Term Rewriting System R:
[X, Y, Z, X1, X2]
dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(ndbl(activate(X)), ndbls(activate(Y)))
dbls(X) -> ndbls(X)
sel(0, cons(X, Y)) -> activate(X)
sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z))
sel(X1, X2) -> nsel(X1, X2)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(nsel(activate(X), activate(Z)), nindx(activate(Y), activate(Z)))
indx(X1, X2) -> nindx(X1, X2)
from(X) -> cons(activate(X), nfrom(ns(activate(X))))
from(X) -> nfrom(X)
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
sel1(0, cons(X, Y)) -> activate(X)
sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(activate(X)))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
s(X) -> ns(X)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(ndbls(X)) -> dbls(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(nindx(X1, X2)) -> indx(X1, X2)
activate(nfrom(X)) -> from(X)
activate(X) -> X

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

DBL(s(X)) -> S(ns(ndbl(activate(X))))
DBL(s(X)) -> ACTIVATE(X)
DBLS(cons(X, Y)) -> ACTIVATE(X)
DBLS(cons(X, Y)) -> ACTIVATE(Y)
SEL(0, cons(X, Y)) -> ACTIVATE(X)
SEL(s(X), cons(Y, Z)) -> SEL(activate(X), activate(Z))
SEL(s(X), cons(Y, Z)) -> ACTIVATE(X)
SEL(s(X), cons(Y, Z)) -> ACTIVATE(Z)
INDX(cons(X, Y), Z) -> ACTIVATE(X)
INDX(cons(X, Y), Z) -> ACTIVATE(Z)
INDX(cons(X, Y), Z) -> ACTIVATE(Y)
FROM(X) -> ACTIVATE(X)
DBL1(s(X)) -> DBL1(activate(X))
DBL1(s(X)) -> ACTIVATE(X)
SEL1(0, cons(X, Y)) -> ACTIVATE(X)
SEL1(s(X), cons(Y, Z)) -> SEL1(activate(X), activate(Z))
SEL1(s(X), cons(Y, Z)) -> ACTIVATE(X)
SEL1(s(X), cons(Y, Z)) -> ACTIVATE(Z)
QUOTE(s(X)) -> QUOTE(activate(X))
QUOTE(s(X)) -> ACTIVATE(X)
QUOTE(dbl(X)) -> DBL1(X)
QUOTE(sel(X, Y)) -> SEL1(X, Y)
ACTIVATE(ns(X)) -> S(X)
ACTIVATE(ndbl(X)) -> DBL(X)
ACTIVATE(ndbls(X)) -> DBLS(X)
ACTIVATE(nsel(X1, X2)) -> SEL(X1, X2)
ACTIVATE(nindx(X1, X2)) -> INDX(X1, X2)
ACTIVATE(nfrom(X)) -> FROM(X)

Furthermore, R contains one SCC.


   R
DPs
       →DP Problem 1
Polynomial Ordering


Dependency Pairs:

INDX(cons(X, Y), Z) -> ACTIVATE(Y)
INDX(cons(X, Y), Z) -> ACTIVATE(Z)
FROM(X) -> ACTIVATE(X)
ACTIVATE(nfrom(X)) -> FROM(X)
INDX(cons(X, Y), Z) -> ACTIVATE(X)
ACTIVATE(nindx(X1, X2)) -> INDX(X1, X2)
SEL(0, cons(X, Y)) -> ACTIVATE(X)
ACTIVATE(nsel(X1, X2)) -> SEL(X1, X2)
DBLS(cons(X, Y)) -> ACTIVATE(Y)
ACTIVATE(ndbls(X)) -> DBLS(X)
DBLS(cons(X, Y)) -> ACTIVATE(X)


