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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 QDPOrderProof (⇔)
    ↳16 QDP
    ↳17 QDPOrderProof (⇔)
    ↳18 QDP
    ↳19 QDPOrderProof (⇔)
    ↳20 QDP
    ↳21 PisEmptyProof (⇔)
    ↳22 TRUE
  ↳23 QDP
    ↳24 QDPOrderProof (⇔)
    ↳25 QDP
    ↳26 QDPOrderProof (⇔)
    ↳27 QDP
    ↳28 QDPOrderProof (⇔)
    ↳29 QDP
    ↳30 QDPOrderProof (⇔)
    ↳31 QDP
    ↳32 PisEmptyProof (⇔)
    ↳33 TRUE
  ↳34 QDP
    ↳35 QDPOrderProof (⇔)
    ↳36 QDP
    ↳37 QDPOrderProof (⇔)
    ↳38 QDP
    ↳39 PisEmptyProof (⇔)
    ↳40 TRUE
  ↳41 QDP
    ↳42 QDPOrderProof (⇔)
    ↳43 QDP
    ↳44 QDPOrderProof (⇔)
    ↳45 QDP
    ↳46 PisEmptyProof (⇔)
    ↳47 TRUE
  ↳48 QDP
    ↳49 QDPOrderProof (⇔)
    ↳50 QDP
    ↳51 QDPOrderProof (⇔)
    ↳52 QDP
    ↳53 QDPOrderProof (⇔)
    ↳54 QDP
    ↳55 QDPOrderProof (⇔)
    ↳56 QDP
    ↳57 PisEmptyProof (⇔)
    ↳58 TRUE
  ↳59 QDP
    ↳60 QDPOrderProof (⇔)
    ↳61 QDP
    ↳62 QDPOrderProof (⇔)
    ↳63 QDP
    ↳64 QDPOrderProof (⇔)
    ↳65 QDP
    ↳66 QDPOrderProof (⇔)
    ↳67 QDP
    ↳68 QDPOrderProof (⇔)
    ↳69 QDP
    ↳70 QDPOrderProof (⇔)
    ↳71 QDP
    ↳72 QDPOrderProof (⇔)
    ↳73 QDP
    ↳74 QDPOrderProof (⇔)
    ↳75 QDP
    ↳76 DependencyGraphProof (⇔)
    ↳77 TRUE

(0) Obligation:

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

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(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(eq(0, 0)) → MARK(true)
ACTIVE(eq(s(X), s(Y))) → MARK(eq(X, Y))
ACTIVE(eq(s(X), s(Y))) → EQ(X, Y)
ACTIVE(eq(X, Y)) → MARK(false)
ACTIVE(inf(X)) → MARK(cons(X, inf(s(X))))
ACTIVE(inf(X)) → CONS(X, inf(s(X)))
ACTIVE(inf(X)) → INF(s(X))
ACTIVE(inf(X)) → S(X)
ACTIVE(take(0, X)) → MARK(nil)
ACTIVE(take(s(X), cons(Y, L))) → MARK(cons(Y, take(X, L)))
ACTIVE(take(s(X), cons(Y, L))) → CONS(Y, take(X, L))
ACTIVE(take(s(X), cons(Y, L))) → TAKE(X, L)
ACTIVE(length(nil)) → MARK(0)
ACTIVE(length(cons(X, L))) → MARK(s(length(L)))
ACTIVE(length(cons(X, L))) → S(length(L))
ACTIVE(length(cons(X, L))) → LENGTH(L)
MARK(eq(X1, X2)) → ACTIVE(eq(X1, X2))
MARK(0) → ACTIVE(0)
MARK(true) → ACTIVE(true)
MARK(s(X)) → ACTIVE(s(X))
MARK(false) → ACTIVE(false)
MARK(inf(X)) → ACTIVE(inf(mark(X)))
MARK(inf(X)) → INF(mark(X))
MARK(inf(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
MARK(take(X1, X2)) → TAKE(mark(X1), mark(X2))
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)
MARK(nil) → ACTIVE(nil)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(length(X)) → LENGTH(mark(X))
MARK(length(X)) → MARK(X)
EQ(mark(X1), X2) → EQ(X1, X2)
EQ(X1, mark(X2)) → EQ(X1, X2)
EQ(active(X1), X2) → EQ(X1, X2)
EQ(X1, active(X2)) → EQ(X1, X2)
S(mark(X)) → S(X)
S(active(X)) → S(X)
INF(mark(X)) → INF(X)
INF(active(X)) → INF(X)
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)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(active(X1), X2) → TAKE(X1, X2)
TAKE(X1, active(X2)) → TAKE(X1, X2)
LENGTH(mark(X)) → LENGTH(X)
LENGTH(active(X)) → LENGTH(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 7 SCCs with 19 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

