Termination w.r.t. Q proof of Transformed_CSR_04_Ex6_15_AEL02

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 UsableRulesProof (⇔, 0 ms)

        ↳7 QDP

        ↳8 QDPSizeChangeProof (⇔, 0 ms)

        ↳9 YES

    ↳10 QDP

        ↳11 UsableRulesProof (⇔, 0 ms)

        ↳12 QDP

        ↳13 QDPSizeChangeProof (⇔, 0 ms)

        ↳14 YES

    ↳15 QDP

        ↳16 QDPOrderProof (⇔, 33 ms)

        ↳17 QDP

        ↳18 DependencyGraphProof (⇔, 0 ms)

        ↳19 AND

            ↳20 QDP

                ↳21 UsableRulesProof (⇔, 0 ms)

                ↳22 QDP

                ↳23 QDPSizeChangeProof (⇔, 0 ms)

                ↳24 YES

            ↳25 QDP

                ↳26 QDPSizeChangeProof (⇔, 0 ms)

                ↳27 YES

    ↳28 QDP

        ↳29 QDPOrderProof (⇔, 94 ms)

        ↳30 QDP

        ↳31 DependencyGraphProof (⇔, 0 ms)

        ↳32 AND

            ↳33 QDP

                ↳34 QDPSizeChangeProof (⇔, 0 ms)

                ↳35 YES

            ↳36 QDP

                ↳37 TransformationProof (⇔, 0 ms)

                ↳38 QDP

                ↳39 TransformationProof (⇔, 0 ms)

                ↳40 QDP

                ↳41 DependencyGraphProof (⇔, 0 ms)

                ↳42 QDP

                ↳43 TransformationProof (⇔, 0 ms)

                ↳44 QDP

                ↳45 DependencyGraphProof (⇔, 0 ms)

                ↳46 QDP

                ↳47 TransformationProof (⇔, 0 ms)

                ↳48 QDP

                ↳49 DependencyGraphProof (⇔, 0 ms)

                ↳50 QDP

                ↳51 TransformationProof (⇔, 0 ms)

                ↳52 QDP

                ↳53 DependencyGraphProof (⇔, 0 ms)

                ↳54 QDP

                ↳55 TransformationProof (⇔, 0 ms)

                ↳56 QDP

                ↳57 QDPOrderProof (⇔, 83 ms)

                ↳58 QDP

                ↳59 PisEmptyProof (⇔, 0 ms)

                ↳60 YES

    ↳61 QDP

        ↳62 QDPSizeChangeProof (⇔, 0 ms)

        ↳63 YES

    ↳64 QDP

        ↳65 TransformationProof (⇔, 0 ms)

        ↳66 QDP

        ↳67 TransformationProof (⇔, 0 ms)

        ↳68 QDP

        ↳69 DependencyGraphProof (⇔, 0 ms)

        ↳70 QDP

        ↳71 TransformationProof (⇔, 0 ms)

        ↳72 QDP

        ↳73 DependencyGraphProof (⇔, 0 ms)

        ↳74 QDP

        ↳75 TransformationProof (⇔, 0 ms)

        ↳76 QDP

        ↳77 TransformationProof (⇔, 0 ms)

        ↳78 QDP

        ↳79 DependencyGraphProof (⇔, 0 ms)

        ↳80 QDP

        ↳81 TransformationProof (⇔, 0 ms)

        ↳82 QDP

        ↳83 DependencyGraphProof (⇔, 0 ms)

        ↳84 QDP

        ↳85 QDPOrderProof (⇔, 19 ms)

        ↳86 QDP

        ↳87 QDPOrderProof (⇔, 0 ms)

        ↳88 QDP

        ↳89 NonTerminationLoopProof (⇒, 0 ms)

        ↳90 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) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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

UNQUOTE(s1(X)) → UNQUOTE(X)

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

(8) 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

(9) YES

(10) 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.

(11) UsableRulesProof (EQUIVALENT transformation)

(12) Obligation:

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

UNQUOTE1(cons1(X, Z)) → UNQUOTE1(Z)

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

(13) 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

(14) YES

(15) 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.

