Term Rewriting System R:
[X, Y, Z, X1, X2]
sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
s(X) -> ns(X)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(nnil) -> nil
activate(nsel(X1, X2)) -> sel(activate(X1), activate(X2))
activate(X) -> X

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

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, nfirst(X, activate(Z)))
FIRST(s(X), cons(Y, Z)) -> ACTIVATE(Z)
FROM(X) -> CONS(X, nfrom(ns(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)
QUOTE(ns(X)) -> QUOTE(activate(X))
QUOTE(ns(X)) -> ACTIVATE(X)
QUOTE(nsel(X, Z)) -> SEL1(activate(X), activate(Z))
QUOTE(nsel(X, Z)) -> ACTIVATE(X)
QUOTE(nsel(X, Z)) -> ACTIVATE(Z)
QUOTE1(ncons(X, Z)) -> QUOTE(activate(X))
QUOTE1(ncons(X, Z)) -> ACTIVATE(X)
QUOTE1(ncons(X, Z)) -> QUOTE1(activate(Z))
QUOTE1(ncons(X, Z)) -> ACTIVATE(Z)
QUOTE1(nfirst(X, Z)) -> FIRST1(activate(X), activate(Z))
QUOTE1(nfirst(X, Z)) -> ACTIVATE(X)
QUOTE1(nfirst(X, Z)) -> ACTIVATE(Z)
UNQUOTE(01) -> 0'
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(nfirst(X1, X2)) -> FIRST(activate(X1), activate(X2))
ACTIVATE(nfirst(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(nfirst(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nfrom(X)) -> FROM(activate(X))
ACTIVATE(nfrom(X)) -> ACTIVATE(X)
ACTIVATE(ns(X)) -> S(activate(X))
ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(n0) -> 0'
ACTIVATE(ncons(X1, X2)) -> CONS(activate(X1), X2)
ACTIVATE(ncons(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(nnil) -> NIL
ACTIVATE(nsel(X1, X2)) -> SEL(activate(X1), activate(X2))
ACTIVATE(nsel(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(nsel(X1, X2)) -> ACTIVATE(X2)

Furthermore, R contains five SCCs.


   R
DPs
       →DP Problem 1
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Nar
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pairs:

ACTIVATE(nsel(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nsel(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(ncons(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(nfrom(X)) -> ACTIVATE(X)
ACTIVATE(nfirst(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nfirst(X1, X2)) -> ACTIVATE(X1)


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
s(X) -> ns(X)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(nnil) -> nil
activate(nsel(X1, X2)) -> sel(activate(X1), activate(X2))
activate(X) -> X


Strategy:

innermost




The following dependency pair can be strictly oriented:

ACTIVATE(nfrom(X)) -> ACTIVATE(X)


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(n__cons(x1, x2))=  x1  
  POL(n__from(x1))=  1 + x1  
  POL(n__sel(x1, x2))=  x1 + x2  
  POL(n__s(x1))=  x1  
  POL(ACTIVATE(x1))=  x1  
  POL(n__first(x1, x2))=  x1 + x2  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 6
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Nar
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pairs:

ACTIVATE(nsel(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nsel(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(ncons(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(ns(X)) -> ACTIVATE(X)
ACTIVATE(nfirst(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nfirst(X1, X2)) -> ACTIVATE(X1)


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
s(X) -> ns(X)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(nnil) -> nil
activate(nsel(X1, X2)) -> sel(activate(X1), activate(X2))
activate(X) -> X


Strategy:

innermost




The following dependency pairs can be strictly oriented:

ACTIVATE(nfirst(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nfirst(X1, X2)) -> ACTIVATE(X1)


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(n__cons(x1, x2))=  x1  
  POL(n__sel(x1, x2))=  x1 + x2  
  POL(n__s(x1))=  x1  
  POL(ACTIVATE(x1))=  x1  
  POL(n__first(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 6
Polo
             ...
               →DP Problem 7
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Nar
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pairs:

ACTIVATE(nsel(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nsel(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(ncons(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(ns(X)) -> ACTIVATE(X)


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
s(X) -> ns(X)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(nnil) -> nil
activate(nsel(X1, X2)) -> sel(activate(X1), activate(X2))
activate(X) -> X


Strategy:

innermost




The following dependency pairs can be strictly oriented:

