Termination w.r.t. Q proof of /tmp/tmpD4uZx8/Ex4_7_56_Bor03

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 QDPOrderProof (⇔)

    ↳11 QDP

    ↳12 PisEmptyProof (⇔)

    ↳13 TRUE

  ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 QDPOrderProof (⇔)

    ↳18 QDP

    ↳19 PisEmptyProof (⇔)

    ↳20 TRUE

  ↳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 PisEmptyProof (⇔)

    ↳36 TRUE

  ↳37 QDP

    ↳38 QDPOrderProof (⇔)

    ↳39 QDP

(0) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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(from(X)) → MARK(cons(X, from(s(X))))
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(after(0, XS)) → MARK(XS)
ACTIVE(after(s(N), cons(X, XS))) → MARK(after(N, XS))
ACTIVE(after(s(N), cons(X, XS))) → AFTER(N, XS)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(from(X)) → FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(after(X1, X2)) → ACTIVE(after(mark(X1), mark(X2)))
MARK(after(X1, X2)) → AFTER(mark(X1), mark(X2))
MARK(after(X1, X2)) → MARK(X1)
MARK(after(X1, X2)) → MARK(X2)
MARK(0) → ACTIVE(0)
FROM(mark(X)) → FROM(X)
FROM(active(X)) → FROM(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)
S(mark(X)) → S(X)
S(active(X)) → S(X)
AFTER(mark(X1), X2) → AFTER(X1, X2)
AFTER(X1, mark(X2)) → AFTER(X1, X2)
AFTER(active(X1), X2) → AFTER(X1, X2)
AFTER(X1, active(X2)) → AFTER(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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 9 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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.

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

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

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

[AFTER₁, active₁]

Status:

AFTER₁: [1]
mark₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

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

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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.

AFTER(mark(X1), X2) → AFTER(X1, X2)

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

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

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

AFTER(active(X1), X2) → AFTER(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

AFTER(active(X1), X2) → AFTER(X1, X2)

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

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

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(11) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) TRUE

(14) 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(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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.

S(active(X)) → S(X)

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

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

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(16) 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(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(mark(X)) → S(X)

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

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

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE

(21) 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(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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.

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)

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

[CONS₁, active₁]

Status:

CONS₁: [1]
mark₁: [1]
active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(23) 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(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

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

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

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

CONS(active(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(active(X1), X2) → CONS(X1, X2)

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

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

trivial

Status:

active₁: [1]

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(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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:

FROM(active(X)) → FROM(X)
FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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.

FROM(active(X)) → FROM(X)

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

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

trivial

Status:

active₁: [1]

The following usable rules [FROCOS05] were oriented: none

(32) Obligation:

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

FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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.

FROM(mark(X)) → FROM(X)

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

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

trivial

Status:

mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(34) Obligation:

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

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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

(35) PisEmptyProof (EQUIVALENT transformation)

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

(36) TRUE

(37) Obligation:

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

MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(from(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(after(0, XS)) → MARK(XS)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(after(s(N), cons(X, XS))) → MARK(after(N, XS))
MARK(s(X)) → MARK(X)
MARK(after(X1, X2)) → ACTIVE(after(mark(X1), mark(X2)))
MARK(after(X1, X2)) → MARK(X1)
MARK(after(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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

(38) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
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
from(x1) = from
ACTIVE(x1) = x1
mark(x1) = mark(x1)
cons(x1, x2) = cons
s(x1) = s
after(x1, x2) = after
0 = 0
active(x1) = active(x1)

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

[MARK, from, after] > [mark₁, cons, s, active₁]
0 > [mark₁, cons, s, active₁]

Status:

MARK: []
from: []
mark₁: [1]
cons: []
s: []
after: []
0: []
active₁: [1]

The following usable rules [FROCOS05] were oriented:

from(active(X)) → from(X)
from(mark(X)) → from(X)
s(active(X)) → s(X)
s(mark(X)) → s(X)
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)
after(X1, mark(X2)) → after(X1, X2)
after(mark(X1), X2) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

(39) Obligation:

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

MARK(from(X)) → ACTIVE(from(mark(X)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(from(X)) → MARK(X)
ACTIVE(after(0, XS)) → MARK(XS)
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(after(s(N), cons(X, XS))) → MARK(after(N, XS))
MARK(s(X)) → MARK(X)
MARK(after(X1, X2)) → ACTIVE(after(mark(X1), mark(X2)))
MARK(after(X1, X2)) → MARK(X1)
MARK(after(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

active(from(X)) → mark(cons(X, from(s(X))))
active(after(0, XS)) → mark(XS)
active(after(s(N), cons(X, XS))) → mark(after(N, XS))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(s(X)) → active(s(mark(X)))
mark(after(X1, X2)) → active(after(mark(X1), mark(X2)))
mark(0) → active(0)
from(mark(X)) → from(X)
from(active(X)) → from(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)
s(mark(X)) → s(X)
s(active(X)) → s(X)
after(mark(X1), X2) → after(X1, X2)
after(X1, mark(X2)) → after(X1, X2)
after(active(X1), X2) → after(X1, X2)
after(X1, active(X2)) → after(X1, X2)

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