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 PisEmptyProof (⇔)
    ↳25 TRUE
  ↳26 QDP
    ↳27 QDPOrderProof (⇔)
    ↳28 QDP
    ↳29 QDPOrderProof (⇔)
    ↳30 QDP
    ↳31 PisEmptyProof (⇔)
    ↳32 TRUE
  ↳33 QDP

(0) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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) → CONS(0, zeros)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(tail(X)) → TAIL(active(X))
ACTIVE(tail(X)) → ACTIVE(X)
CONS(mark(X1), X2) → CONS(X1, X2)
TAIL(mark(X)) → TAIL(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(tail(X)) → TAIL(proper(X))
PROPER(tail(X)) → PROPER(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
TAIL(ok(X)) → TAIL(X)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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 7 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

TAIL(ok(X)) → TAIL(X)
TAIL(mark(X)) → TAIL(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

TAIL(mark(X)) → TAIL(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


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

Lexicographic Path Order [LPO].
Precedence:
trivial


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(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


CONS(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)  =  x1
ok(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


CONS(ok(X1), ok(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)
ok(x1)  =  ok(x1)

Lexicographic Path Order [LPO].
Precedence:
ok1 > CONS1


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(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(tail(X)) → PROPER(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

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

PROPER(tail(X)) → PROPER(X)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.


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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE

(26) Obligation:

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

ACTIVE(tail(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(27) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(28) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(cons(X1, X2)) → ACTIVE(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:
[ACTIVE1, cons2]


The following usable rules [FROCOS05] were oriented: none

(30) Obligation:

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

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(31) PisEmptyProof (EQUIVALENT transformation)

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

(32) TRUE

(33) Obligation:

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

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

active(zeros) → mark(cons(0, zeros))
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
tail(mark(X)) → mark(tail(X))
proper(zeros) → ok(zeros)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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