Term Rewriting System R:
[X, XS, N, Y, YS]
from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)
minus(X, 0) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) -> nil
zWquot(nil, XS) -> nil
zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), zWquot(XS, YS))

Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

FROM(X) -> FROM(s(X))
SEL(s(N), cons(X, XS)) -> SEL(N, XS)
MINUS(s(X), s(Y)) -> MINUS(X, Y)
QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))
QUOT(s(X), s(Y)) -> MINUS(X, Y)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> QUOT(X, Y)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ZWQUOT(XS, YS)

Furthermore, R contains five SCCs.


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


Dependency Pair:

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


Rules:


from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)
minus(X, 0) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) -> nil
zWquot(nil, XS) -> nil
zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), zWquot(XS, YS))





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
Inst
           →DP Problem 6
Instantiation Transformation
       →DP Problem 2
Remaining
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pair:

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


Rules:


from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)
minus(X, 0) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) -> nil
zWquot(nil, XS) -> nil
zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), zWquot(XS, YS))





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
Inst
           →DP Problem 6
Inst
             ...
               →DP Problem 7
Instantiation Transformation
       →DP Problem 2
Remaining
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pair:

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


Rules:


from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)
minus(X, 0) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) -> nil
zWquot(nil, XS) -> nil
zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), zWquot(XS, YS))





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
Inst
           →DP Problem 6
Inst
             ...
               →DP Problem 8
Instantiation Transformation
       →DP Problem 2
Remaining
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pair:

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


Rules:


from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)
minus(X, 0) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) -> nil
zWquot(nil, XS) -> nil
zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), zWquot(XS, YS))





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
Inst
           →DP Problem 6
Inst
             ...
               →DP Problem 9
Instantiation Transformation
       →DP Problem 2
Remaining
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pair:

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


Rules:


from(X) -> cons(X, from(s(X)))
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, XS)
minus(X, 0) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
zWquot(XS, nil) -> nil
zWquot(nil, XS) -> nil
zWquot(cons(X, XS), cons(Y, YS)) -> cons(quot(X, Y), zWquot(XS, YS))





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
Inst
       →DP Problem 2
Remaining Obligation(s)
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)




The following remains to be proven:


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




The following remains to be proven:


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




The following remains to be proven:


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




The following remains to be proven:


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




The following remains to be proven:

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