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

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 QDPSizeChangeProof (⇔, 0 ms)
        ↳7 YES
    ↳8 QDP
        ↳9 QDPSizeChangeProof (⇔, 0 ms)
        ↳10 YES
    ↳11 QDP
        ↳12 QDPOrderProof (⇔, 15 ms)
        ↳13 QDP
        ↳14 DependencyGraphProof (⇔, 0 ms)
        ↳15 QDP
        ↳16 QDPOrderProof (⇔, 13 ms)
        ↳17 QDP
        ↳18 QDPSizeChangeProof (⇔, 0 ms)
        ↳19 YES
    ↳20 QDP
        ↳21 QDPOrderProof (⇔, 127 ms)
        ↳22 QDP
        ↳23 DependencyGraphProof (⇔, 0 ms)
        ↳24 AND
            ↳25 QDP
                ↳26 Narrowing (⇔, 0 ms)
                ↳27 QDP
                ↳28 Narrowing (⇔, 0 ms)
                ↳29 QDP
                ↳30 DependencyGraphProof (⇔, 0 ms)
                ↳31 QDP
                ↳32 Narrowing (⇔, 0 ms)
                ↳33 QDP
                ↳34 DependencyGraphProof (⇔, 0 ms)
                ↳35 QDP
                ↳36 Narrowing (⇔, 0 ms)
                ↳37 QDP
                ↳38 DependencyGraphProof (⇔, 0 ms)
                ↳39 QDP
                ↳40 Narrowing (⇔, 0 ms)
                ↳41 QDP
                ↳42 DependencyGraphProof (⇔, 0 ms)
                ↳43 QDP
                ↳44 Narrowing (⇔, 0 ms)
                ↳45 QDP
                ↳46 QDPOrderProof (⇔, 63 ms)
                ↳47 QDP
                ↳48 PisEmptyProof (⇔, 0 ms)
                ↳49 YES
            ↳50 QDP
                ↳51 QDPSizeChangeProof (⇔, 0 ms)
                ↳52 YES
    ↳53 QDP
        ↳54 QDPSizeChangeProof (⇔, 0 ms)
        ↳55 YES
    ↳56 QDP
        ↳57 NonLoopProof (⇔, 398 ms)
        ↳58 NO

(0) Obligation:

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

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → 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:

SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
FIRST(0, Z) → NIL
FIRST(s(X), cons(Y, Z)) → CONS(Y, n__first(X, activate(Z)))
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
FROM(X) → CONS(X, n__from(s(X)))
FROM(X) → S(X)
SEL1(s(X), cons(Y, Z)) → SEL1(X, activate(Z))
SEL1(s(X), cons(Y, Z)) → ACTIVATE(Z)
SEL1(0, cons(X, Z)) → QUOTE(X)
FIRST1(s(X), cons(Y, Z)) → QUOTE(Y)
FIRST1(s(X), cons(Y, Z)) → FIRST1(X, activate(Z))
FIRST1(s(X), cons(Y, Z)) → ACTIVATE(Z)
QUOTE1(n__cons(X, Z)) → QUOTE(activate(X))
QUOTE1(n__cons(X, Z)) → ACTIVATE(X)
QUOTE1(n__cons(X, Z)) → QUOTE1(activate(Z))
QUOTE1(n__cons(X, Z)) → ACTIVATE(Z)
QUOTE(n__s(X)) → QUOTE(activate(X))
QUOTE(n__s(X)) → ACTIVATE(X)
QUOTE(n__sel(X, Z)) → SEL1(activate(X), activate(Z))
QUOTE(n__sel(X, Z)) → ACTIVATE(X)
QUOTE(n__sel(X, Z)) → ACTIVATE(Z)
QUOTE1(n__first(X, Z)) → FIRST1(activate(X), activate(Z))
QUOTE1(n__first(X, Z)) → ACTIVATE(X)
QUOTE1(n__first(X, Z)) → ACTIVATE(Z)
UNQUOTE(01) → 01
UNQUOTE(s1(X)) → S(unquote(X))
UNQUOTE(s1(X)) → UNQUOTE(X)
UNQUOTE1(nil1) → NIL
UNQUOTE1(cons1(X, Z)) → FCONS(unquote(X), unquote1(Z))
UNQUOTE1(cons1(X, Z)) → UNQUOTE(X)
UNQUOTE1(cons1(X, Z)) → UNQUOTE1(Z)
FCONS(X, Z) → CONS(X, Z)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
ACTIVATE(n__from(X)) → FROM(X)
ACTIVATE(n__0) → 01
ACTIVATE(n__cons(X1, X2)) → CONS(X1, X2)
ACTIVATE(n__nil) → NIL
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__sel(X1, X2)) → SEL(X1, X2)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → 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 6 SCCs with 27 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

