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)

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 eight SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Inst
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳Remaining
       →DP Problem 6
         ↳Remaining
       →DP Problem 7
         ↳Remaining
       →DP Problem 8
         ↳Remaining


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)





The following dependency pair can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(SEL(x1, x2))=  x1  
  POL(cons(x1, x2))=  0  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 9
             ↳Dependency Graph
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Inst
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳Remaining
       →DP Problem 6
         ↳Remaining
       →DP Problem 7
         ↳Remaining
       →DP Problem 8
         ↳Remaining


Dependency Pair:


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)





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polynomial Ordering
       →DP Problem 3
         ↳Inst
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳Remaining
       →DP Problem 6
         ↳Remaining
       →DP Problem 7
         ↳Remaining
       →DP Problem 8
         ↳Remaining


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)





The following dependency pair can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(cons(x1, x2))=  0  
  POL(FIRST(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 10
             ↳Dependency Graph
       →DP Problem 3
         ↳Inst
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳Remaining
       →DP Problem 6
         ↳Remaining
       →DP Problem 7
         ↳Remaining
       →DP Problem 8
         ↳Remaining


Dependency Pair:


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)





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Instantiation Transformation
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳Remaining
       →DP Problem 6
         ↳Remaining
       →DP Problem 7
         ↳Remaining
       →DP Problem 8
         ↳Remaining


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)





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
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Inst
           →DP Problem 11
             ↳Instantiation Transformation
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳Remaining
       →DP Problem 6
         ↳Remaining
       →DP Problem 7
         ↳Remaining
       →DP Problem 8
         ↳Remaining


Dependency Pair:

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





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

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Inst
           →DP Problem 11
             ↳Inst
             ...
               →DP Problem 12
                 ↳Instantiation Transformation
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳Remaining
       →DP Problem 6
         ↳Remaining
       →DP Problem 7
         ↳Remaining
       →DP Problem 8
         ↳Remaining


Dependency Pair:

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





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

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Inst
           →DP Problem 11
             ↳Inst
             ...
               →DP Problem 13
                 ↳Instantiation Transformation
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳Remaining
       →DP Problem 6
         ↳Remaining
       →DP Problem 7
         ↳Remaining
       →DP Problem 8
         ↳Remaining


Dependency Pair:

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





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

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Inst
           →DP Problem 11
             ↳Inst
             ...
               →DP Problem 14
                 ↳Instantiation Transformation
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳Remaining
       →DP Problem 6
         ↳Remaining
       →DP Problem 7
         ↳Remaining
       →DP Problem 8
         ↳Remaining


Dependency Pair:

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





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

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Inst
       →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:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Inst
       →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:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Inst
       →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:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Inst
       →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:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Inst
       →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:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Inst
       →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:

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