Rules:


dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(ndbl(activate(X)), ndbls(activate(Y)))
dbls(X) -> ndbls(X)
sel(0, cons(X, Y)) -> activate(X)
sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z))
sel(X1, X2) -> nsel(X1, X2)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(nsel(activate(X), activate(Z)), nindx(activate(Y), activate(Z)))
indx(X1, X2) -> nindx(X1, X2)
from(X) -> cons(activate(X), nfrom(ns(activate(X))))
from(X) -> nfrom(X)
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
sel1(0, cons(X, Y)) -> activate(X)
sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(activate(X)))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
s(X) -> ns(X)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(ndbls(X)) -> dbls(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(nindx(X1, X2)) -> indx(X1, X2)
activate(nfrom(X)) -> from(X)
activate(X) -> X


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(FROM(x1))=  x1  
  POL(n__from(x1))=  1 + x1  
  POL(n__indx(x1, x2))=  x1 + x2  
  POL(n__sel(x1, x2))=  x2  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(SEL(x1, x2))=  x2  
  POL(n__dbls(x1))=  x1  
  POL(INDX(x1, x2))=  x1 + x2  
  POL(ACTIVATE(x1))=  x1  
  POL(DBLS(x1))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 2
Dependency Graph


Dependency Pairs:

INDX(cons(X, Y), Z) -> ACTIVATE(Y)
INDX(cons(X, Y), Z) -> ACTIVATE(Z)
FROM(X) -> ACTIVATE(X)
INDX(cons(X, Y), Z) -> ACTIVATE(X)
ACTIVATE(nindx(X1, X2)) -> INDX(X1, X2)
SEL(0, cons(X, Y)) -> ACTIVATE(X)
ACTIVATE(nsel(X1, X2)) -> SEL(X1, X2)
DBLS(cons(X, Y)) -> ACTIVATE(Y)
ACTIVATE(ndbls(X)) -> DBLS(X)
DBLS(cons(X, Y)) -> ACTIVATE(X)


Rules:


dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(ndbl(activate(X)), ndbls(activate(Y)))
dbls(X) -> ndbls(X)
sel(0, cons(X, Y)) -> activate(X)
sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z))
sel(X1, X2) -> nsel(X1, X2)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(nsel(activate(X), activate(Z)), nindx(activate(Y), activate(Z)))
indx(X1, X2) -> nindx(X1, X2)
from(X) -> cons(activate(X), nfrom(ns(activate(X))))
from(X) -> nfrom(X)
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
sel1(0, cons(X, Y)) -> activate(X)
sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(activate(X)))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
s(X) -> ns(X)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(ndbls(X)) -> dbls(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(nindx(X1, X2)) -> indx(X1, X2)
activate(nfrom(X)) -> from(X)
activate(X) -> X


Strategy:

innermost




Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
DPs
       →DP Problem 1
Polo
           →DP Problem 2
DGraph
             ...
               →DP Problem 3
Polynomial Ordering


Dependency Pairs:

DBLS(cons(X, Y)) -> ACTIVATE(Y)
INDX(cons(X, Y), Z) -> ACTIVATE(Z)
INDX(cons(X, Y), Z) -> ACTIVATE(X)
ACTIVATE(nindx(X1, X2)) -> INDX(X1, X2)
SEL(0, cons(X, Y)) -> ACTIVATE(X)
ACTIVATE(nsel(X1, X2)) -> SEL(X1, X2)
DBLS(cons(X, Y)) -> ACTIVATE(X)
ACTIVATE(ndbls(X)) -> DBLS(X)
INDX(cons(X, Y), Z) -> ACTIVATE(Y)


Rules:


dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(ndbl(activate(X)), ndbls(activate(Y)))
dbls(X) -> ndbls(X)
sel(0, cons(X, Y)) -> activate(X)
sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z))
sel(X1, X2) -> nsel(X1, X2)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(nsel(activate(X), activate(Z)), nindx(activate(Y), activate(Z)))
indx(X1, X2) -> nindx(X1, X2)
from(X) -> cons(activate(X), nfrom(ns(activate(X))))
from(X) -> nfrom(X)
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
sel1(0, cons(X, Y)) -> activate(X)
sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(activate(X)))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
s(X) -> ns(X)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(ndbls(X)) -> dbls(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(nindx(X1, X2)) -> indx(X1, X2)
activate(nfrom(X)) -> from(X)
activate(X) -> X