LENGTH(active(X)) → LENGTH(X)
LENGTH(mark(X)) → LENGTH(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(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.


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

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

Status:
active1: [1]


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

LENGTH(mark(X)) → LENGTH(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

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

Status:
mark1: [1]


The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

TAKE(X1, mark(X2)) → TAKE(X1, X2)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(active(X1), X2) → TAKE(X1, X2)
TAKE(X1, active(X2)) → TAKE(X1, X2)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(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.


TAKE(X1, mark(X2)) → TAKE(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAKE(x1, x2)  =  TAKE(x2)
mark(x1)  =  mark(x1)
active(x1)  =  x1

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

Status:
TAKE1: [1]
mark1: [1]


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(active(X1), X2) → TAKE(X1, X2)
TAKE(X1, active(X2)) → TAKE(X1, X2)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(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.


TAKE(mark(X1), X2) → TAKE(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAKE(x1, x2)  =  TAKE(x1, x2)
mark(x1)  =  mark(x1)
active(x1)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:
[TAKE2, mark1]

Status:
TAKE2: [2,1]
mark1: [1]


The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

TAKE(active(X1), X2) → TAKE(X1, X2)
TAKE(X1, active(X2)) → TAKE(X1, X2)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TAKE(X1, active(X2)) → TAKE(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TAKE(x1, x2)  =  x2
active(x1)  =  active(x1)

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

Status:
active1: [1]


The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

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

TAKE(active(X1), X2) → TAKE(X1, X2)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TAKE(active(X1), X2) → TAKE(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:
[TAKE2, active1]

Status:
active1: [1]
TAKE2: [2,1]


The following usable rules [FROCOS05] were oriented: none

(20) Obligation:

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

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE

(23) Obligation:

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

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(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(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.


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)  =  CONS(x2)
mark(x1)  =  mark(x1)
active(x1)  =  x1

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

Status:
CONS1: [1]
mark1: [1]


The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

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

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(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(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.


CONS(mark(X1), X2) → CONS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CONS(x1, x2)  =  CONS(x1, x2)
mark(x1)  =  mark(x1)
active(x1)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:
[CONS2, mark1]

Status:
CONS2: [2,1]
mark1: [1]


The following usable rules [FROCOS05] were oriented: none

(27) Obligation:

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

CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(28) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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)  =  x2
active(x1)  =  active(x1)

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

Status:
active1: [1]


The following usable rules [FROCOS05] were oriented: none

(29) Obligation:

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

CONS(active(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


CONS(active(X1), X2) → CONS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:
[CONS2, active1]

Status:
active1: [1]
CONS2: [2,1]


The following usable rules [FROCOS05] were oriented: none

(31) Obligation:

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

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(32) PisEmptyProof (EQUIVALENT transformation)

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

(33) TRUE

(34) Obligation:

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

INF(active(X)) → INF(X)
INF(mark(X)) → INF(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

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

Status:
active1: [1]


The following usable rules [FROCOS05] were oriented: none

(36) Obligation:

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

INF(mark(X)) → INF(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(37) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

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

Status:
mark1: [1]


The following usable rules [FROCOS05] were oriented: none

(38) Obligation:

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

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(39) PisEmptyProof (EQUIVALENT transformation)

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

(40) TRUE

(41) Obligation:

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(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.