ACTIVATE(nsel(X1, X2)) -> ACTIVATE(X2)
ACTIVATE(nsel(X1, X2)) -> ACTIVATE(X1)


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(n__cons(x1, x2))=  x1  
  POL(n__sel(x1, x2))=  1 + x1 + x2  
  POL(n__s(x1))=  x1  
  POL(ACTIVATE(x1))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 6
Polo
             ...
               →DP Problem 8
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Nar
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pairs:

ACTIVATE(ncons(X1, X2)) -> ACTIVATE(X1)
ACTIVATE(ns(X)) -> ACTIVATE(X)


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
s(X) -> ns(X)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(nnil) -> nil
activate(nsel(X1, X2)) -> sel(activate(X1), activate(X2))
activate(X) -> X


Strategy:

innermost




The following dependency pair can be strictly oriented:

ACTIVATE(ncons(X1, X2)) -> ACTIVATE(X1)


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(n__cons(x1, x2))=  1 + x1  
  POL(n__s(x1))=  x1  
  POL(ACTIVATE(x1))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 6
Polo
             ...
               →DP Problem 9
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Nar
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pair:

ACTIVATE(ns(X)) -> ACTIVATE(X)


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
s(X) -> ns(X)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(nnil) -> nil
activate(nsel(X1, X2)) -> sel(activate(X1), activate(X2))
activate(X) -> X


Strategy:

innermost




The following dependency pair can be strictly oriented:

ACTIVATE(ns(X)) -> ACTIVATE(X)


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(n__s(x1))=  1 + x1  
  POL(ACTIVATE(x1))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 6
Polo
             ...
               →DP Problem 10
Dependency Graph
       →DP Problem 2
Polo
       →DP Problem 3
Nar
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pair:


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
s(X) -> ns(X)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(nnil) -> nil
activate(nsel(X1, X2)) -> sel(activate(X1), activate(X2))
activate(X) -> X


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polynomial Ordering
       →DP Problem 3
Nar
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pair:

UNQUOTE(s1(X)) -> UNQUOTE(X)


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
s(X) -> ns(X)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(nnil) -> nil
activate(nsel(X1, X2)) -> sel(activate(X1), activate(X2))
activate(X) -> X


Strategy:

innermost




The following dependency pair can be strictly oriented:

UNQUOTE(s1(X)) -> UNQUOTE(X)


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(s1(x1))=  1 + x1  
  POL(UNQUOTE(x1))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
           →DP Problem 11
Dependency Graph
       →DP Problem 3
Nar
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pair:


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
s(X) -> ns(X)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(nnil) -> nil
activate(nsel(X1, X2)) -> sel(activate(X1), activate(X2))
activate(X) -> X


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Narrowing Transformation
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pair:

QUOTE(ns(X)) -> QUOTE(activate(X))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
s(X) -> ns(X)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(nnil) -> nil
activate(nsel(X1, X2)) -> sel(activate(X1), activate(X2))
activate(X) -> X


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

QUOTE(ns(X)) -> QUOTE(activate(X))
eight new Dependency Pairs are created:

QUOTE(ns(nfirst(X1', X2'))) -> QUOTE(first(activate(X1'), activate(X2')))
QUOTE(ns(nfrom(X''))) -> QUOTE(from(activate(X'')))
QUOTE(ns(ns(X''))) -> QUOTE(s(activate(X'')))
QUOTE(ns(n0)) -> QUOTE(0)
QUOTE(ns(ncons(X1', X2'))) -> QUOTE(cons(activate(X1'), X2'))
QUOTE(ns(nnil)) -> QUOTE(nil)
QUOTE(ns(nsel(X1', X2'))) -> QUOTE(sel(activate(X1'), activate(X2')))
QUOTE(ns(X'')) -> QUOTE(X'')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 12
Rewriting Transformation
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pairs:

QUOTE(ns(X'')) -> QUOTE(X'')
QUOTE(ns(nsel(X1', X2'))) -> QUOTE(sel(activate(X1'), activate(X2')))
QUOTE(ns(nnil)) -> QUOTE(nil)
QUOTE(ns(ncons(X1', X2'))) -> QUOTE(cons(activate(X1'), X2'))
QUOTE(ns(n0)) -> QUOTE(0)
QUOTE(ns(ns(X''))) -> QUOTE(s(activate(X'')))
QUOTE(ns(nfrom(X''))) -> QUOTE(from(activate(X'')))
QUOTE(ns(nfirst(X1', X2'))) -> QUOTE(first(activate(X1'), activate(X2')))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
s(X) -> ns(X)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(nnil) -> nil
activate(nsel(X1, X2)) -> sel(activate(X1), activate(X2))
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOTE(ns(n0)) -> QUOTE(0)
one new Dependency Pair is created:

