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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 UsableRulesProof (⇔)
    ↳7 QDP
    ↳8 QDPSizeChangeProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 UsableRulesProof (⇔)
    ↳12 QDP
    ↳13 QDPSizeChangeProof (⇔)
    ↳14 TRUE
  ↳15 QDP
    ↳16 QDPOrderProof (⇔)
    ↳17 QDP
    ↳18 DependencyGraphProof (⇔)
    ↳19 QDP
    ↳20 QDPOrderProof (⇔)
    ↳21 QDP
    ↳22 DependencyGraphProof (⇔)
    ↳23 QDP
    ↳24 QDPSizeChangeProof (⇔)
    ↳25 TRUE
  ↳26 QDP
    ↳27 QDPOrderProof (⇔)
    ↳28 QDP
    ↳29 DependencyGraphProof (⇔)
    ↳30 AND
      ↳31 QDP
        ↳32 QDPSizeChangeProof (⇔)
        ↳33 TRUE
      ↳34 QDP
        ↳35 Narrowing (⇔)
        ↳36 QDP
        ↳37 Narrowing (⇔)
        ↳38 QDP
        ↳39 DependencyGraphProof (⇔)
        ↳40 QDP
        ↳41 Narrowing (⇔)
        ↳42 QDP
        ↳43 DependencyGraphProof (⇔)
        ↳44 QDP
        ↳45 Narrowing (⇔)
        ↳46 QDP
        ↳47 DependencyGraphProof (⇔)
        ↳48 QDP
        ↳49 Narrowing (⇔)
        ↳50 QDP
        ↳51 DependencyGraphProof (⇔)
        ↳52 QDP
        ↳53 Narrowing (⇔)
        ↳54 QDP
        ↳55 QDPOrderProof (⇔)
        ↳56 QDP
        ↳57 QDPOrderProof (⇔)
        ↳58 QDP
        ↳59 NonMonReductionPairProof (⇔)
        ↳60 QDP
        ↳61 PisEmptyProof (⇔)
        ↳62 TRUE
  ↳63 QDP
    ↳64 QDPSizeChangeProof (⇔)
    ↳65 TRUE
  ↳66 QDP
    ↳67 Narrowing (⇔)
    ↳68 QDP
    ↳69 Narrowing (⇔)
    ↳70 QDP
    ↳71 DependencyGraphProof (⇔)
    ↳72 QDP
    ↳73 Narrowing (⇔)
    ↳74 QDP
    ↳75 DependencyGraphProof (⇔)
    ↳76 QDP
    ↳77 Narrowing (⇔)
    ↳78 QDP
    ↳79 Narrowing (⇔)
    ↳80 QDP
    ↳81 DependencyGraphProof (⇔)
    ↳82 QDP
    ↳83 Narrowing (⇔)
    ↳84 QDP
    ↳85 DependencyGraphProof (⇔)
    ↳86 QDP
    ↳87 QDPOrderProof (⇔)
    ↳88 QDP
    ↳89 NonTerminationProof (⇔)
    ↳90 FALSE

(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) TRUE

(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)

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.

(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) TRUE

(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].


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: Matrix interpretation [MATRO]:

POL(SEL(x1, x2)) =
/0\
\0/
 +
/00\
\00/
·x1 +
/01\
\01/
·x2

POL(s(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(cons(x1, x2)) =
/1\
\0/
 +
/00\
\11/
·x1 +
/00\
\01/
·x2

POL(ACTIVATE(x1)) =
/0\
\0/
 +
/01\
\01/
·x1

POL(n__first(x1, x2)) =
/0\
\1/
 +
/10\
\00/
·x1 +
/00\
\01/
·x2

POL(FIRST(x1, x2)) =
/0\
\1/
 +
/00\
\00/
·x1 +
/01\
\01/
·x2

POL(n__sel(x1, x2)) =
/1\
\0/
 +
/00\
\00/
·x1 +
/01\
\01/
·x2

POL(activate(x1)) =
/1\
\0/
 +
/10\
\01/
·x1

POL(first(x1, x2)) =
/1\
\1/
 +
/10\
\00/
·x1 +
/00\
\01/
·x2

POL(0) =
/1\
\0/

POL(nil) =
/1\
\0/

POL(from(x1)) =
/1\
\0/
 +
/01\
\11/
·x1

POL(n__from(x1)) =
/0\
\0/
 +
/01\
\11/
·x1

POL(n__0) =
/0\
\0/

POL(sel(x1, x2)) =
/1\
\0/
 +
/00\
\00/
·x1 +
/01\
\01/
·x2

POL(n__s(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(n__cons(x1, x2)) =
/1\
\0/
 +
/00\
\11/
·x1 +
/00\
\01/
·x2

POL(n__nil) =
/0\
\0/

The following usable rules [FROCOS05] were oriented:

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

(17) 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.