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

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

Status:
active1: [1]


The following usable rules [FROCOS05] were oriented: none

(43) Obligation:

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

S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(44) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

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

Status:
mark1: [1]


The following usable rules [FROCOS05] were oriented: none

(45) Obligation:

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

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(46) PisEmptyProof (EQUIVALENT transformation)

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

(47) TRUE

(48) Obligation:

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

EQ(X1, mark(X2)) → EQ(X1, X2)
EQ(mark(X1), X2) → EQ(X1, X2)
EQ(active(X1), X2) → EQ(X1, X2)
EQ(X1, active(X2)) → EQ(X1, X2)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(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.


EQ(X1, mark(X2)) → EQ(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
EQ(x1, x2)  =  EQ(x2)
mark(x1)  =  mark(x1)
active(x1)  =  x1

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

Status:
mark1: [1]
EQ1: [1]


The following usable rules [FROCOS05] were oriented: none

(50) Obligation:

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

EQ(mark(X1), X2) → EQ(X1, X2)
EQ(active(X1), X2) → EQ(X1, X2)
EQ(X1, active(X2)) → EQ(X1, X2)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(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.


EQ(mark(X1), X2) → EQ(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
EQ(x1, x2)  =  EQ(x1, x2)
mark(x1)  =  mark(x1)
active(x1)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:
[EQ2, mark1]

Status:
mark1: [1]
EQ2: [2,1]


The following usable rules [FROCOS05] were oriented: none

(52) Obligation:

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

EQ(active(X1), X2) → EQ(X1, X2)
EQ(X1, active(X2)) → EQ(X1, X2)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(53) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


EQ(X1, active(X2)) → EQ(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
EQ(x1, x2)  =  x2
active(x1)  =  active(x1)

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

Status:
active1: [1]


The following usable rules [FROCOS05] were oriented: none

(54) Obligation:

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

EQ(active(X1), X2) → EQ(X1, X2)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(55) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


EQ(active(X1), X2) → EQ(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Quasi-Precedence:
[EQ2, active1]

Status:
active1: [1]
EQ2: [2,1]


The following usable rules [FROCOS05] were oriented: none

(56) Obligation:

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

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(57) PisEmptyProof (EQUIVALENT transformation)

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

(58) TRUE

(59) Obligation:

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

ACTIVE(eq(s(X), s(Y))) → MARK(eq(X, Y))
MARK(eq(X1, X2)) → ACTIVE(eq(X1, X2))
ACTIVE(inf(X)) → MARK(cons(X, inf(s(X))))
MARK(s(X)) → ACTIVE(s(X))
ACTIVE(take(s(X), cons(Y, L))) → MARK(cons(Y, take(X, L)))
MARK(inf(X)) → ACTIVE(inf(mark(X)))
ACTIVE(length(cons(X, L))) → MARK(s(length(L)))
MARK(inf(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(length(X)) → MARK(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(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.


MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  ACTIVE
eq(x1, x2)  =  eq
s(x1)  =  s
MARK(x1)  =  x1
inf(x1)  =  x1
cons(x1, x2)  =  cons
take(x1, x2)  =  take(x1, x2)
mark(x1)  =  x1
length(x1)  =  x1
true  =  true
active(x1)  =  active(x1)
false  =  false
nil  =  nil
0  =  0

Lexicographic path order with status [LPO].
Quasi-Precedence:
take2 > [ACTIVE, eq, s, cons, active1]
true > [ACTIVE, eq, s, cons, active1]
false > [ACTIVE, eq, s, cons, active1]
0 > nil > [ACTIVE, eq, s, cons, active1]

Status:
eq: []
active1: [1]
cons: []
true: []
false: []
s: []
take2: [1,2]
0: []
nil: []
ACTIVE: []


