Termination w.r.t. Q proof of /tmp/tmpnCODj_/Ex1_GL02a

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

    ↳8 PisEmptyProof (⇔)

    ↳9 TRUE

  ↳10 QDP

    ↳11 QDPOrderProof (⇔)

    ↳12 QDP

    ↳13 PisEmptyProof (⇔)

    ↳14 TRUE

  ↳15 QDP

    ↳16 QDPOrderProof (⇔)

    ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 PisEmptyProof (⇔)

    ↳21 TRUE

  ↳22 QDP

    ↳23 QDPOrderProof (⇔)

    ↳24 QDP

    ↳25 PisEmptyProof (⇔)

    ↳26 TRUE

  ↳27 QDP

    ↳28 QDPOrderProof (⇔)

    ↳29 QDP

    ↳30 PisEmptyProof (⇔)

    ↳31 TRUE

  ↳32 QDP

    ↳33 QDPOrderProof (⇔)

    ↳34 QDP

    ↳35 QDPOrderProof (⇔)

    ↳36 QDP

    ↳37 PisEmptyProof (⇔)

    ↳38 TRUE

  ↳39 QDP

    ↳40 QDPOrderProof (⇔)

    ↳41 QDP

    ↳42 QDPOrderProof (⇔)

    ↳43 QDP

(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)
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) = LENGTH(x1)
active(x1) = active(x1)
mark(x1) = mark(x1)
eq(x1, x2) = eq
0 = 0
true = true
s(x1) = s
false = false
inf(x1) = inf
cons(x1, x2) = x2
take(x1, x2) = take
nil = nil
length(x1) = length

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

LENGTH₁ > [s, false, take, nil]
eq > true > [active₁, mark₁, 0, inf, length] > [s, false, take, nil]

Status:

LENGTH₁: [1]
active₁: [1]
mark₁: [1]
eq: []
0: []
true: []
s: []
false: []
inf: []
take: []
nil: []
length: []

The following usable rules [FROCOS05] were oriented:

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)

(7) 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.

(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:

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.

(11) 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)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(active(X1), X2) → TAKE(X1, X2)
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) = TAKE(x1, x2)
mark(x1) = mark(x1)
active(x1) = active(x1)
eq(x1, x2) = eq(x2)
0 = 0
true = true
s(x1) = x1
false = false
inf(x1) = inf
cons(x1, x2) = cons
take(x1, x2) = take
nil = nil
length(x1) = length

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

[TAKE₂, mark₁, active₁, eq₁] > [0, true, length] > [take, nil] > cons
[TAKE₂, mark₁, active₁, eq₁] > false > cons
[TAKE₂, mark₁, active₁, eq₁] > inf > cons

Status:

TAKE₂: [1,2]
mark₁: [1]
active₁: [1]
eq₁: [1]
0: []
true: []
false: []
inf: []
cons: []
take: []
nil: []
length: []

The following usable rules [FROCOS05] were oriented:

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)

(12) 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.

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

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.

(16) 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)
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(x2)
mark(x1) = mark(x1)
active(x1) = active(x1)
eq(x1, x2) = eq(x1, x2)
0 = 0
true = true
s(x1) = x1
false = false
inf(x1) = inf
cons(x1, x2) = x2
take(x1, x2) = take
nil = nil
length(x1) = length

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

length > [CONS₁, mark₁, active₁] > [eq₂, true] > false
length > [CONS₁, mark₁, active₁] > 0 > nil
length > [CONS₁, mark₁, active₁] > inf
length > [CONS₁, mark₁, active₁] > take > nil

Status:

CONS₁: [1]
mark₁: [1]
active₁: [1]
eq₂: [1,2]
0: []
true: []
false: []
inf: []
take: []
nil: []
length: []

The following usable rules [FROCOS05] were oriented:

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)

(17) 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)

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.

(18) 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) = CONS(x1, x2)
mark(x1) = mark(x1)
active(x1) = active(x1)
eq(x1, x2) = x2
0 = 0
true = true
s(x1) = x1
false = false
inf(x1) = inf
cons(x1, x2) = cons
take(x1, x2) = take
nil = nil
length(x1) = length

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

inf > [cons, take, nil] > [CONS₂, mark₁, active₁, length] > 0 > true > false

Status:

CONS₂: [2,1]
mark₁: [1]
active₁: [1]
0: []
true: []
false: []
inf: []
cons: []
take: []
nil: []
length: []

The following usable rules [FROCOS05] were oriented:

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)

(19) 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.

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) 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.