QUOTE(ns(n0)) -> QUOTE(n0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 12
Rw
             ...
               →DP Problem 13
Rewriting Transformation
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pairs:

QUOTE(ns(nsel(X1', X2'))) -> QUOTE(sel(activate(X1'), activate(X2')))
QUOTE(ns(nnil)) -> QUOTE(nil)
QUOTE(ns(ncons(X1', X2'))) -> QUOTE(cons(activate(X1'), X2'))
QUOTE(ns(ns(X''))) -> QUOTE(s(activate(X'')))
QUOTE(ns(nfrom(X''))) -> QUOTE(from(activate(X'')))
QUOTE(ns(nfirst(X1', X2'))) -> QUOTE(first(activate(X1'), activate(X2')))
QUOTE(ns(X'')) -> QUOTE(X'')


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
s(X) -> ns(X)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(nnil) -> nil
activate(nsel(X1, X2)) -> sel(activate(X1), activate(X2))
activate(X) -> X


Strategy:

innermost




On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

QUOTE(ns(nnil)) -> QUOTE(nil)
one new Dependency Pair is created:

QUOTE(ns(nnil)) -> QUOTE(nnil)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
           →DP Problem 12
Rw
             ...
               →DP Problem 14
Forward Instantiation Transformation
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pairs:

QUOTE(ns(X'')) -> QUOTE(X'')
QUOTE(ns(ncons(X1', X2'))) -> QUOTE(cons(activate(X1'), X2'))
QUOTE(ns(ns(X''))) -> QUOTE(s(activate(X'')))
QUOTE(ns(nfrom(X''))) -> QUOTE(from(activate(X'')))
QUOTE(ns(nfirst(X1', X2'))) -> QUOTE(first(activate(X1'), activate(X2')))
QUOTE(ns(nsel(X1', X2'))) -> QUOTE(sel(activate(X1'), activate(X2')))


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, activate(Z))
sel(0, cons(X, Z)) -> X
sel(X1, X2) -> nsel(X1, X2)
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, nfirst(X, activate(Z)))
first(X1, X2) -> nfirst(X1, X2)
from(X) -> cons(X, nfrom(ns(X)))
from(X) -> nfrom(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(n0) -> 01
quote(ns(X)) -> s1(quote(activate(X)))
quote(nsel(X, Z)) -> sel1(activate(X), activate(Z))
quote1(ncons(X, Z)) -> cons1(quote(activate(X)), quote1(activate(Z)))
quote1(nnil) -> nil1
quote1(nfirst(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)
s(X) -> ns(X)
0 -> n0
cons(X1, X2) -> ncons(X1, X2)
nil -> nnil
activate(nfirst(X1, X2)) -> first(activate(X1), activate(X2))
activate(nfrom(X)) -> from(activate(X))
activate(ns(X)) -> s(activate(X))
activate(n0) -> 0
activate(ncons(X1, X2)) -> cons(activate(X1), X2)
activate(nnil) -> nil
activate(nsel(X1, X2)) -> sel(activate(X1), activate(X2))
activate(X) -> X


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

QUOTE(ns(X'')) -> QUOTE(X'')
six new Dependency Pairs are created:

QUOTE(ns(ns(nfirst(X1''', X2''')))) -> QUOTE(ns(nfirst(X1''', X2''')))
QUOTE(ns(ns(nfrom(X'''')))) -> QUOTE(ns(nfrom(X'''')))
QUOTE(ns(ns(ns(X'''')))) -> QUOTE(ns(ns(X'''')))
QUOTE(ns(ns(ncons(X1''', X2''')))) -> QUOTE(ns(ncons(X1''', X2''')))
QUOTE(ns(ns(nsel(X1''', X2''')))) -> QUOTE(ns(nsel(X1''', X2''')))
QUOTE(ns(ns(X''''))) -> QUOTE(ns(X''''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Nar
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)




The following remains to be proven:

Innermost Termination of R could not be shown.
Duration:
0:38 minutes