The following usable rules [FROCOS05] were oriented:

eq(active(X1), X2) → eq(X1, X2)
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)

(61) Obligation:

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

ACTIVE(eq(s(X), s(Y))) → MARK(eq(X, Y))
MARK(eq(X1, X2)) → ACTIVE(eq(X1, X2))
ACTIVE(inf(X)) → MARK(cons(X, inf(s(X))))
MARK(s(X)) → ACTIVE(s(X))
ACTIVE(take(s(X), cons(Y, L))) → MARK(cons(Y, take(X, L)))
MARK(inf(X)) → ACTIVE(inf(mark(X)))
ACTIVE(length(cons(X, L))) → MARK(s(length(L)))
MARK(inf(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(length(X)) → MARK(X)

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(62) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(length(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  ACTIVE
eq(x1, x2)  =  eq
s(x1)  =  s
MARK(x1)  =  x1
inf(x1)  =  x1
cons(x1, x2)  =  cons
take(x1, x2)  =  x1
mark(x1)  =  x1
length(x1)  =  length(x1)
true  =  true
active(x1)  =  active(x1)
false  =  false
nil  =  nil
0  =  0

Lexicographic path order with status [LPO].
Quasi-Precedence:
true > [length1, active1] > nil > [ACTIVE, eq, s, cons]
false > [length1, active1] > nil > [ACTIVE, eq, s, cons]
0 > [length1, active1] > nil > [ACTIVE, eq, s, cons]

Status:
eq: []
active1: [1]
cons: []
true: []
false: []
s: []
length1: [1]
0: []
nil: []
ACTIVE: []


The following usable rules [FROCOS05] were oriented:

eq(active(X1), X2) → eq(X1, X2)
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)

(63) Obligation:

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

ACTIVE(eq(s(X), s(Y))) → MARK(eq(X, Y))
MARK(eq(X1, X2)) → ACTIVE(eq(X1, X2))
ACTIVE(inf(X)) → MARK(cons(X, inf(s(X))))
MARK(s(X)) → ACTIVE(s(X))
ACTIVE(take(s(X), cons(Y, L))) → MARK(cons(Y, take(X, L)))
MARK(inf(X)) → ACTIVE(inf(mark(X)))
ACTIVE(length(cons(X, L))) → MARK(s(length(L)))
MARK(inf(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(64) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(take(s(X), cons(Y, L))) → MARK(cons(Y, take(X, L)))
ACTIVE(length(cons(X, L))) → MARK(s(length(L)))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  ACTIVE(x1)
eq(x1, x2)  =  eq
s(x1)  =  s
MARK(x1)  =  MARK(x1)
inf(x1)  =  x1
cons(x1, x2)  =  cons
take(x1, x2)  =  take
mark(x1)  =  x1
length(x1)  =  length
true  =  true
active(x1)  =  x1
false  =  false
nil  =  nil
0  =  0

Lexicographic path order with status [LPO].
Quasi-Precedence:
take > [ACTIVE1, eq, MARK1, length, true] > s > cons
take > [ACTIVE1, eq, MARK1, length, true] > false > cons
take > [ACTIVE1, eq, MARK1, length, true] > 0 > cons
take > nil > 0 > cons

Status:
eq: []
MARK1: [1]
cons: []
true: []
false: []
s: []
length: []
0: []
ACTIVE1: [1]
take: []
nil: []


The following usable rules [FROCOS05] were oriented:

mark(true) → active(true)
mark(false) → active(false)
mark(nil) → active(nil)
eq(active(X1), X2) → eq(X1, X2)
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)
inf(active(X)) → inf(X)
inf(mark(X)) → inf(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
active(eq(0, 0)) → mark(true)
mark(length(X)) → active(length(mark(X)))
mark(s(X)) → active(s(X))
active(inf(X)) → mark(cons(X, inf(s(X))))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(X1, X2))
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
active(eq(s(X), s(Y))) → mark(eq(X, Y))
mark(inf(X)) → active(inf(mark(X)))
active(length(cons(X, L))) → mark(s(length(L)))
take(X1, mark(X2)) → take(X1, X2)
take(mark(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
active(take(0, X)) → mark(nil)
length(active(X)) → length(X)
length(mark(X)) → length(X)
active(eq(X, Y)) → mark(false)
mark(0) → active(0)
active(length(nil)) → mark(0)

