Term Rewriting System R:
[X, XS, N, Y, YS, X1, X2]
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))
SEL(s(N), cons(X, XS)) -> ACTIVATE(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)) -> ACTIVATE(XS)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)
ACTIVATE(nfrom(X)) -> FROM(X)
ACTIVATE(nzWquot(X1, X2)) -> ZWQUOT(X1, X2)

Furthermore, R contains four SCCs.


   R
DPs
       →DP Problem 1
Argument Filtering and Ordering
       →DP Problem 2
Nar
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


Dependency Pair:

MINUS(s(X), s(Y)) -> MINUS(X, Y)


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X


Strategy:

innermost




The following dependency pair can be strictly oriented:

MINUS(s(X), s(Y)) -> MINUS(X, Y)


There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
MINUS(x1, x2) -> MINUS(x1, x2)
s(x1) -> s(x1)


   R
DPs
       →DP Problem 1
AFS
           →DP Problem 5
Dependency Graph
       →DP Problem 2
Nar
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


Dependency Pair:


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Narrowing Transformation
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining


Dependency Pair:

QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))


Rules:


from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
sel(0, cons(X, XS)) -> X
sel(s(N), cons(X, XS)) -> sel(N, activate(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), nzWquot(activate(XS), activate(YS)))
zWquot(X1, X2) -> nzWquot(X1, X2)
activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X


Strategy:

innermost




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

QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))
two new Dependency Pairs are created:

QUOT(s(X''), s(0)) -> QUOT(0, s(0))
QUOT(s(s(X'')), s(s(Y''))) -> QUOT(minus(X'', Y''), s(s(Y'')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
       →DP Problem 3
Remaining Obligation(s)
       →DP Problem 4
Remaining Obligation(s)




The following remains to be proven:

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