Term Rewriting System R:
[X, Y, Z]
sel(s(X), cons(Y, Z)) -> sel(X, Z)
sel(0, cons(X, Z)) -> X
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
sel1(0, cons(X, Z)) -> quote(X)
first1(0, Z) -> nil1
first1(s(X), cons(Y, Z)) -> cons1(quote(Y), first1(X, Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(sel(X, Z)) -> sel1(X, Z)
quote1(cons(X, Z)) -> cons1(quote(X), quote1(Z))
quote1(nil) -> nil1
quote1(first(X, Z)) -> first1(X, 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)

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, Z)
FIRST(s(X), cons(Y, Z)) -> FIRST(X, Z)
FROM(X) -> FROM(s(X))
SEL1(s(X), cons(Y, Z)) -> SEL1(X, Z)
SEL1(0, cons(X, Z)) -> QUOTE(X)
FIRST1(s(X), cons(Y, Z)) -> QUOTE(Y)
FIRST1(s(X), cons(Y, Z)) -> FIRST1(X, Z)
QUOTE(s(X)) -> QUOTE(X)
QUOTE(sel(X, Z)) -> SEL1(X, Z)
QUOTE1(cons(X, Z)) -> QUOTE(X)
QUOTE1(cons(X, Z)) -> QUOTE1(Z)
QUOTE1(first(X, Z)) -> FIRST1(X, Z)
UNQUOTE(s1(X)) -> UNQUOTE(X)
UNQUOTE1(cons1(X, Z)) -> FCONS(unquote(X), unquote1(Z))
UNQUOTE1(cons1(X, Z)) -> UNQUOTE(X)
UNQUOTE1(cons1(X, Z)) -> UNQUOTE1(Z)

Furthermore, R contains nine SCCs. 


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


Dependency Pair:

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


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, Z)
sel(0, cons(X, Z)) -> X
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
sel1(0, cons(X, Z)) -> quote(X)
first1(0, Z) -> nil1
first1(s(X), cons(Y, Z)) -> cons1(quote(Y), first1(X, Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(sel(X, Z)) -> sel1(X, Z)
quote1(cons(X, Z)) -> cons1(quote(X), quote1(Z))
quote1(nil) -> nil1
quote1(first(X, Z)) -> first1(X, 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)


Strategy:

innermost




As we are in the innermost case, we can delete all 20 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 10
             ↳Size-Change Principle
       →DP Problem 2
         ↳UsableRules
       →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
         ↳Usable Rules (Innermost)
       →DP Problem 3
         ↳UsableRules


Dependency Pair:

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


Rules:


sel(s(X), cons(Y, Z)) -> sel(X, Z)
sel(0, cons(X, Z)) -> X
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
sel1(0, cons(X, Z)) -> quote(X)
first1(0, Z) -> nil1
first1(s(X), cons(Y, Z)) -> cons1(quote(Y), first1(X, Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(sel(X, Z)) -> sel1(X, Z)
quote1(cons(X, Z)) -> cons1(quote(X), quote1(Z))
quote1(nil) -> nil1
quote1(first(X, Z)) -> first1(X, 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)


Strategy:

innermost




As we are in the innermost case, we can delete all 20 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 11
             ↳Size-Change Principle
       →DP Problem 3
         ↳UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. FIRST(s(X), cons(Y, Z)) -> FIRST(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:


sel(s(X), cons(Y, Z)) -> sel(X, Z)
sel(0, cons(X, Z)) -> X
first(0, Z) -> nil
first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
from(X) -> cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) -> sel1(X, Z)
sel1(0, cons(X, Z)) -> quote(X)
first1(0, Z) -> nil1
first1(s(X), cons(Y, Z)) -> cons1(quote(Y), first1(X, Z))
quote(0) -> 01
quote(s(X)) -> s1(quote(X))
quote(sel(X, Z)) -> sel1(X, Z)
quote1(cons(X, Z)) -> cons1(quote(X), quote1(Z))
quote1(nil) -> nil1
quote1(first(X, Z)) -> first1(X, 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)


Strategy:

innermost




As we are in the innermost case, we can delete all 20 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
           →DP Problem 12
             ↳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