(65) Obligation:

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

ACTIVE(eq(s(X), s(Y))) → MARK(eq(X, Y))
MARK(eq(X1, X2)) → ACTIVE(eq(X1, X2))
ACTIVE(inf(X)) → MARK(cons(X, inf(s(X))))
MARK(s(X)) → ACTIVE(s(X))
MARK(inf(X)) → ACTIVE(inf(mark(X)))
MARK(inf(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(66) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(inf(X)) → MARK(cons(X, inf(s(X))))
MARK(inf(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
eq(x1, x2)  =  eq
s(x1)  =  s
MARK(x1)  =  x1
inf(x1)  =  inf(x1)
cons(x1, x2)  =  cons(x1)
mark(x1)  =  x1
true  =  true
active(x1)  =  x1
false  =  false
nil  =  nil
0  =  0
take(x1, x2)  =  take(x1, x2)
length(x1)  =  length

Lexicographic path order with status [LPO].
Quasi-Precedence:
eq > [true, 0]
eq > false
inf1 > [s, length] > cons1
inf1 > [s, length] > [true, 0]
[nil, take2] > cons1
[nil, take2] > [true, 0]

Status:
eq: []
inf1: [1]
cons1: [1]
true: []
false: []
s: []
take2: [2,1]
length: []
0: []
nil: []


The following usable rules [FROCOS05] were oriented:

mark(true) → active(true)
mark(false) → active(false)
mark(nil) → active(nil)
eq(active(X1), X2) → eq(X1, X2)
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)
inf(active(X)) → inf(X)
inf(mark(X)) → inf(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
active(eq(0, 0)) → mark(true)
take(X1, mark(X2)) → take(X1, X2)
take(mark(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
mark(length(X)) → active(length(mark(X)))
mark(s(X)) → active(s(X))
active(inf(X)) → mark(cons(X, inf(s(X))))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(X1, X2))
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
active(eq(s(X), s(Y))) → mark(eq(X, Y))
mark(inf(X)) → active(inf(mark(X)))
active(length(cons(X, L))) → mark(s(length(L)))
active(take(0, X)) → mark(nil)
length(active(X)) → length(X)
length(mark(X)) → length(X)
active(eq(X, Y)) → mark(false)
mark(0) → active(0)
active(length(nil)) → mark(0)

(67) Obligation:

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

ACTIVE(eq(s(X), s(Y))) → MARK(eq(X, Y))
MARK(eq(X1, X2)) → ACTIVE(eq(X1, X2))
MARK(s(X)) → ACTIVE(s(X))
MARK(inf(X)) → ACTIVE(inf(mark(X)))
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(68) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(inf(X)) → ACTIVE(inf(mark(X)))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  ACTIVE
eq(x1, x2)  =  eq
s(x1)  =  s
MARK(x1)  =  x1
inf(x1)  =  inf(x1)
mark(x1)  =  mark(x1)
cons(x1, x2)  =  cons
true  =  true
active(x1)  =  active(x1)
false  =  false
nil  =  nil
0  =  0
take(x1, x2)  =  take
length(x1)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:
[true, 0] > active1 > [ACTIVE, eq, s, inf1, cons] > [mark1, take]
[true, 0] > active1 > nil > [mark1, take]
false > active1 > [ACTIVE, eq, s, inf1, cons] > [mark1, take]
false > active1 > nil > [mark1, take]