Strategy:

innermost




The following dependency pair can be strictly oriented:

ACTIVATE(nindx(X1, X2)) -> INDX(X1, X2)


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(n__indx(x1, x2))=  1 + x1 + x2  
  POL(n__sel(x1, x2))=  x2  
  POL(0)=  0  
  POL(cons(x1, x2))=  x1 + x2  
  POL(SEL(x1, x2))=  x2  
  POL(n__dbls(x1))=  x1  
  POL(INDX(x1, x2))=  x1 + x2  
  POL(ACTIVATE(x1))=  x1  
  POL(DBLS(x1))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 2
DGraph
             ...
               →DP Problem 4
Dependency Graph


Dependency Pairs:

DBLS(cons(X, Y)) -> ACTIVATE(Y)
INDX(cons(X, Y), Z) -> ACTIVATE(Z)
INDX(cons(X, Y), Z) -> ACTIVATE(X)
SEL(0, cons(X, Y)) -> ACTIVATE(X)
ACTIVATE(nsel(X1, X2)) -> SEL(X1, X2)
DBLS(cons(X, Y)) -> ACTIVATE(X)
ACTIVATE(ndbls(X)) -> DBLS(X)
INDX(cons(X, Y), Z) -> ACTIVATE(Y)


Rules:


dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(ndbl(activate(X)), ndbls(activate(Y)))
dbls(X) -> ndbls(X)
sel(0, cons(X, Y)) -> activate(X)
sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z))
sel(X1, X2) -> nsel(X1, X2)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(nsel(activate(X), activate(Z)), nindx(activate(Y), activate(Z)))
indx(X1, X2) -> nindx(X1, X2)
from(X) -> cons(activate(X), nfrom(ns(activate(X))))
from(X) -> nfrom(X)
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
sel1(0, cons(X, Y)) -> activate(X)
sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(activate(X)))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
s(X) -> ns(X)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(ndbls(X)) -> dbls(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(nindx(X1, X2)) -> indx(X1, X2)
activate(nfrom(X)) -> from(X)
activate(X) -> X


Strategy:

innermost




Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
DPs
       →DP Problem 1
Polo
           →DP Problem 2
DGraph
             ...
               →DP Problem 5
Polynomial Ordering


Dependency Pairs:

SEL(0, cons(X, Y)) -> ACTIVATE(X)
ACTIVATE(nsel(X1, X2)) -> SEL(X1, X2)
DBLS(cons(X, Y)) -> ACTIVATE(X)
ACTIVATE(ndbls(X)) -> DBLS(X)
DBLS(cons(X, Y)) -> ACTIVATE(Y)


Rules:


dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(ndbl(activate(X)), ndbls(activate(Y)))
dbls(X) -> ndbls(X)
sel(0, cons(X, Y)) -> activate(X)
sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z))
sel(X1, X2) -> nsel(X1, X2)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(nsel(activate(X), activate(Z)), nindx(activate(Y), activate(Z)))
indx(X1, X2) -> nindx(X1, X2)
from(X) -> cons(activate(X), nfrom(ns(activate(X))))
from(X) -> nfrom(X)
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
sel1(0, cons(X, Y)) -> activate(X)
sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(activate(X)))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
s(X) -> ns(X)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(ndbls(X)) -> dbls(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(nindx(X1, X2)) -> indx(X1, X2)
activate(nfrom(X)) -> from(X)
activate(X) -> X


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(n__sel(x1, x2))=  1 + x2  
  POL(0)=  0  
  POL(SEL(x1, x2))=  x2  
  POL(cons(x1, x2))=  x1 + x2  
  POL(n__dbls(x1))=  x1  
  POL(ACTIVATE(x1))=  x1  
  POL(DBLS(x1))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 2
DGraph
             ...
               →DP Problem 6
Dependency Graph


Dependency Pairs:

