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 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 QDPOrderProof (⇔)
    ↳16 QDP
    ↳17 PisEmptyProof (⇔)
    ↳18 TRUE
  ↳19 QDP
    ↳20 QDPOrderProof (⇔)
    ↳21 QDP
    ↳22 QDPOrderProof (⇔)
    ↳23 QDP
    ↳24 QDPOrderProof (⇔)
    ↳25 QDP
    ↳26 QDPOrderProof (⇔)
    ↳27 QDP
    ↳28 PisEmptyProof (⇔)
    ↳29 TRUE
  ↳30 QDP
    ↳31 QDPOrderProof (⇔)
    ↳32 QDP
    ↳33 QDPOrderProof (⇔)
    ↳34 QDP
    ↳35 QDPOrderProof (⇔)
    ↳36 QDP
    ↳37 QDPOrderProof (⇔)
    ↳38 QDP
    ↳39 PisEmptyProof (⇔)
    ↳40 TRUE
  ↳41 QDP
    ↳42 QDPOrderProof (⇔)
    ↳43 QDP
    ↳44 QDPOrderProof (⇔)
    ↳45 QDP

(0) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(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(zeros) → MARK(cons(0, zeros))
ACTIVE(zeros) → CONS(0, zeros)
ACTIVE(and(tt, X)) → MARK(X)
ACTIVE(length(nil)) → MARK(0)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
ACTIVE(length(cons(N, L))) → S(length(L))
ACTIVE(length(cons(N, L))) → LENGTH(L)
MARK(zeros) → ACTIVE(zeros)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(0) → ACTIVE(0)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
MARK(and(X1, X2)) → AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(tt) → ACTIVE(tt)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(length(X)) → LENGTH(mark(X))
MARK(length(X)) → MARK(X)
MARK(nil) → ACTIVE(nil)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(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)
AND(mark(X1), X2) → AND(X1, X2)
AND(X1, mark(X2)) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)
LENGTH(mark(X)) → LENGTH(X)
LENGTH(active(X)) → LENGTH(X)
S(mark(X)) → S(X)
S(active(X)) → S(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(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 5 SCCs with 11 less nodes.

(4) Complex Obligation (AND)

(5) 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(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(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.


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

Recursive Path Order [RPO].
Precedence:
[S1, active1]


The following usable rules [FROCOS05] were oriented: none

(7) 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(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(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.


S(mark(X)) → S(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive Path Order [RPO].
Precedence:
mark1 > S1


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(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(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:

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

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(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.


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)  =  LENGTH(x1)
active(x1)  =  active(x1)
mark(x1)  =  x1

Recursive Path Order [RPO].
Precedence:
[LENGTH1, active1]


The following usable rules [FROCOS05] were oriented: none

(14) 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(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(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.


LENGTH(mark(X)) → LENGTH(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive Path Order [RPO].
Precedence:
mark1 > LENGTH1


The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) Obligation:

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

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

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

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

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

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
[AND2, mark1]


The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

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

AND(active(X1), X2) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(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.


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

Recursive Path Order [RPO].
Precedence:
[AND1, active1]


The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

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

AND(X1, active(X2)) → AND(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(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.


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

Recursive Path Order [RPO].
Precedence:
active1 > AND1


The following usable rules [FROCOS05] were oriented: none

(27) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) TRUE

(30) Obligation:

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

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(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(31) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(32) 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(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(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.


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

Recursive Path Order [RPO].
Precedence:
[CONS2, mark1]


The following usable rules [FROCOS05] were oriented: none

(34) 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(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(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.


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)
active(x1)  =  active(x1)

Recursive Path Order [RPO].
Precedence:
[CONS1, active1]


The following usable rules [FROCOS05] were oriented: none

(36) Obligation:

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

CONS(X1, active(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(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.


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)
active(x1)  =  active(x1)

Recursive Path Order [RPO].
Precedence:
active1 > CONS1


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(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(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:

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(and(tt, X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(and(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(length(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(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.


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

Recursive Path Order [RPO].
Precedence:
nil > [MARK, cons, mark, and, tt, zeros, length] > 0
nil > [MARK, cons, mark, and, tt, zeros, length] > s


The following usable rules [FROCOS05] were oriented:

s(active(X)) → s(X)
s(mark(X)) → s(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
and(X1, mark(X2)) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
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)

(43) Obligation:

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

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(and(tt, X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(and(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(length(X)) → MARK(X)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(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.


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

Recursive Path Order [RPO].
Precedence:
nil > mark > tt > cons
nil > mark > active > s > [MARK, and, zeros, 0, length] > cons


The following usable rules [FROCOS05] were oriented:

length(active(X)) → length(X)
length(mark(X)) → length(X)
and(X1, mark(X2)) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
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)

(45) Obligation:

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

ACTIVE(and(tt, X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(and(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(length(X)) → MARK(X)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(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)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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