Termination w.r.t. Q proof of /tmp/tmpUR8uhy/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 QDPOrderProof (⇔)

    ↳13 QDP

    ↳14 PisEmptyProof (⇔)

    ↳15 TRUE

  ↳16 QDP

    ↳17 QDPOrderProof (⇔)

    ↳18 QDP

    ↳19 QDPOrderProof (⇔)

    ↳20 QDP

    ↳21 PisEmptyProof (⇔)

    ↳22 TRUE

  ↳23 QDP

    ↳24 QDPOrderProof (⇔)

    ↳25 QDP

    ↳26 QDPOrderProof (⇔)

    ↳27 QDP

    ↳28 QDPOrderProof (⇔)

    ↳29 QDP

    ↳30 QDPOrderProof (⇔)

    ↳31 QDP

    ↳32 PisEmptyProof (⇔)

    ↳33 TRUE

  ↳34 QDP

    ↳35 QDPOrderProof (⇔)

    ↳36 QDP

    ↳37 QDPOrderProof (⇔)

    ↳38 QDP

    ↳39 PisEmptyProof (⇔)

    ↳40 TRUE

  ↳41 QDP

(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)

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) = x1

Recursive Path Order [RPO].
Precedence:

mark₁ > AFTER₁

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)
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.

(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) = AFTER(x1, x2)
mark(x1) = mark(x1)
active(x1) = x1

Recursive Path Order [RPO].
Precedence:

mark₁ > AFTER₂

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)
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.

(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) = AFTER(x1)
active(x1) = active(x1)

Recursive Path Order [RPO].
Precedence:

active₁ > AFTER₁

The following usable rules [FROCOS05] were oriented: none

(11) Obligation:

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

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.

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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)
active(x1) = active(x1)

Recursive Path Order [RPO].
Precedence:

active₁ > AFTER₁

The following usable rules [FROCOS05] were oriented: none

(13) 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.

(14) PisEmptyProof (EQUIVALENT transformation)

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

(15) TRUE

(16) 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.

(17) 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:

active₁ > S₁

The following usable rules [FROCOS05] were oriented: none

(18) 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.

(19) 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

(20) 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.

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE

(23) Obligation:

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

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

The TRS R consists of the following rules:

active(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(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:

mark₁ > CONS₁

The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

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

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

The TRS R consists of the following rules:

active(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(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:

mark₁ > CONS₂

The following usable rules [FROCOS05] were oriented: none

(27) Obligation:

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

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

The TRS R consists of the following rules:

active(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) 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:

active₁ > CONS₁

The following usable rules [FROCOS05] were oriented: none

(29) 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(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.

(30) 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

(31) 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.

(32) PisEmptyProof (EQUIVALENT transformation)

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

(33) TRUE

(34) Obligation:

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

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.

(35) 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) = FROM(x1)
active(x1) = active(x1)
mark(x1) = x1

Recursive Path Order [RPO].
Precedence:

active₁ > FROM₁

The following usable rules [FROCOS05] were oriented: none

(36) 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.

(37) 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: Recursive Path Order [RPO].
Precedence:

mark₁ > FROM₁

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(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.

(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(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.