(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)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( SEL(x₁, x₂) ) = x₂ + 1

POL( activate(x₁) ) = 2x₁

POL( n__first(x₁, x₂) ) = x₂

POL( first(x₁, x₂) ) = 2x₂

POL( n__from(x₁) ) = x₁ + 1

POL( from(x₁) ) = 2x₁ + 2

POL( n__0 ) = 0

POL( 0 ) = 0

POL( n__cons(x₁, x₂) ) = x₁ + x₂

POL( cons(x₁, x₂) ) = x₁ + 2x₂

POL( n__nil ) = 0

POL( nil ) = 0

POL( n__s(x₁) ) = 0

POL( s(x₁) ) = 0

POL( n__sel(x₁, x₂) ) = 2x₂ + 2

POL( sel(x₁, x₂) ) = 2x₂ + 2

POL( ACTIVATE(x₁) ) = x₁ + 1

POL( FIRST(x₁, x₂) ) = x₂ + 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)) → ACTIVATE(Z)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
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) DependencyGraphProof (EQUIVALENT transformation)

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

(19) Complex Obligation (AND)

(20) Obligation:

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

FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__first(X1, X2)) → FIRST(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.

(21) UsableRulesProof (EQUIVALENT transformation)

(22) Obligation:

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

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

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

(23) 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:

ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
The graph contains the following edges 1 > 1, 1 > 2
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
The graph contains the following edges 2 > 1

(24) YES

(25) 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.

(26) 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

(27) YES

(28) 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.

(29) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

SEL1(0, cons(X, Z)) → QUOTE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(SEL1(x₁, x₂)) = 5A + 4A · x₁ + 5A · x₂

POL(0) = 3A

POL(cons(x₁, x₂)) = -I + 3A · x₁ + 0A · x₂

POL(QUOTE(x₁)) = 5A + 3A · x₁

POL(n__s(x₁)) = -I + 0A · x₁

POL(activate(x₁)) = -I + 0A · x₁

POL(n__sel(x₁, x₂)) = -I + 3A · x₁ + 2A · x₂

POL(s(x₁)) = -I + 0A · x₁

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

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

POL(n__from(x₁)) = -I + 5A · x₁

POL(from(x₁)) = -I + 5A · x₁

POL(n__0) = 3A

POL(n__cons(x₁, x₂)) = -I + 3A · x₁ + 0A · x₂

POL(n__nil) = 0A

POL(nil) = 0A

POL(sel(x₁, x₂)) = -I + 3A · x₁ + 2A · 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

(30) Obligation:

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

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.

(31) DependencyGraphProof (EQUIVALENT transformation)

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

(32) Complex Obligation (AND)

(33) 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.

(34) 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

(35) YES

(36) 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.

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

(38) 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.

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

(40) 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.

(41) DependencyGraphProof (EQUIVALENT transformation)

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

(42) 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.

(43) TransformationProof (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) → QUOTE(n__s(n__0)) → QUOTE(n__0)

(44) 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.

(45) DependencyGraphProof (EQUIVALENT transformation)

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

(46) 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.

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

(48) 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.

(49) DependencyGraphProof (EQUIVALENT transformation)

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

(50) 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.

(51) TransformationProof (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) → QUOTE(n__s(n__nil)) → QUOTE(n__nil)

(52) 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.

(53) DependencyGraphProof (EQUIVALENT transformation)

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

(54) 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.

(55) TransformationProof (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)) → QUOTE(n__s(n__s(x0))) → QUOTE(n__s(x0))

(56) 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.

(57) 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:

[n_{_first₂}, first₂] > activate₁ > 0 > nil > n_{_nil}
[n_{_first₂}, first₂] > activate₁ > 0 > n_₀
[n_{_first₂}, first₂] > activate₁ > from₁ > [n_{_s₁}, s₁] > n_{_from₁}
[n_{_first₂}, first₂] > activate₁ > from₁ > [n_{_s₁}, s₁] > [cons₂, n_{_cons₂}]
[n_{_sel₂}, sel₂] > activate₁ > 0 > nil > n_{_nil}
[n_{_sel₂}, sel₂] > activate₁ > 0 > n_₀
[n_{_sel₂}, sel₂] > activate₁ > from₁ > [n_{_s₁}, s₁] > n_{_from₁}
[n_{_sel₂}, sel₂] > activate₁ > from₁ > [n_{_s₁}, s₁] > [cons₂, n_{_cons₂}]

Status:

n_{_s₁}: multiset
n_{_first₂}: [1,2]
first₂: [1,2]
n_{_sel₂}: [1,2]
sel₂: [1,2]
n_{_from₁}: multiset
cons₂: multiset
s₁: multiset
0: multiset
nil: multiset
activate₁: multiset
n_{_cons₂}: multiset
from₁: multiset
n_₀: multiset
n_{_nil}: 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

(58) 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.

