil: []
cons: []
from₁: [1]

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))))
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)

(57) 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.

(58) 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: Combined order from the following AFS and order.
AND(x1, x2) = x2
mark(x1) = mark(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) = s
first(x1, x2) = first
nil = nil
cons(x1, x2) = cons
from(x1) = from(x1)

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

if₂ > [mark₁, and₁, true] > 0 > [false, s, nil, cons]
add₁ > [mark₁, and₁, true] > 0 > [false, s, nil, cons]
first > [mark₁, and₁, true] > 0 > [false, s, nil, cons]
from₁ > [mark₁, and₁, true] > 0 > [false, s, nil, cons]

Status:

mark₁: [1]
and₁: [1]
true: []
false: []
if₂: [1,2]
add₁: [1]
0: []
s: []
first: []
nil: []
cons: []
from₁: [1]

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))))
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)

(59) Obligation:

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

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.

(60) 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)
active(x1) = active(x1)
and(x1, x2) = x2
true = true
mark(x1) = mark(x1)
false = false
if(x1, x2, x3) = if(x2, x3)
add(x1, x2) = add(x2)
0 = 0
s(x1) = s
first(x1, x2) = first
nil = nil
cons(x1, x2) = cons
from(x1) = from

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

[active₁, mark₁, if₂, first, from] > AND₂ > [false, s]
[active₁, mark₁, if₂, first, from] > true > [false, s]
[active₁, mark₁, if₂, first, from] > add₁ > [false, s]
[active₁, mark₁, if₂, first, from] > 0 > nil > [false, s]
[active₁, mark₁, if₂, first, from] > cons > [false, s]

Status:

AND₂: [2,1]
active₁: [1]
true: []
mark₁: [1]
false: []
if₂: [2,1]
add₁: [1]
0: []
s: []
first: []
nil: []
cons: []
from: []

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))))
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)

(61) 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.

(62) PisEmptyProof (EQUIVALENT transformation)

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

(63) TRUE

(64) 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.

(65) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(and(true, X)) → MARK(X)
MARK(and(X1, X2)) → MARK(X1)
ACTIVE(if(true, X, Y)) → MARK(X)
MARK(if(X1, X2, X3)) → MARK(X1)
ACTIVE(if(false, X, Y)) → MARK(Y)
MARK(add(X1, X2)) → MARK(X1)
ACTIVE(add(0, X)) → MARK(X)
ACTIVE(add(s(X), Y)) → MARK(s(add(X, Y)))
MARK(first(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
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) = ACTIVE(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
active(x1) = x1
nil = nil

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

true > [MARK₁, and₂, ACTIVE₁, s₁, first₂] > nil > cons
if₃ > [MARK₁, and₂, ACTIVE₁, s₁, first₂] > nil > cons
add₂ > [MARK₁, and₂, ACTIVE₁, s₁, first₂] > nil > cons
false > [MARK₁, and₂, ACTIVE₁, s₁, first₂] > nil > cons
0 > nil > cons
from > [MARK₁, and₂, ACTIVE₁, s₁, first₂] > nil > cons

Status:

MARK₁: [1]
and₂: [2,1]
ACTIVE₁: [1]
true: []
if₃: [1,2,3]
add₂: [1,2]
false: []
s₁: [1]
0: []
first₂: [1,2]
cons: []
from: []
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))))
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)

(66) Obligation:

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

MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))
MARK(add(X1, X2)) → ACTIVE(add(mark(X1), X2))
MARK(s(X)) → ACTIVE(s(X))
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
MARK(from(X)) → ACTIVE(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.

(67) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 7 less nodes.

(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) QDPOrderProof (EQUIVALENT transformation)

(14) Obligation:

(15) QDPOrderProof (EQUIVALENT transformation)

(16) Obligation:

(17) PisEmptyProof (EQUIVALENT transformation)

(18) TRUE

(19) Obligation:

(20) QDPOrderProof (EQUIVALENT transformation)

(21) Obligation:

(22) QDPOrderProof (EQUIVALENT transformation)

(23) Obligation:

(24) QDPOrderProof (EQUIVALENT transformation)

(25) Obligation:

(26) QDPOrderProof (EQUIVALENT transformation)

(27) Obligation:

(28) PisEmptyProof (EQUIVALENT transformation)

(29) TRUE

(30) Obligation:

(31) QDPOrderProof (EQUIVALENT transformation)

(32) Obligation:

(33) PisEmptyProof (EQUIVALENT transformation)

(34) TRUE

(35) Obligation:

(36) QDPOrderProof (EQUIVALENT transformation)

(37) Obligation:

(38) QDPOrderProof (EQUIVALENT transformation)

(39) Obligation:

(40) QDPOrderProof (EQUIVALENT transformation)

(41) Obligation:

(42) QDPOrderProof (EQUIVALENT transformation)

(43) Obligation:

(44) PisEmptyProof (EQUIVALENT transformation)

(45) TRUE

(46) Obligation:

(47) QDPOrderProof (EQUIVALENT transformation)

(48) Obligation:

(49) QDPOrderProof (EQUIVALENT transformation)

(50) Obligation:

(51) QDPOrderProof (EQUIVALENT transformation)

(52) Obligation:

(53) PisEmptyProof (EQUIVALENT transformation)

(54) TRUE

(55) Obligation:

(56) QDPOrderProof (EQUIVALENT transformation)

(57) Obligation:

(58) QDPOrderProof (EQUIVALENT transformation)

(59) Obligation:

(60) QDPOrderProof (EQUIVALENT transformation)

(61) Obligation:

(62) PisEmptyProof (EQUIVALENT transformation)

(63) TRUE

(64) Obligation:

(65) QDPOrderProof (EQUIVALENT transformation)

(66) Obligation:

(67) DependencyGraphProof (EQUIVALENT transformation)

(68) TRUE