Termination Proof using AProVE

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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)

Termination of R to be shown.

     ↳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))
DBL1(s(X)) -> DBL1(X)
SEL1(s(X), cons(Y, Z)) -> SEL1(X, Z)
QUOTE(s(X)) -> QUOTE(X)
QUOTE(dbl(X)) -> DBL1(X)
QUOTE(sel(X, Y)) -> SEL1(X, Y)

Furthermore, R contains eight SCCs.

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

       →DP Problem 5

         ↳Remaining Obligation(s)

       →DP Problem 6

         ↳Remaining Obligation(s)

       →DP Problem 7

         ↳Remaining Obligation(s)

       →DP Problem 8

         ↳Remaining Obligation(s)

The following remains to be proven:

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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
DBL1(s(X)) -> DBL1(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
SEL1(s(X), cons(Y, Z)) -> SEL1(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
DBLS(cons(X, Y)) -> DBLS(Y)

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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
INDX(cons(X, Y), Z) -> INDX(Y, 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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
QUOTE(s(X)) -> QUOTE(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

       →DP Problem 5

         ↳Remaining Obligation(s)

       →DP Problem 6

         ↳Remaining Obligation(s)

       →DP Problem 7

         ↳Remaining Obligation(s)

       →DP Problem 8

         ↳Remaining Obligation(s)

The following remains to be proven:

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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
DBL1(s(X)) -> DBL1(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
SEL1(s(X), cons(Y, Z)) -> SEL1(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
DBLS(cons(X, Y)) -> DBLS(Y)

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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
INDX(cons(X, Y), Z) -> INDX(Y, 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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
QUOTE(s(X)) -> QUOTE(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

       →DP Problem 5

         ↳Remaining Obligation(s)

       →DP Problem 6

         ↳Remaining Obligation(s)

       →DP Problem 7

         ↳Remaining Obligation(s)

       →DP Problem 8

         ↳Remaining Obligation(s)

The following remains to be proven:

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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
DBL1(s(X)) -> DBL1(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
SEL1(s(X), cons(Y, Z)) -> SEL1(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
DBLS(cons(X, Y)) -> DBLS(Y)

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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
INDX(cons(X, Y), Z) -> INDX(Y, 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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
QUOTE(s(X)) -> QUOTE(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

       →DP Problem 5

         ↳Remaining Obligation(s)

       →DP Problem 6

         ↳Remaining Obligation(s)

       →DP Problem 7

         ↳Remaining Obligation(s)

       →DP Problem 8

         ↳Remaining Obligation(s)

The following remains to be proven:

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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
DBL1(s(X)) -> DBL1(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
SEL1(s(X), cons(Y, Z)) -> SEL1(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
DBLS(cons(X, Y)) -> DBLS(Y)

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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
INDX(cons(X, Y), Z) -> INDX(Y, 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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
QUOTE(s(X)) -> QUOTE(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

       →DP Problem 5

         ↳Remaining Obligation(s)

       →DP Problem 6

         ↳Remaining Obligation(s)

       →DP Problem 7

         ↳Remaining Obligation(s)

       →DP Problem 8

         ↳Remaining Obligation(s)

The following remains to be proven:

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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
DBL1(s(X)) -> DBL1(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
SEL1(s(X), cons(Y, Z)) -> SEL1(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
DBLS(cons(X, Y)) -> DBLS(Y)

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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
INDX(cons(X, Y), Z) -> INDX(Y, 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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
QUOTE(s(X)) -> QUOTE(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

       →DP Problem 5

         ↳Remaining Obligation(s)

       →DP Problem 6

         ↳Remaining Obligation(s)

       →DP Problem 7

         ↳Remaining Obligation(s)

       →DP Problem 8

         ↳Remaining Obligation(s)

The following remains to be proven:

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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
DBL1(s(X)) -> DBL1(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
SEL1(s(X), cons(Y, Z)) -> SEL1(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
DBLS(cons(X, Y)) -> DBLS(Y)

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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
INDX(cons(X, Y), Z) -> INDX(Y, 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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
QUOTE(s(X)) -> QUOTE(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

       →DP Problem 5

         ↳Remaining Obligation(s)

       →DP Problem 6

         ↳Remaining Obligation(s)

       →DP Problem 7

         ↳Remaining Obligation(s)

       →DP Problem 8

         ↳Remaining Obligation(s)

The following remains to be proven:

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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
DBL1(s(X)) -> DBL1(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
SEL1(s(X), cons(Y, Z)) -> SEL1(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
DBLS(cons(X, Y)) -> DBLS(Y)

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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
INDX(cons(X, Y), Z) -> INDX(Y, 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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
QUOTE(s(X)) -> QUOTE(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

       →DP Problem 5

         ↳Remaining Obligation(s)

       →DP Problem 6

         ↳Remaining Obligation(s)

       →DP Problem 7

         ↳Remaining Obligation(s)

       →DP Problem 8

         ↳Remaining Obligation(s)

The following remains to be proven:

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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
DBL1(s(X)) -> DBL1(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
SEL1(s(X), cons(Y, Z)) -> SEL1(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
DBLS(cons(X, Y)) -> DBLS(Y)

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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
INDX(cons(X, Y), Z) -> INDX(Y, 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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)
Dependency Pair:
QUOTE(s(X)) -> QUOTE(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)))
dbl1(0) -> 01
dbl1(s(X)) -> s1(s1(dbl1(X)))
sel1(0, cons(X, Y)) -> X
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(dbl(X)) -> dbl1(X)
quote(sel(X, Y)) -> sel1(X, Y)

Termination of R could not be shown.
Duration:
0:00 minutes