(59) PisEmptyProof (EQUIVALENT transformation)

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

(60) YES

(61) 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.

(62) 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

(63) YES

(64) 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.

(65) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule QUOTE1(n__cons(X, Z)) → QUOTE1(activate(Z)) at position [0] we obtained the following new rules [LPAR04]:

QUOTE1(n__cons(y0, n__first(x0, x1))) → QUOTE1(first(x0, x1)) → QUOTE1(n__cons(y0, n__first(x0, x1))) → QUOTE1(first(x0, x1))
QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(from(x0)) → QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(from(x0))
QUOTE1(n__cons(y0, n__0)) → QUOTE1(0) → QUOTE1(n__cons(y0, n__0)) → QUOTE1(0)
QUOTE1(n__cons(y0, n__cons(x0, x1))) → QUOTE1(cons(x0, x1)) → QUOTE1(n__cons(y0, n__cons(x0, x1))) → QUOTE1(cons(x0, x1))
QUOTE1(n__cons(y0, n__nil)) → QUOTE1(nil) → QUOTE1(n__cons(y0, n__nil)) → QUOTE1(nil)
QUOTE1(n__cons(y0, n__s(x0))) → QUOTE1(s(x0)) → QUOTE1(n__cons(y0, n__s(x0))) → QUOTE1(s(x0))
QUOTE1(n__cons(y0, n__sel(x0, x1))) → QUOTE1(sel(x0, x1)) → QUOTE1(n__cons(y0, n__sel(x0, x1))) → QUOTE1(sel(x0, x1))
QUOTE1(n__cons(y0, x0)) → QUOTE1(x0) → QUOTE1(n__cons(y0, x0)) → QUOTE1(x0)

(66) Obligation:

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

QUOTE1(n__cons(y0, n__first(x0, x1))) → QUOTE1(first(x0, x1))
QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(from(x0))
QUOTE1(n__cons(y0, n__0)) → QUOTE1(0)
QUOTE1(n__cons(y0, n__cons(x0, x1))) → QUOTE1(cons(x0, x1))
QUOTE1(n__cons(y0, n__nil)) → QUOTE1(nil)
QUOTE1(n__cons(y0, n__s(x0))) → QUOTE1(s(x0))
QUOTE1(n__cons(y0, n__sel(x0, x1))) → QUOTE1(sel(x0, x1))
QUOTE1(n__cons(y0, x0)) → QUOTE1(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.

(67) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(from(x0)) at position [0] we obtained the following new rules [LPAR04]:

QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(cons(x0, n__from(s(x0)))) → QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(cons(x0, n__from(s(x0))))
QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(n__from(x0)) → QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(n__from(x0))

(68) Obligation:

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

QUOTE1(n__cons(y0, n__first(x0, x1))) → QUOTE1(first(x0, x1))
QUOTE1(n__cons(y0, n__0)) → QUOTE1(0)
QUOTE1(n__cons(y0, n__cons(x0, x1))) → QUOTE1(cons(x0, x1))
QUOTE1(n__cons(y0, n__nil)) → QUOTE1(nil)
QUOTE1(n__cons(y0, n__s(x0))) → QUOTE1(s(x0))
QUOTE1(n__cons(y0, n__sel(x0, x1))) → QUOTE1(sel(x0, x1))
QUOTE1(n__cons(y0, x0)) → QUOTE1(x0)
QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(cons(x0, n__from(s(x0))))
QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(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.

(69) DependencyGraphProof (EQUIVALENT transformation)

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

(70) Obligation:

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

QUOTE1(n__cons(y0, n__first(x0, x1))) → QUOTE1(first(x0, x1))
QUOTE1(n__cons(y0, n__0)) → QUOTE1(0)
QUOTE1(n__cons(y0, n__cons(x0, x1))) → QUOTE1(cons(x0, x1))
QUOTE1(n__cons(y0, n__nil)) → QUOTE1(nil)
QUOTE1(n__cons(y0, n__s(x0))) → QUOTE1(s(x0))
QUOTE1(n__cons(y0, n__sel(x0, x1))) → QUOTE1(sel(x0, x1))
QUOTE1(n__cons(y0, x0)) → QUOTE1(x0)
QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(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.