(18) DependencyGraphProof (EQUIVALENT transformation)

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

(19) 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.

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 1   
POL(ACTIVATE(x1)) = x1   
POL(SEL(x1, x2)) = 1 + x2   
POL(activate(x1)) = 1 + x1   
POL(cons(x1, x2)) = 1 + x1 + x2   
POL(first(x1, x2)) = 1 + x2   
POL(from(x1)) = 1 + x1   
POL(n__0) = 0   
POL(n__cons(x1, x2)) = x1 + x2   
POL(n__first(x1, x2)) = x2   
POL(n__from(x1)) = x1   
POL(n__nil) = 1   
POL(n__s(x1)) = 0   
POL(n__sel(x1, x2)) = 1 + x2   
POL(nil) = 1   
POL(s(x1)) = 0   
POL(sel(x1, x2)) = 1 + x2   

The following usable rules [FROCOS05] were oriented:

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

(21) 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))

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.

(22) DependencyGraphProof (EQUIVALENT transformation)

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

(23) 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.

(24) 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

(25) TRUE

(26) 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.

(27) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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)) = -I + 0A·x1 + 0A·x2

POL(0) = 1A

POL(cons(x1, x2)) = -I + 1A·x1 + 0A·x2

POL(QUOTE(x1)) = 0A + 0A·x1

POL(n__s(x1)) = -I + 0A·x1

POL(activate(x1)) = -I + 0A·x1

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

POL(s(x1)) = -I + 0A·x1

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

POL(nil) = 0A

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

POL(from(x1)) = -I + 1A·x1

POL(n__from(x1)) = -I + 1A·x1

POL(n__0) = 1A

POL(sel(x1, x2)) = -I + 0A·x1 + 0A·x2

POL(n__cons(x1, x2)) = -I + 1A·x1 + 0A·x2

POL(n__nil) = 0A

The following usable rules [FROCOS05] were oriented:

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

(28) 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.

(29) DependencyGraphProof (EQUIVALENT transformation)

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

(30) Complex Obligation (AND)

(31) 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.

(32) 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

(33) TRUE

(34) 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.

(35) 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)

(36) 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.

(37) 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))

(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__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.

(39) DependencyGraphProof (EQUIVALENT transformation)

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

(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))))

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) 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)

(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__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.

(43) DependencyGraphProof (EQUIVALENT transformation)

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

(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))))

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) 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))

(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__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.

(47) DependencyGraphProof (EQUIVALENT transformation)

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

(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))))

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) 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)

(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__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.

(51) DependencyGraphProof (EQUIVALENT transformation)

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

(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))))

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) 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))

(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__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.

(55) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


QUOTE(n__s(n__from(x0))) → QUOTE(cons(x0, n__from(s(x0))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(QUOTE(x1)) =
/0A\
|-I|
\-I/
 +
/0A0A0A\
|-I-I-I|
\-I-I-I/
·x1

POL(n__s(x1)) =
/-I\
|-I|
\-I/
 +
/0A0A0A\
|0A0A0A|
\0A0A0A/
·x1

POL(n__first(x1, x2)) =
/0A\
|-I|
\0A/
 +
/-I-I-I\
|-I0A0A|
\-I-I0A/
·x1 +
/-I0A0A\
|-I0A0A|
\-I-I0A/
·x2

POL(first(x1, x2)) =
/0A\
|0A|
\0A/
 +
/-I-I-I\
|-I0A0A|
\-I-I0A/
·x1 +
/-I0A0A\
|-I0A0A|
\-I-I0A/
·x2