SEL(0, cons(X, Y)) -> ACTIVATE(X)
DBLS(cons(X, Y)) -> ACTIVATE(X)
ACTIVATE(ndbls(X)) -> DBLS(X)
DBLS(cons(X, Y)) -> ACTIVATE(Y)


Rules:


dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(ndbl(activate(X)), ndbls(activate(Y)))
dbls(X) -> ndbls(X)
sel(0, cons(X, Y)) -> activate(X)
sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z))
sel(X1, X2) -> nsel(X1, X2)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(nsel(activate(X), activate(Z)), nindx(activate(Y), activate(Z)))
indx(X1, X2) -> nindx(X1, X2)
from(X) -> cons(activate(X), nfrom(ns(activate(X))))
from(X) -> nfrom(X)
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
sel1(0, cons(X, Y)) -> activate(X)
sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(activate(X)))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
s(X) -> ns(X)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(ndbls(X)) -> dbls(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(nindx(X1, X2)) -> indx(X1, X2)
activate(nfrom(X)) -> from(X)
activate(X) -> X


Strategy:

innermost




Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
DPs
       →DP Problem 1
Polo
           →DP Problem 2
DGraph
             ...
               →DP Problem 7
Polynomial Ordering


Dependency Pairs:

DBLS(cons(X, Y)) -> ACTIVATE(Y)
DBLS(cons(X, Y)) -> ACTIVATE(X)
ACTIVATE(ndbls(X)) -> DBLS(X)


Rules:


dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(ndbl(activate(X)), ndbls(activate(Y)))
dbls(X) -> ndbls(X)
sel(0, cons(X, Y)) -> activate(X)
sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z))
sel(X1, X2) -> nsel(X1, X2)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(nsel(activate(X), activate(Z)), nindx(activate(Y), activate(Z)))
indx(X1, X2) -> nindx(X1, X2)
from(X) -> cons(activate(X), nfrom(ns(activate(X))))
from(X) -> nfrom(X)
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
sel1(0, cons(X, Y)) -> activate(X)
sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(activate(X)))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
s(X) -> ns(X)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(ndbls(X)) -> dbls(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(nindx(X1, X2)) -> indx(X1, X2)
activate(nfrom(X)) -> from(X)
activate(X) -> X


Strategy:

innermost




The following dependency pairs can be strictly oriented:

DBLS(cons(X, Y)) -> ACTIVATE(Y)
DBLS(cons(X, Y)) -> ACTIVATE(X)


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(cons(x1, x2))=  1 + x1 + x2  
  POL(n__dbls(x1))=  x1  
  POL(ACTIVATE(x1))=  x1  
  POL(DBLS(x1))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 2
DGraph
             ...
               →DP Problem 8
Dependency Graph


Dependency Pair:

ACTIVATE(ndbls(X)) -> DBLS(X)


Rules:


dbl(0) -> 0
dbl(s(X)) -> s(ns(ndbl(activate(X))))
dbl(X) -> ndbl(X)
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(ndbl(activate(X)), ndbls(activate(Y)))
dbls(X) -> ndbls(X)
sel(0, cons(X, Y)) -> activate(X)
sel(s(X), cons(Y, Z)) -> sel(activate(X), activate(Z))
sel(X1, X2) -> nsel(X1, X2)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(nsel(activate(X), activate(Z)), nindx(activate(Y), activate(Z)))
indx(X1, X2) -> nindx(X1, X2)
from(X) -> cons(activate(X), nfrom(ns(activate(X))))
from(X) -> nfrom(X)
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(activate(X))))
sel1(0, cons(X, Y)) -> activate(X)
sel1(s(X), cons(Y, Z)) -> sel1(activate(X), activate(Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(activate(X)))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
s(X) -> ns(X)
activate(ns(X)) -> s(X)
activate(ndbl(X)) -> dbl(X)
activate(ndbls(X)) -> dbls(X)
activate(nsel(X1, X2)) -> sel(X1, X2)
activate(nindx(X1, X2)) -> indx(X1, X2)
activate(nfrom(X)) -> from(X)
activate(X) -> X


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes