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)

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))
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. 


   R
     ↳DPs
       →DP Problem 1
         ↳Usable Rules (Innermost)
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules


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)


Strategy:

innermost




As we are in the innermost case, we can delete all 17 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 9
             ↳Size-Change Principle
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. DBL(s(X)) -> DBL(X)
and get the following Size-Change Graph(s):
{1} , {1}
1>1

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳Usable Rules (Innermost)
       →DP Problem 3
         ↳UsableRules


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)


Strategy:

innermost




As we are in the innermost case, we can delete all 17 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 10
             ↳Size-Change Principle
       →DP Problem 3
         ↳UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. SEL(s(X), cons(Y, Z)) -> SEL(X, Z)
and get the following Size-Change Graph(s):
{1} , {1}
1>1
2>2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
cons(x1, x2) -> cons(x1, x2)
s(x1) -> s(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳Usable Rules (Innermost)


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)


Strategy:

innermost




As we are in the innermost case, we can delete all 17 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
           →DP Problem 11
             ↳Non Termination


Dependency Pair:

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


Rule:

none


Strategy:

innermost




Found an infinite P-chain over R:
P =

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

R = none

s = FROM(X)
evaluates to t =FROM(s(X))

Thus, s starts an infinite chain as s matches t.

Innermost Termination of R could not be shown.
Duration:
0:01 minutes