POL(n__sel(x1, x2)) =
/-I\
|0A|
\-I/
 +
/0A-I-I\
|0A-I1A|
\0A-I0A/
·x1 +
/-I0A-I\
|-I0A-I|
\-I0A-I/
·x2

POL(sel(x1, x2)) =
/-I\
|0A|
\-I/
 +
/0A-I-I\
|0A-I1A|
\0A-I0A/
·x1 +
/-I0A-I\
|-I0A-I|
\-I0A-I/
·x2

POL(n__from(x1)) =
/1A\
|-I|
\-I/
 +
/1A1A1A\
|0A0A0A|
\-I0A0A/
·x1

POL(cons(x1, x2)) =
/0A\
|-I|
\-I/
 +
/-I0A-I\
|0A0A0A|
\0A0A0A/
·x1 +
/-I-I-I\
|-I0A0A|
\-I0A0A/
·x2

POL(s(x1)) =
/-I\
|-I|
\-I/
 +
/0A0A0A\
|0A0A0A|
\0A0A0A/
·x1

POL(activate(x1)) =
/0A\
|-I|
\-I/
 +
/0A0A0A\
|-I0A0A|
\-I0A0A/
·x1

POL(0) =
/0A\
|0A|
\0A/

POL(nil) =
/0A\
|0A|
\0A/

POL(from(x1)) =
/1A\
|-I|
\-I/
 +
/1A1A1A\
|0A0A0A|
\0A0A0A/
·x1

POL(n__0) =
/0A\
|-I|
\0A/

POL(n__cons(x1, x2)) =
/0A\
|-I|
\-I/
 +
/-I-I-I\
|-I-I0A|
\0A0A0A/
·x1 +
/-I-I-I\
|-I0A-I|
\-I0A0A/
·x2

POL(n__nil) =
/0A\
|0A|
\-I/

The following usable rules [FROCOS05] were oriented:

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

(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__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].


The following pairs can be oriented strictly and are deleted.


QUOTE(n__s(n__sel(x0, x1))) → QUOTE(sel(x0, x1))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(QUOTE(x1)) =
/1A\
|-I|
\0A/
 +
/-I-I0A\
|-I-I-I|
\-I-I0A/
·x1

POL(n__s(x1)) =
/0A\
|0A|
\2A/
 +
/-I0A0A\
|0A-I0A|
\0A-I0A/
·x1

POL(n__first(x1, x2)) =
/0A\
|-I|
\0A/
 +
/3A0A0A\
|3A1A3A|
\-I0A-I/
·x1 +
/0A-I0A\
|-I0A0A|
\-I0A0A/
·x2

POL(first(x1, x2)) =
/0A\
|0A|
\2A/
 +
/3A1A0A\
|3A1A3A|
\3A0A-I/
·x1 +
/0A-I0A\
|-I0A0A|
\-I0A0A/
·x2

POL(n__sel(x1, x2)) =
/0A\
|0A|
\-I/
 +
/2A2A0A\
|1A-I0A|
\0A1A-I/
·x1 +
/1A1A1A\
|0A0A0A|
\-I-I-I/
·x2

POL(sel(x1, x2)) =
/0A\
|0A|
\-I/
 +
/2A2A0A\
|1A0A0A|
\1A1A-I/
·x1 +
/1A1A1A\
|0A0A0A|
\-I0A0A/
·x2

POL(activate(x1)) =
/0A\
|-I|
\2A/
 +
/0A0A-I\
|-I0A0A|
\-I0A0A/
·x1

POL(0) =
/0A\
|0A|
\0A/

POL(nil) =
/0A\
|0A|
\3A/

POL(s(x1)) =
/0A\
|1A|
\2A/
 +
/0A0A0A\
|0A-I0A|
\0A-I0A/
·x1

POL(cons(x1, x2)) =
/0A\
|0A|
\0A/
 +
/0A0A0A\
|-I0A0A|
\-I0A0A/
·x1 +
/-I-I0A\
|0A-I0A|
\-I0A-I/
·x2

POL(from(x1)) =
/2A\
|3A|
\3A/
 +
/1A0A1A\
|1A0A1A|
\1A1A1A/
·x1

POL(n__from(x1)) =
/0A\
|3A|
\0A/
 +
/-I-I-I\
|1A0A1A|
\-I1A0A/
·x1

POL(n__0) =
/0A\
|0A|
\-I/

POL(n__cons(x1, x2)) =
/-I\
|-I|
\0A/
 +
/0A0A0A\
|-I-I-I|
\-I0A0A/
·x1 +
/-I-I0A\
|0A-I0A|
\-I0A-I/
·x2

POL(n__nil) =
/0A\
|-I|
\3A/

The following usable rules [FROCOS05] were oriented:

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

(58) 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(x0)) → QUOTE(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.

