Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 PisEmptyProof (⇔)
        ↳9 TRUE
    ↳10 QDP
        ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 QDPOrderProof (⇔)
        ↳14 QDP
        ↳15 PisEmptyProof (⇔)
        ↳16 TRUE
    ↳17 QDP
        ↳18 QDPOrderProof (⇔)
        ↳19 QDP
        ↳20 PisEmptyProof (⇔)
        ↳21 TRUE
    ↳22 QDP
        ↳23 QDPOrderProof (⇔)
        ↳24 QDP
        ↳25 QDPOrderProof (⇔)
        ↳26 QDP
        ↳27 PisEmptyProof (⇔)
        ↳28 TRUE
    ↳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
        ↳42 QDPOrderProof (⇔)
        ↳43 QDP
        ↳44 QDPOrderProof (⇔)
        ↳45 QDP
        ↳46 PisEmptyProof (⇔)
        ↳47 TRUE

(0) Obligation:

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

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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(first(s(X), cons(Y, Z))) → CONS(Y, first(X, Z))
ACTIVE(first(s(X), cons(Y, Z))) → FIRST(X, Z)
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))
ACTIVE(from(X)) → S(X)
ACTIVE(first(X1, X2)) → FIRST(active(X1), X2)
ACTIVE(first(X1, X2)) → ACTIVE(X1)
ACTIVE(first(X1, X2)) → FIRST(X1, active(X2))
ACTIVE(first(X1, X2)) → ACTIVE(X2)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(from(X)) → FROM(active(X))
ACTIVE(from(X)) → ACTIVE(X)
FIRST(mark(X1), X2) → FIRST(X1, X2)
FIRST(X1, mark(X2)) → FIRST(X1, X2)
S(mark(X)) → S(X)
CONS(mark(X1), X2) → CONS(X1, X2)
FROM(mark(X)) → FROM(X)
PROPER(first(X1, X2)) → FIRST(proper(X1), proper(X2))
PROPER(first(X1, X2)) → PROPER(X1)
PROPER(first(X1, X2)) → PROPER(X2)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(from(X)) → FROM(proper(X))
PROPER(from(X)) → PROPER(X)
FIRST(ok(X1), ok(X2)) → FIRST(X1, X2)
S(ok(X)) → S(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
FROM(ok(X)) → FROM(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(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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 7 SCCs with 16 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

FROM(ok(X)) → FROM(X)
FROM(mark(X)) → FROM(X)

The TRS R consists of the following rules:

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.


FROM(ok(X)) → FROM(X)
FROM(mark(X)) → FROM(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(FROM(x1)) = x1   
POL(mark(x1)) = 1 + x1   
POL(ok(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(11) 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: Polynomial interpretation [POLO]:

POL(CONS(x1, x2)) = x2   
POL(mark(x1)) = 0   
POL(ok(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(12) 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(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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: Polynomial interpretation [POLO]:

POL(CONS(x1, x2)) = x1   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

S(ok(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


S(ok(X)) → S(X)
S(mark(X)) → S(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(S(x1)) = x1   
POL(mark(x1)) = 1 + x1   
POL(ok(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) Obligation:

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

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

The TRS R consists of the following rules:

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FIRST(mark(X1), X2) → FIRST(X1, X2)
FIRST(ok(X1), ok(X2)) → FIRST(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(FIRST(x1, x2)) = x1   
POL(mark(x1)) = 1 + x1   
POL(ok(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(24) Obligation:

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

FIRST(X1, mark(X2)) → FIRST(X1, X2)

The TRS R consists of the following rules:

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(25) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FIRST(X1, mark(X2)) → FIRST(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(FIRST(x1, x2)) = x2   
POL(mark(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(26) Obligation:

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

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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) PisEmptyProof (EQUIVALENT transformation)

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

(28) TRUE

(29) Obligation:

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

PROPER(first(X1, X2)) → PROPER(X2)
PROPER(first(X1, X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(from(X)) → PROPER(X)

The TRS R consists of the following rules:

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(first(X1, X2)) → PROPER(X2)
PROPER(first(X1, X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(from(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(PROPER(x1)) = x1   
POL(cons(x1, x2)) = 1 + x1 + x2   
POL(first(x1, x2)) = 1 + x1 + x2   
POL(from(x1)) = 1 + x1   
POL(s(x1)) = 1 + x1   

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(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(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:

ACTIVE(first(X1, X2)) → ACTIVE(X2)
ACTIVE(first(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(from(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(first(X1, X2)) → ACTIVE(X2)
ACTIVE(first(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = x1   
POL(cons(x1, x2)) = 1 + x1   
POL(first(x1, x2)) = 1 + x1 + x2   
POL(from(x1)) = x1   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(36) Obligation:

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

ACTIVE(from(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(37) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(from(X)) → ACTIVE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(ACTIVE(x1)) = x1   
POL(from(x1)) = 1 + x1   

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(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(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:

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

The TRS R consists of the following rules:

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(42) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TOP(mark(X)) → TOP(proper(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 1   
POL(TOP(x1)) = x1   
POL(active(x1)) = x1   
POL(cons(x1, x2)) = x1   
POL(first(x1, x2)) = x1 + x2   
POL(from(x1)) = 1 + x1   
POL(mark(x1)) = 1 + x1   
POL(nil) = 0   
POL(ok(x1)) = x1   
POL(proper(x1)) = x1   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented:

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))

(43) Obligation:

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

TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(44) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


TOP(ok(X)) → TOP(active(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 1   
POL(TOP(x1)) = x1   
POL(active(x1)) = x1   
POL(cons(x1, x2)) = x2   
POL(first(x1, x2)) = x1   
POL(from(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(ok(x1)) = 1 + x1   
POL(s(x1)) = x1   

The following usable rules [FROCOS05] were oriented:

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))

(45) Obligation:

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

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(s(X)) → s(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
s(mark(X)) → mark(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(s(X)) → s(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
s(ok(X)) → ok(s(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(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.

(46) PisEmptyProof (EQUIVALENT transformation)

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

(47) TRUE