UNQUOTE(s1(X)) → UNQUOTE(X)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(6) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • UNQUOTE(s1(X)) → UNQUOTE(X)
    The graph contains the following edges 1 > 1

(7) YES

(8) Obligation:

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

UNQUOTE1(cons1(X, Z)) → UNQUOTE1(Z)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(9) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • UNQUOTE1(cons1(X, Z)) → UNQUOTE1(Z)
    The graph contains the following edges 1 > 1

(10) YES

(11) Obligation:

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

SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__sel(X1, X2)) → SEL(X1, X2)
SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( SEL(x1, x2) ) = x2 + 2


POL( activate(x1) ) = x1


POL( n__first(x1, x2) ) = x2 + 2


POL( first(x1, x2) ) = x2 + 2


POL( n__from(x1) ) = 2x1


POL( from(x1) ) = 2x1


POL( n__0 ) = 0


POL( 0 ) = 0


POL( n__cons(x1, x2) ) = x1 + x2


POL( cons(x1, x2) ) = x1 + x2


POL( n__nil ) = 0


POL( nil ) = 0


POL( n__s(x1) ) = 0


POL( s(x1) ) = 0


POL( n__sel(x1, x2) ) = 2x2


POL( sel(x1, x2) ) = 2x2


POL( ACTIVATE(x1) ) = x1 + 2


POL( FIRST(x1, x2) ) = x2 + 2



The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X
first(0, Z) → nil
first(X1, X2) → n__first(X1, X2)
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
cons(X1, X2) → n__cons(X1, X2)
sel(0, cons(X, Z)) → X
sel(X1, X2) → n__sel(X1, X2)
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
nil → n__nil
from(X) → cons(X, n__from(s(X)))
from(X) → n__from(X)
s(X) → n__s(X)
0 → n__0

(13) Obligation:

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

SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__sel(X1, X2)) → SEL(X1, X2)
SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(14) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(15) Obligation:

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

ACTIVATE(n__sel(X1, X2)) → SEL(X1, X2)
SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


ACTIVATE(n__sel(X1, X2)) → SEL(X1, X2)
SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( SEL(x1, x2) ) = x2 + 2


POL( activate(x1) ) = x1 + 1


POL( n__first(x1, x2) ) = max{0, x2 - 1}


POL( first(x1, x2) ) = x2


POL( n__from(x1) ) = 2x1


POL( from(x1) ) = 2x1 + 1


POL( n__0 ) = 1


POL( 0 ) = 1


POL( n__cons(x1, x2) ) = 2x1 + 2x2 + 1


POL( cons(x1, x2) ) = 2x1 + 2x2 + 1


POL( n__nil ) = 0


POL( nil ) = 0


POL( n__s(x1) ) = 0


POL( s(x1) ) = 0


POL( n__sel(x1, x2) ) = 2x2 + 1


POL( sel(x1, x2) ) = 2x2 + 2


POL( ACTIVATE(x1) ) = 2x1 + 1



The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X
first(0, Z) → nil
first(X1, X2) → n__first(X1, X2)
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
cons(X1, X2) → n__cons(X1, X2)
sel(0, cons(X, Z)) → X
sel(X1, X2) → n__sel(X1, X2)
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
nil → n__nil
from(X) → cons(X, n__from(s(X)))
from(X) → n__from(X)
s(X) → n__s(X)
0 → n__0

(17) Obligation:

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

SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(18) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
    The graph contains the following edges 1 > 1

(19) YES

(20) Obligation:

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

SEL1(0, cons(X, Z)) → QUOTE(X)
QUOTE(n__s(X)) → QUOTE(activate(X))
QUOTE(n__sel(X, Z)) → SEL1(activate(X), activate(Z))
SEL1(s(X), cons(Y, Z)) → SEL1(X, activate(Z))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


QUOTE(n__sel(X, Z)) → SEL1(activate(X), activate(Z))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(SEL1(x1, x2)) = -I + 0A·x1 + 0A·x2

POL(0) = 0A

POL(cons(x1, x2)) = -I + 0A·x1 + 0A·x2

POL(QUOTE(x1)) = -I + 0A·x1

POL(n__s(x1)) = -I + 0A·x1

POL(activate(x1)) = -I + 0A·x1

POL(n__sel(x1, x2)) = -I + 1A·x1 + 1A·x2

POL(s(x1)) = -I + 0A·x1

POL(n__first(x1, x2)) = -I + 0A·x1 + 5A·x2

POL(first(x1, x2)) = -I + 0A·x1 + 5A·x2

POL(n__from(x1)) = -I + 0A·x1

POL(from(x1)) = -I + 0A·x1

POL(n__0) = 0A

POL(n__cons(x1, x2)) = -I + 0A·x1 + 0A·x2

POL(n__nil) = 0A

POL(nil) = 0A

POL(sel(x1, x2)) = -I + 1A·x1 + 1A·x2

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X
first(0, Z) → nil
first(X1, X2) → n__first(X1, X2)
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
cons(X1, X2) → n__cons(X1, X2)
sel(0, cons(X, Z)) → X
sel(X1, X2) → n__sel(X1, X2)
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
nil → n__nil
from(X) → cons(X, n__from(s(X)))
from(X) → n__from(X)
s(X) → n__s(X)
0 → n__0

(22) Obligation:

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

SEL1(0, cons(X, Z)) → QUOTE(X)
QUOTE(n__s(X)) → QUOTE(activate(X))
SEL1(s(X), cons(Y, Z)) → SEL1(X, activate(Z))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(23) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(24) Complex Obligation (AND)

(25) Obligation:

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

QUOTE(n__s(X)) → QUOTE(activate(X))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(26) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule QUOTE(n__s(X)) → QUOTE(activate(X)) at position [0] we obtained the following new rules [LPAR04]:

QUOTE(n__s(n__first(x0, x1))) → QUOTE(first(x0, x1))
QUOTE(n__s(n__from(x0))) → QUOTE(from(x0))
QUOTE(n__s(n__0)) → QUOTE(0)
QUOTE(n__s(n__cons(x0, x1))) → QUOTE(cons(x0, x1))
QUOTE(n__s(n__nil)) → QUOTE(nil)
QUOTE(n__s(n__s(x0))) → QUOTE(s(x0))
QUOTE(n__s(n__sel(x0, x1))) → QUOTE(sel(x0, x1))
QUOTE(n__s(x0)) → QUOTE(x0)

(27) Obligation:

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

QUOTE(n__s(n__first(x0, x1))) → QUOTE(first(x0, x1))
QUOTE(n__s(n__from(x0))) → QUOTE(from(x0))
QUOTE(n__s(n__0)) → QUOTE(0)
QUOTE(n__s(n__cons(x0, x1))) → QUOTE(cons(x0, x1))
QUOTE(n__s(n__nil)) → QUOTE(nil)
QUOTE(n__s(n__s(x0))) → QUOTE(s(x0))
QUOTE(n__s(n__sel(x0, x1))) → QUOTE(sel(x0, x1))
QUOTE(n__s(x0)) → QUOTE(x0)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(28) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule QUOTE(n__s(n__from(x0))) → QUOTE(from(x0)) at position [0] we obtained the following new rules [LPAR04]:

QUOTE(n__s(n__from(x0))) → QUOTE(cons(x0, n__from(s(x0))))
QUOTE(n__s(n__from(x0))) → QUOTE(n__from(x0))

(29) Obligation:

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

QUOTE(n__s(n__first(x0, x1))) → QUOTE(first(x0, x1))
QUOTE(n__s(n__0)) → QUOTE(0)
QUOTE(n__s(n__cons(x0, x1))) → QUOTE(cons(x0, x1))
QUOTE(n__s(n__nil)) → QUOTE(nil)
QUOTE(n__s(n__s(x0))) → QUOTE(s(x0))
QUOTE(n__s(n__sel(x0, x1))) → QUOTE(sel(x0, x1))
QUOTE(n__s(x0)) → QUOTE(x0)
QUOTE(n__s(n__from(x0))) → QUOTE(cons(x0, n__from(s(x0))))
QUOTE(n__s(n__from(x0))) → QUOTE(n__from(x0))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(30) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(31) Obligation:

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

QUOTE(n__s(n__first(x0, x1))) → QUOTE(first(x0, x1))
QUOTE(n__s(n__0)) → QUOTE(0)
QUOTE(n__s(n__cons(x0, x1))) → QUOTE(cons(x0, x1))
QUOTE(n__s(n__nil)) → QUOTE(nil)
QUOTE(n__s(n__s(x0))) → QUOTE(s(x0))
QUOTE(n__s(n__sel(x0, x1))) → QUOTE(sel(x0, x1))
QUOTE(n__s(x0)) → QUOTE(x0)
QUOTE(n__s(n__from(x0))) → QUOTE(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(32) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule QUOTE(n__s(n__0)) → QUOTE(0) at position [0] we obtained the following new rules [LPAR04]:

QUOTE(n__s(n__0)) → QUOTE(n__0)

(33) Obligation:

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

QUOTE(n__s(n__first(x0, x1))) → QUOTE(first(x0, x1))
QUOTE(n__s(n__cons(x0, x1))) → QUOTE(cons(x0, x1))
QUOTE(n__s(n__nil)) → QUOTE(nil)
QUOTE(n__s(n__s(x0))) → QUOTE(s(x0))
QUOTE(n__s(n__sel(x0, x1))) → QUOTE(sel(x0, x1))
QUOTE(n__s(x0)) → QUOTE(x0)
QUOTE(n__s(n__from(x0))) → QUOTE(cons(x0, n__from(s(x0))))
QUOTE(n__s(n__0)) → QUOTE(n__0)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(34) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(35) Obligation:

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

QUOTE(n__s(n__first(x0, x1))) → QUOTE(first(x0, x1))
QUOTE(n__s(n__cons(x0, x1))) → QUOTE(cons(x0, x1))
QUOTE(n__s(n__nil)) → QUOTE(nil)
QUOTE(n__s(n__s(x0))) → QUOTE(s(x0))
QUOTE(n__s(n__sel(x0, x1))) → QUOTE(sel(x0, x1))
QUOTE(n__s(x0)) → QUOTE(x0)
QUOTE(n__s(n__from(x0))) → QUOTE(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(36) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule QUOTE(n__s(n__cons(x0, x1))) → QUOTE(cons(x0, x1)) at position [0] we obtained the following new rules [LPAR04]:

QUOTE(n__s(n__cons(x0, x1))) → QUOTE(n__cons(x0, x1))

(37) Obligation:

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

QUOTE(n__s(n__first(x0, x1))) → QUOTE(first(x0, x1))
QUOTE(n__s(n__nil)) → QUOTE(nil)
QUOTE(n__s(n__s(x0))) → QUOTE(s(x0))
QUOTE(n__s(n__sel(x0, x1))) → QUOTE(sel(x0, x1))
QUOTE(n__s(x0)) → QUOTE(x0)
QUOTE(n__s(n__from(x0))) → QUOTE(cons(x0, n__from(s(x0))))
QUOTE(n__s(n__cons(x0, x1))) → QUOTE(n__cons(x0, x1))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(38) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(39) Obligation:

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

QUOTE(n__s(n__first(x0, x1))) → QUOTE(first(x0, x1))
QUOTE(n__s(n__nil)) → QUOTE(nil)
QUOTE(n__s(n__s(x0))) → QUOTE(s(x0))
QUOTE(n__s(n__sel(x0, x1))) → QUOTE(sel(x0, x1))
QUOTE(n__s(x0)) → QUOTE(x0)
QUOTE(n__s(n__from(x0))) → QUOTE(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(40) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule QUOTE(n__s(n__nil)) → QUOTE(nil) at position [0] we obtained the following new rules [LPAR04]:

QUOTE(n__s(n__nil)) → QUOTE(n__nil)

(41) Obligation:

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

QUOTE(n__s(n__first(x0, x1))) → QUOTE(first(x0, x1))
QUOTE(n__s(n__s(x0))) → QUOTE(s(x0))
QUOTE(n__s(n__sel(x0, x1))) → QUOTE(sel(x0, x1))
QUOTE(n__s(x0)) → QUOTE(x0)
QUOTE(n__s(n__from(x0))) → QUOTE(cons(x0, n__from(s(x0))))
QUOTE(n__s(n__nil)) → QUOTE(n__nil)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(42) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(43) Obligation:

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

QUOTE(n__s(n__first(x0, x1))) → QUOTE(first(x0, x1))
QUOTE(n__s(n__s(x0))) → QUOTE(s(x0))
QUOTE(n__s(n__sel(x0, x1))) → QUOTE(sel(x0, x1))
QUOTE(n__s(x0)) → QUOTE(x0)
QUOTE(n__s(n__from(x0))) → QUOTE(cons(x0, n__from(s(x0))))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(44) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule QUOTE(n__s(n__s(x0))) → QUOTE(s(x0)) at position [0] we obtained the following new rules [LPAR04]:

QUOTE(n__s(n__s(x0))) → QUOTE(n__s(x0))

(45) Obligation:

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

QUOTE(n__s(n__first(x0, x1))) → QUOTE(first(x0, x1))
QUOTE(n__s(n__sel(x0, x1))) → QUOTE(sel(x0, x1))
QUOTE(n__s(x0)) → QUOTE(x0)
QUOTE(n__s(n__from(x0))) → QUOTE(cons(x0, n__from(s(x0))))
QUOTE(n__s(n__s(x0))) → QUOTE(n__s(x0))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(46) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


QUOTE(n__s(n__first(x0, x1))) → QUOTE(first(x0, x1))
QUOTE(n__s(n__sel(x0, x1))) → QUOTE(sel(x0, x1))
QUOTE(n__s(x0)) → QUOTE(x0)
QUOTE(n__s(n__from(x0))) → QUOTE(cons(x0, n__from(s(x0))))
QUOTE(n__s(n__s(x0))) → QUOTE(n__s(x0))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOTE(x1)  =  x1
n__s(x1)  =  n__s(x1)
n__first(x1, x2)  =  n__first(x1, x2)
first(x1, x2)  =  first(x1, x2)
n__sel(x1, x2)  =  n__sel(x1, x2)
sel(x1, x2)  =  sel(x1, x2)
n__from(x1)  =  n__from(x1)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)
0  =  0
nil  =  nil
activate(x1)  =  activate(x1)
n__cons(x1, x2)  =  n__cons(x1, x2)
from(x1)  =  from(x1)
n__0  =  n__0
n__nil  =  n__nil

Recursive path order with status [RPO].
Quasi-Precedence:
[nfirst2, first2, nil, nnil] > activate1 > 0 > n0
[nfirst2, first2, nil, nnil] > activate1 > from1 > [ns1, s1] > [nfrom1, cons2, ncons2]
[nsel2, sel2] > activate1 > 0 > n0
[nsel2, sel2] > activate1 > from1 > [ns1, s1] > [nfrom1, cons2, ncons2]

Status:
ns1: multiset
nfirst2: [1,2]
first2: [1,2]
nsel2: [1,2]
sel2: [1,2]
nfrom1: multiset
cons2: [2,1]
s1: multiset
0: multiset
nil: multiset
activate1: multiset
ncons2: [2,1]
from1: [1]
n0: multiset
nnil: multiset


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
first(X1, X2) → n__first(X1, X2)
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
sel(X1, X2) → n__sel(X1, X2)
s(X) → n__s(X)
cons(X1, X2) → n__cons(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(X) → X
activate(n__sel(X1, X2)) → sel(X1, X2)
nil → n__nil
from(X) → cons(X, n__from(s(X)))
from(X) → n__from(X)
0 → n__0

(47) Obligation:

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

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(48) PisEmptyProof (EQUIVALENT transformation)

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

(49) YES

(50) Obligation:

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

SEL1(s(X), cons(Y, Z)) → SEL1(X, activate(Z))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(51) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • SEL1(s(X), cons(Y, Z)) → SEL1(X, activate(Z))
    The graph contains the following edges 1 > 1

(52) YES

(53) Obligation:

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

FIRST1(s(X), cons(Y, Z)) → FIRST1(X, activate(Z))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(54) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • FIRST1(s(X), cons(Y, Z)) → FIRST1(X, activate(Z))
    The graph contains the following edges 1 > 1

(55) YES

(56) Obligation:

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

QUOTE1(n__cons(X, Z)) → QUOTE1(activate(Z))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(Z))
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
0 → n__0
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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

(57) NonLoopProof (EQUIVALENT transformation)

By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 0,
σ' = [ ], and μ' = [x1 / x0, x0 / n__s(x0)] on the rule
QUOTE1(n__cons(x0, n__from(n__s(x0))))[ ]n[ ] → QUOTE1(n__cons(x0, n__from(n__s(x0))))[ ]n[x1 / x0, x0 / n__s(x0)]
This rule is correct for the QDP as the following derivation shows:

intermediate steps: Equivalent (Simplify mu) - Instantiate mu - Instantiation
QUOTE1(n__cons(x1, n__from(y0)))[ ]n[ ] → QUOTE1(n__cons(y0, n__from(n__s(y0))))[ ]n[ ]
    by Rewrite t
        QUOTE1(n__cons(x1, n__from(y0)))[ ]n[ ] → QUOTE1(from(y0))[ ]n[ ]
            by Narrowing at position: [0]
                intermediate steps: Instantiation - Instantiation
                QUOTE1(n__cons(X, Z))[ ]n[ ] → QUOTE1(activate(Z))[ ]n[ ]
                    by OriginalRule from TRS P

                intermediate steps: Instantiation
                activate(n__from(X))[ ]n[ ] → from(X)[ ]n[ ]
                    by OriginalRule from TRS R

(58) NO