(59) NonMonReductionPairProof (EQUIVALENT transformation)

Using the following max-polynomial ordering, we can orient the general usable rules and all rules from P weakly and some rules from P strictly:
Polynomial interpretation with max [POLO,NEGPOLO,MAXPOLO]:

POL(0) = 0   
POL(01) = 1   
POL(QUOTE(x1)) = x1   
POL(activate(x1)) = 0   
POL(cons(x1, x2)) = max(0, -x1 - x2)   
POL(cons1(x1, x2)) = 0   
POL(fcons(x1, x2)) = max(0, -x1 - x2)   
POL(first(x1, x2)) = max(0, -x1 - x2)   
POL(first1(x1, x2)) = 0   
POL(from(x1)) = 0   
POL(n__0) = 0   
POL(n__cons(x1, x2)) = 0   
POL(n__first(x1, x2)) = 0   
POL(n__from(x1)) = 1   
POL(n__nil) = 0   
POL(n__s(x1)) = 1 + x1   
POL(n__sel(x1, x2)) = x2   
POL(nil) = 0   
POL(nil1) = 0   
POL(quote(x1)) = x1   
POL(quote1(x1)) = 1   
POL(s(x1)) = 0   
POL(s1(x1)) = x1   
POL(sel(x1, x2)) = max(0, -x1 - x2)   
POL(sel1(x1, x2)) = 0   
POL(unquote(x1)) = 0   
POL(unquote1(x1)) = 0   

The following pairs can be oriented strictly and are deleted.

QUOTE(n__s(n__first(x0, x1))) → QUOTE(first(x0, x1))
QUOTE(n__s(x0)) → QUOTE(x0)
QUOTE(n__s(n__s(x0))) → QUOTE(n__s(x0))
The remaining pairs can at least be oriented weakly.
none
The following rules are usable:

first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
first(X1, X2) → n__first(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
nil → n__nil

(60) 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.

(61) PisEmptyProof (EQUIVALENT transformation)

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

(62) TRUE

(63) 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.

(64) 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

(65) TRUE

(66) 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.

(67) Narrowing (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__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)

(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__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.

(69) Narrowing (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(n__from(x0))

(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))))
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.

(71) DependencyGraphProof (EQUIVALENT transformation)

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

(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__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.

(73) Narrowing (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)

(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))))
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.

(75) DependencyGraphProof (EQUIVALENT transformation)

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

(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__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.

(77) Narrowing (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))

(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__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.

(79) Narrowing (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)

(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))
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.

(81) DependencyGraphProof (EQUIVALENT transformation)

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

(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__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.

(83) Narrowing (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))

(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))
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.

(85) DependencyGraphProof (EQUIVALENT transformation)

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

(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, 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.

(87) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


QUOTE1(n__cons(y0, n__sel(x0, x1))) → QUOTE1(sel(x0, x1))
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 interpretation [POLO]:

POL(0) = 1   
POL(QUOTE1(x1)) = x1   
POL(activate(x1)) = 1 + x1   
POL(cons(x1, x2)) = 1 + x1 + x2   
POL(first(x1, x2)) = 1 + x2   
POL(from(x1)) = 1 + x1   
POL(n__0) = 0   
POL(n__cons(x1, x2)) = 1 + x1 + x2   
POL(n__first(x1, x2)) = x2   
POL(n__from(x1)) = x1   
POL(n__nil) = 0   
POL(n__s(x1)) = 0   
POL(n__sel(x1, x2)) = x2   
POL(nil) = 1   
POL(s(x1)) = 0   
POL(sel(x1, x2)) = x2   

The following usable rules [FROCOS05] were oriented:

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

(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) NonTerminationProof (EQUIVALENT 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.



(90) FALSE