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

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
AFS
       →DP Problem 3
AFS
       →DP Problem 4
AFS


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





The following dependency pair can be strictly oriented:

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


There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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
AFS
       →DP Problem 3
AFS
       →DP Problem 4
AFS


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





Using the Dependency Graph resulted in no new DP problems.


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


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





The following dependency pair can be strictly oriented:

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


The following usable rules w.r.t. to the AFS can be oriented:

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


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
QUOT > 0
minus > 0
s > 0

resulting in one new DP problem.
Used Argument Filtering System:
QUOT(x1, x2) -> QUOT(x1, x2)
s(x1) -> s(x1)
minus(x1, x2) -> x1


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
           →DP Problem 6
Dependency Graph
       →DP Problem 3
AFS
       →DP Problem 4
AFS


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





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(XS)
ACTIVATE(nzWquot(X1, X2)) -> ZWQUOT(X1, X2)


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





The following dependency pairs can be strictly oriented:

ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(YS)
ZWQUOT(cons(X, XS), cons(Y, YS)) -> ACTIVATE(XS)
ACTIVATE(nzWquot(X1, X2)) -> ZWQUOT(X1, X2)


There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
cons > ACTIVATE > ZWQUOT

resulting in one new DP problem.
Used Argument Filtering System:
ZWQUOT(x1, x2) -> ZWQUOT(x1, x2)
ACTIVATE(x1) -> ACTIVATE(x1)
cons(x1, x2) -> cons(x1, x2)
nzWquot(x1, x2) -> nzWquot(x1, x2)


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
           →DP Problem 7
Dependency Graph
       →DP Problem 4
AFS


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





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))


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





The following dependency pair can be strictly oriented:

SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))


The following usable rules w.r.t. to the AFS can be oriented:

activate(nfrom(X)) -> from(X)
activate(nzWquot(X1, X2)) -> zWquot(X1, X2)
activate(X) -> X
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)
from(X) -> cons(X, nfrom(s(X)))
from(X) -> nfrom(X)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))
minus(X, 0) -> 0
minus(s(X), s(Y)) -> minus(X, Y)


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
SEL > {zWquot, activate, from} > nil > 0
SEL > {zWquot, activate, from} > nfrom > 0
SEL > {zWquot, activate, from} > cons > 0
SEL > {zWquot, activate, from} > s > 0
quot > 0
minus > 0
nzWquot > 0

resulting in one new DP problem.
Used Argument Filtering System:
SEL(x1, x2) -> SEL(x1, x2)
s(x1) -> s(x1)
cons(x1, x2) -> cons(x1, x2)
activate(x1) -> activate(x1)
nfrom(x1) -> nfrom(x1)
from(x1) -> from(x1)
nzWquot(x1, x2) -> x1
zWquot(x1, x2) -> zWquot(x1)
quot(x1, x2) -> x1
minus(x1, x2) -> x1


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
AFS
           →DP Problem 8
Dependency Graph


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





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:01 minutes