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 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
    ↳46 PisEmptyProof (⇔)
    ↳47 TRUE
  ↳48 QDP

(0) Obligation:

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

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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(f(X)) → MARK(cons(X, f(g(X))))
ACTIVE(f(X)) → CONS(X, f(g(X)))
ACTIVE(f(X)) → F(g(X))
ACTIVE(f(X)) → G(X)
ACTIVE(g(0)) → MARK(s(0))
ACTIVE(g(0)) → S(0)
ACTIVE(g(s(X))) → MARK(s(s(g(X))))
ACTIVE(g(s(X))) → S(s(g(X)))
ACTIVE(g(s(X))) → S(g(X))
ACTIVE(g(s(X))) → G(X)
ACTIVE(sel(0, cons(X, Y))) → MARK(X)
ACTIVE(sel(s(X), cons(Y, Z))) → MARK(sel(X, Z))
ACTIVE(sel(s(X), cons(Y, Z))) → SEL(X, Z)
MARK(f(X)) → ACTIVE(f(mark(X)))
MARK(f(X)) → F(mark(X))
MARK(f(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(g(X)) → ACTIVE(g(mark(X)))
MARK(g(X)) → G(mark(X))
MARK(g(X)) → MARK(X)
MARK(0) → ACTIVE(0)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2)))
MARK(sel(X1, X2)) → SEL(mark(X1), mark(X2))
MARK(sel(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → MARK(X2)
F(mark(X)) → F(X)
F(active(X)) → F(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)
G(mark(X)) → G(X)
G(active(X)) → G(X)
S(mark(X)) → S(X)
S(active(X)) → S(X)
SEL(mark(X1), X2) → SEL(X1, X2)
SEL(X1, mark(X2)) → SEL(X1, X2)
SEL(active(X1), X2) → SEL(X1, X2)
SEL(X1, active(X2)) → SEL(X1, X2)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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 6 SCCs with 14 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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.


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

Lexicographic Path Order [LPO].
Precedence:
mark1 > SEL1

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

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

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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.


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

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

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

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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.


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

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(11) Obligation:

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

SEL(mark(X1), X2) → SEL(X1, X2)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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.


SEL(mark(X1), X2) → SEL(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:
trivial

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(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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

Lexicographic Path Order [LPO].
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(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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: Combined order from the following AFS and order.
S(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

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(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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:

G(active(X)) → G(X)
G(mark(X)) → G(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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.


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

Lexicographic Path Order [LPO].
Precedence:
active1 > G1

The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

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

G(mark(X)) → G(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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.


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

Lexicographic Path Order [LPO].
Precedence:
trivial

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(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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:

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(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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.


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

Lexicographic Path Order [LPO].
Precedence:
mark1 > CONS1

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(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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.


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)  =  x2
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(34) 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(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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.


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

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(36) Obligation:

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

CONS(mark(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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.


CONS(mark(X1), X2) → CONS(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:
trivial

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(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(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:

F(active(X)) → F(X)
F(mark(X)) → F(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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.


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

Lexicographic Path Order [LPO].
Precedence:
active1 > F1

The following usable rules [FROCOS05] were oriented: none

(43) Obligation:

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

F(mark(X)) → F(X)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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.


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

Lexicographic Path Order [LPO].
Precedence:
trivial

The following usable rules [FROCOS05] were oriented: none

(45) Obligation:

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

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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

(46) PisEmptyProof (EQUIVALENT transformation)

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

(47) TRUE

(48) Obligation:

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

MARK(f(X)) → ACTIVE(f(mark(X)))
ACTIVE(f(X)) → MARK(cons(X, f(g(X))))
MARK(f(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(g(0)) → MARK(s(0))
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(g(s(X))) → MARK(s(s(g(X))))
MARK(cons(X1, X2)) → MARK(X1)
MARK(g(X)) → ACTIVE(g(mark(X)))
ACTIVE(sel(0, cons(X, Y))) → MARK(X)
MARK(g(X)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(sel(X1, X2)) → ACTIVE(sel(mark(X1), mark(X2)))
ACTIVE(sel(s(X), cons(Y, Z))) → MARK(sel(X, Z))
MARK(sel(X1, X2)) → MARK(X1)
MARK(sel(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

active(f(X)) → mark(cons(X, f(g(X))))
active(g(0)) → mark(s(0))
active(g(s(X))) → mark(s(s(g(X))))
active(sel(0, cons(X, Y))) → mark(X)
active(sel(s(X), cons(Y, Z))) → mark(sel(X, Z))
mark(f(X)) → active(f(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(g(X)) → active(g(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(sel(X1, X2)) → active(sel(mark(X1), mark(X2)))
f(mark(X)) → f(X)
f(active(X)) → f(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)
g(mark(X)) → g(X)
g(active(X)) → g(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
sel(mark(X1), X2) → sel(X1, X2)
sel(X1, mark(X2)) → sel(X1, X2)
sel(active(X1), X2) → sel(X1, X2)
sel(X1, active(X2)) → sel(X1, X2)

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