Status:
eq: []
true: []
mark1: [1]
s: []
0: []
take: []
inf1: [1]
active1: [1]
cons: []
false: []
nil: []
ACTIVE: []


The following usable rules [FROCOS05] were oriented:

eq(active(X1), X2) → eq(X1, X2)
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)

(69) Obligation:

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

ACTIVE(eq(s(X), s(Y))) → MARK(eq(X, Y))
MARK(eq(X1, X2)) → ACTIVE(eq(X1, X2))
MARK(s(X)) → ACTIVE(s(X))
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(70) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(s(X)) → ACTIVE(s(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  ACTIVE
eq(x1, x2)  =  eq
s(x1)  =  s(x1)
MARK(x1)  =  x1
cons(x1, x2)  =  cons
active(x1)  =  x1
mark(x1)  =  mark

Lexicographic path order with status [LPO].
Quasi-Precedence:
s1 > [ACTIVE, eq, cons]

Status:
eq: []
cons: []
mark: []
s1: [1]
ACTIVE: []


The following usable rules [FROCOS05] were oriented:

eq(active(X1), X2) → eq(X1, X2)
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)

(71) Obligation:

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

ACTIVE(eq(s(X), s(Y))) → MARK(eq(X, Y))
MARK(eq(X1, X2)) → ACTIVE(eq(X1, X2))
MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(72) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(cons(X1, X2)) → ACTIVE(cons(X1, X2))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
eq(x1, x2)  =  eq
s(x1)  =  s
MARK(x1)  =  MARK
cons(x1, x2)  =  cons
active(x1)  =  active
mark(x1)  =  mark

Lexicographic path order with status [LPO].
Quasi-Precedence:
[eq, MARK, active] > cons

Status:
eq: []
active: []
MARK: []
cons: []
mark: []
s: []


The following usable rules [FROCOS05] were oriented:

cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)

(73) Obligation:

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

ACTIVE(eq(s(X), s(Y))) → MARK(eq(X, Y))
MARK(eq(X1, X2)) → ACTIVE(eq(X1, X2))

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(74) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MARK(eq(X1, X2)) → ACTIVE(eq(X1, X2))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
eq(x1, x2)  =  eq(x1)
s(x1)  =  s(x1)
MARK(x1)  =  MARK(x1)
active(x1)  =  active(x1)
mark(x1)  =  mark(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
active1 > [eq1, s1, MARK1]
mark1 > [eq1, s1, MARK1]

Status:
eq1: [1]
active1: [1]
MARK1: [1]
mark1: [1]
s1: [1]


The following usable rules [FROCOS05] were oriented:

eq(active(X1), X2) → eq(X1, X2)
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)

(75) Obligation:

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

ACTIVE(eq(s(X), s(Y))) → MARK(eq(X, Y))

The TRS R consists of the following rules:

active(eq(0, 0)) → mark(true)
active(eq(s(X), s(Y))) → mark(eq(X, Y))
active(eq(X, Y)) → mark(false)
active(inf(X)) → mark(cons(X, inf(s(X))))
active(take(0, X)) → mark(nil)
active(take(s(X), cons(Y, L))) → mark(cons(Y, take(X, L)))
active(length(nil)) → mark(0)
active(length(cons(X, L))) → mark(s(length(L)))
mark(eq(X1, X2)) → active(eq(X1, X2))
mark(0) → active(0)
mark(true) → active(true)
mark(s(X)) → active(s(X))
mark(false) → active(false)
mark(inf(X)) → active(inf(mark(X)))
mark(cons(X1, X2)) → active(cons(X1, X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(nil) → active(nil)
mark(length(X)) → active(length(mark(X)))
eq(mark(X1), X2) → eq(X1, X2)
eq(X1, mark(X2)) → eq(X1, X2)
eq(active(X1), X2) → eq(X1, X2)
eq(X1, active(X2)) → eq(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
inf(mark(X)) → inf(X)
inf(active(X)) → inf(X)
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)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)

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

(76) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(77) TRUE