(23) 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)
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) = INF(x1)
active(x1) = active(x1)
mark(x1) = mark(x1)
eq(x1, x2) = eq
0 = 0
true = true
s(x1) = s
false = false
inf(x1) = inf
cons(x1, x2) = x2
take(x1, x2) = take
nil = nil
length(x1) = length

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

INF₁ > [s, false, take, nil]
eq > true > [active₁, mark₁, 0, inf, length] > [s, false, take, nil]

Status:

INF₁: [1]
active₁: [1]
mark₁: [1]
eq: []
0: []
true: []
s: []
false: []
inf: []
take: []
nil: []
length: []

The following usable rules [FROCOS05] were oriented:

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)

(24) 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.

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE

(27) 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.

(28) 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)
eq(x1, x2) = eq
0 = 0
true = true
s(x1) = s
false = false
inf(x1) = inf
cons(x1, x2) = x2
take(x1, x2) = take
nil = nil
length(x1) = length

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

S₁ > [s, false, take, nil]
eq > true > [active₁, mark₁, 0, inf, length] > [s, false, take, nil]

Status:

S₁: [1]
active₁: [1]
mark₁: [1]
eq: []
0: []
true: []
s: []
false: []
inf: []
take: []
nil: []
length: []

The following usable rules [FROCOS05] were oriented:

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)

(29) 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.

(30) PisEmptyProof (EQUIVALENT transformation)

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

(31) TRUE

(32) 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.

(33) 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)
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) = EQ(x2)
mark(x1) = mark(x1)
active(x1) = active(x1)
eq(x1, x2) = eq(x1, x2)
0 = 0
true = true
s(x1) = x1
false = false
inf(x1) = inf
cons(x1, x2) = x2
take(x1, x2) = take
nil = nil
length(x1) = length

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

length > [EQ₁, mark₁, active₁] > [eq₂, true] > false
length > [EQ₁, mark₁, active₁] > 0 > nil
length > [EQ₁, mark₁, active₁] > inf
length > [EQ₁, mark₁, active₁] > take > nil

Status:

EQ₁: [1]
mark₁: [1]
active₁: [1]
eq₂: [1,2]
0: []
true: []
false: []
inf: []
take: []
nil: []
length: []

The following usable rules [FROCOS05] were oriented:

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)

(34) 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)

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.

EQ(mark(X1), X2) → EQ(X1, X2)
EQ(active(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) = active(x1)
eq(x1, x2) = x2
0 = 0
true = true
s(x1) = x1
false = false
inf(x1) = inf
cons(x1, x2) = cons
take(x1, x2) = take
nil = nil
length(x1) = length

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

inf > [cons, take, nil] > [EQ₂, mark₁, active₁, length] > 0 > true > false

Status:

EQ₂: [2,1]
mark₁: [1]
active₁: [1]
0: []
true: []
false: []
inf: []
cons: []
take: []
nil: []
length: []

The following usable rules [FROCOS05] were oriented:

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)

(36) 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.

(37) PisEmptyProof (EQUIVALENT transformation)

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

(38) TRUE

(39) 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.

(40) 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))
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
inf(x1) = inf
cons(x1, x2) = cons
take(x1, x2) = take
mark(x1) = mark
length(x1) = length
active(x1) = active
0 = 0
true = true
false = false
nil = nil

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

[eq, MARK, inf, take, length] > false > [mark, active, nil] > 0 > [s, cons]
true > [mark, active, nil] > 0 > [s, cons]

Status:

eq: []
s: []
MARK: []
inf: []
cons: []
take: []
mark: []
length: []
active: []
0: []
true: []
false: []
nil: []

The following usable rules [FROCOS05] were oriented:

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)

(41) 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))))
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(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.

(42) 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))))
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(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 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)
cons(x1, x2) = cons
take(x1, x2) = take(x1, x2)
mark(x1) = x1
length(x1) = length(x1)
active(x1) = x1
0 = 0
true = true
false = false
nil = nil

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

inf₁ > [ACTIVE, eq] > s > cons
inf₁ > [ACTIVE, eq] > true > cons
inf₁ > [ACTIVE, eq] > false > cons
take₂ > [ACTIVE, eq] > s > cons
take₂ > [ACTIVE, eq] > true > cons
take₂ > [ACTIVE, eq] > false > cons
length₁ > [ACTIVE, eq] > s > cons
length₁ > [ACTIVE, eq] > true > cons
length₁ > [ACTIVE, eq] > false > cons
length₁ > 0 > true > cons
length₁ > 0 > nil > cons

Status:

ACTIVE: []
eq: []
s: []
inf₁: [1]
cons: []
take₂: [1,2]
length₁: [1]
0: []
true: []
false: []
nil: []

The following usable rules [FROCOS05] were oriented:

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)

(43) 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.