Term Rewriting System R:
[X, Y, Z]
dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
from(X) -> cons(X, from(s(X)))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

DBL(s(X)) -> DBL(X)
DBLS(cons(X, Y)) -> DBL(X)
DBLS(cons(X, Y)) -> DBLS(Y)
SEL(s(X), cons(Y, Z)) -> SEL(X, Z)
INDX(cons(X, Y), Z) -> SEL(X, Z)
INDX(cons(X, Y), Z) -> INDX(Y, Z)
FROM(X) -> FROM(s(X))

Furthermore, R contains five SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Inst
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳Remaining


Dependency Pair:

DBL(s(X)) -> DBL(X)


Rules:


dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
from(X) -> cons(X, from(s(X)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

DBL(s(X)) -> DBL(X)


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
DBL(x1) -> DBL(x1)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 6
             ↳Dependency Graph
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Inst
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳Remaining


Dependency Pair:


Rules:


dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
from(X) -> cons(X, from(s(X)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Argument Filtering and Ordering
       →DP Problem 3
         ↳Inst
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳Remaining


Dependency Pair:

SEL(s(X), cons(Y, Z)) -> SEL(X, Z)


Rules:


dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
from(X) -> cons(X, from(s(X)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

SEL(s(X), cons(Y, Z)) -> SEL(X, Z)


There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
SEL(x1, x2) -> SEL(x1, x2)
s(x1) -> s(x1)
cons(x1, x2) -> cons(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →DP Problem 7
             ↳Dependency Graph
       →DP Problem 3
         ↳Inst
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳Remaining


Dependency Pair:


Rules:


dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
from(X) -> cons(X, from(s(X)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Instantiation Transformation
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳Remaining


Dependency Pair:

FROM(X) -> FROM(s(X))


Rules:


dbl(0) -> 0
dbl(s(X)) -> s(s(dbl(X)))
dbls(nil) -> nil
dbls(cons(X, Y)) -> cons(dbl(X), dbls(Y))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)
indx(nil, X) -> nil
indx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
from(X) -> cons(X, from(s(X)))


Strategy:

innermost




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

FROM(X) -> FROM(s(X))
one new Dependency Pair is created:

FROM(s(X'')) -> FROM(s(s(X'')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Inst
       →DP Problem 4
         ↳Remaining Obligation(s)
       →DP Problem 5
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Inst
       →DP Problem 4
         ↳Remaining Obligation(s)
       →DP Problem 5
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Inst
       →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:
0:00 minutes