(71) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule QUOTE1(n__cons(y0, n__0)) → QUOTE1(0) at position [0] we obtained the following new rules [LPAR04]:

QUOTE1(n__cons(y0, n__0)) → QUOTE1(n__0) → QUOTE1(n__cons(y0, n__0)) → QUOTE1(n__0)

(72) Obligation:

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

QUOTE1(n__cons(y0, n__first(x0, x1))) → QUOTE1(first(x0, x1))
QUOTE1(n__cons(y0, n__cons(x0, x1))) → QUOTE1(cons(x0, x1))
QUOTE1(n__cons(y0, n__nil)) → QUOTE1(nil)
QUOTE1(n__cons(y0, n__s(x0))) → QUOTE1(s(x0))
QUOTE1(n__cons(y0, n__sel(x0, x1))) → QUOTE1(sel(x0, x1))
QUOTE1(n__cons(y0, x0)) → QUOTE1(x0)
QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(cons(x0, n__from(s(x0))))
QUOTE1(n__cons(y0, n__0)) → QUOTE1(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.

(73) DependencyGraphProof (EQUIVALENT transformation)

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

(74) Obligation:

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

QUOTE1(n__cons(y0, n__first(x0, x1))) → QUOTE1(first(x0, x1))
QUOTE1(n__cons(y0, n__cons(x0, x1))) → QUOTE1(cons(x0, x1))
QUOTE1(n__cons(y0, n__nil)) → QUOTE1(nil)
QUOTE1(n__cons(y0, n__s(x0))) → QUOTE1(s(x0))
QUOTE1(n__cons(y0, n__sel(x0, x1))) → QUOTE1(sel(x0, x1))
QUOTE1(n__cons(y0, x0)) → QUOTE1(x0)
QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(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.

(75) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule QUOTE1(n__cons(y0, n__cons(x0, x1))) → QUOTE1(cons(x0, x1)) at position [0] we obtained the following new rules [LPAR04]:

QUOTE1(n__cons(y0, n__cons(x0, x1))) → QUOTE1(n__cons(x0, x1)) → QUOTE1(n__cons(y0, n__cons(x0, x1))) → QUOTE1(n__cons(x0, x1))

(76) Obligation:

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

QUOTE1(n__cons(y0, n__first(x0, x1))) → QUOTE1(first(x0, x1))
QUOTE1(n__cons(y0, n__nil)) → QUOTE1(nil)
QUOTE1(n__cons(y0, n__s(x0))) → QUOTE1(s(x0))
QUOTE1(n__cons(y0, n__sel(x0, x1))) → QUOTE1(sel(x0, x1))
QUOTE1(n__cons(y0, x0)) → QUOTE1(x0)
QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(cons(x0, n__from(s(x0))))
QUOTE1(n__cons(y0, n__cons(x0, x1))) → QUOTE1(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.

(77) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule QUOTE1(n__cons(y0, n__nil)) → QUOTE1(nil) at position [0] we obtained the following new rules [LPAR04]:

QUOTE1(n__cons(y0, n__nil)) → QUOTE1(n__nil) → QUOTE1(n__cons(y0, n__nil)) → QUOTE1(n__nil)

(78) Obligation:

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

QUOTE1(n__cons(y0, n__first(x0, x1))) → QUOTE1(first(x0, x1))
QUOTE1(n__cons(y0, n__s(x0))) → QUOTE1(s(x0))
QUOTE1(n__cons(y0, n__sel(x0, x1))) → QUOTE1(sel(x0, x1))
QUOTE1(n__cons(y0, x0)) → QUOTE1(x0)
QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(cons(x0, n__from(s(x0))))
QUOTE1(n__cons(y0, n__cons(x0, x1))) → QUOTE1(n__cons(x0, x1))
QUOTE1(n__cons(y0, n__nil)) → QUOTE1(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.

(79) DependencyGraphProof (EQUIVALENT transformation)

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

(80) Obligation:

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

QUOTE1(n__cons(y0, n__first(x0, x1))) → QUOTE1(first(x0, x1))
QUOTE1(n__cons(y0, n__s(x0))) → QUOTE1(s(x0))
QUOTE1(n__cons(y0, n__sel(x0, x1))) → QUOTE1(sel(x0, x1))
QUOTE1(n__cons(y0, x0)) → QUOTE1(x0)
QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(cons(x0, n__from(s(x0))))
QUOTE1(n__cons(y0, n__cons(x0, x1))) → QUOTE1(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.

(81) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule QUOTE1(n__cons(y0, n__s(x0))) → QUOTE1(s(x0)) at position [0] we obtained the following new rules [LPAR04]:

QUOTE1(n__cons(y0, n__s(x0))) → QUOTE1(n__s(x0)) → QUOTE1(n__cons(y0, n__s(x0))) → QUOTE1(n__s(x0))

(82) Obligation:

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

QUOTE1(n__cons(y0, n__first(x0, x1))) → QUOTE1(first(x0, x1))
QUOTE1(n__cons(y0, n__sel(x0, x1))) → QUOTE1(sel(x0, x1))
QUOTE1(n__cons(y0, x0)) → QUOTE1(x0)
QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(cons(x0, n__from(s(x0))))
QUOTE1(n__cons(y0, n__cons(x0, x1))) → QUOTE1(n__cons(x0, x1))
QUOTE1(n__cons(y0, n__s(x0))) → QUOTE1(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.

(83) DependencyGraphProof (EQUIVALENT transformation)

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

(84) Obligation:

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

QUOTE1(n__cons(y0, n__first(x0, x1))) → QUOTE1(first(x0, x1))
QUOTE1(n__cons(y0, n__sel(x0, x1))) → QUOTE1(sel(x0, x1))
QUOTE1(n__cons(y0, x0)) → QUOTE1(x0)
QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(cons(x0, n__from(s(x0))))
QUOTE1(n__cons(y0, n__cons(x0, x1))) → QUOTE1(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.

(85) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

QUOTE1(n__cons(y0, x0)) → QUOTE1(x0)
QUOTE1(n__cons(y0, n__cons(x0, x1))) → QUOTE1(n__cons(x0, x1))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( QUOTE1(x₁) ) = max{0, 2x₁ - 1}

POL( first(x₁, x₂) ) = x₂ + 1

POL( 0 ) = 0

POL( nil ) = 0

POL( s(x₁) ) = 0

POL( cons(x₁, x₂) ) = 2x₁ + x₂ + 1

POL( n__first(x₁, x₂) ) = x₂

POL( activate(x₁) ) = x₁ + 1

POL( sel(x₁, x₂) ) = 2x₂ + 1

POL( n__sel(x₁, x₂) ) = 2x₂

POL( n__from(x₁) ) = 2x₁ + 1

POL( n__s(x₁) ) = 0

POL( n__cons(x₁, x₂) ) = 2x₁ + x₂ + 1

POL( from(x₁) ) = 2x₁ + 2

POL( n__0 ) = 0

POL( n__nil ) = 0

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

(86) Obligation:

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

QUOTE1(n__cons(y0, n__first(x0, x1))) → QUOTE1(first(x0, x1))
QUOTE1(n__cons(y0, n__sel(x0, x1))) → QUOTE1(sel(x0, x1))
QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(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.

(87) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

QUOTE1(n__cons(y0, n__sel(x0, x1))) → QUOTE1(sel(x0, x1))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( QUOTE1(x₁) ) = 2x₁ + 2

POL( first(x₁, x₂) ) = 2x₂

POL( 0 ) = 0

POL( nil ) = 0

POL( s(x₁) ) = 0

POL( cons(x₁, x₂) ) = 2x₁ + 2x₂ + 2

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

POL( activate(x₁) ) = 2x₁ + 2

POL( sel(x₁, x₂) ) = x₂ + 2

POL( n__sel(x₁, x₂) ) = x₂ + 1

POL( n__from(x₁) ) = 2x₁

POL( n__s(x₁) ) = 0

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

POL( from(x₁) ) = 2x₁ + 2

POL( n__0 ) = 0

POL( n__nil ) = 0

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

(88) Obligation:

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

QUOTE1(n__cons(y0, n__first(x0, x1))) → QUOTE1(first(x0, x1))
QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(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.

(89) NonTerminationLoopProof (COMPLETE transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = QUOTE1(cons(X1, n__from(x0))) evaluates to t =QUOTE1(cons(x0, n__from(s(x0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Matcher: [X1 / x0, x0 / s(x0)]
Semiunifier: [ ]

Rewriting sequence

QUOTE1(cons(X1, n__from(x0))) → QUOTE1(n__cons(X1, n__from(x0)))
with rule cons(X1', X2) → n__cons(X1', X2) at position [0] and matcher [X1' / X1, X2 / n__from(x0)]

QUOTE1(n__cons(X1, n__from(x0))) → QUOTE1(cons(x0, n__from(s(x0))))
with rule QUOTE1(n__cons(y0, n__from(x0))) → QUOTE1(cons(x0, n__from(s(x0))))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence

All these steps are and every following step will be a correct step w.r.t to Q.