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
↳Size-Change Principle
→DP Problem 2
↳Neg POLO
→DP Problem 3
↳SCP
→DP Problem 4
↳SCP
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
We number the DPs as follows:
- MINUS(s(X), s(Y)) -> MINUS(X, Y)
and get the following Size-Change Graph(s):
which lead(s) to this/these maximal multigraph(s):
DP: empty set
Oriented Rules: none
We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial
with Argument Filtering System:
s(x1) -> s(x1)
We obtain no new DP problems.
R
↳DPs
→DP Problem 1
↳SCP
→DP Problem 2
↳Negative Polynomial Order
→DP Problem 3
↳SCP
→DP Problem 4
↳SCP
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 using the given order.
QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))
Moreover, the following usable rules (regarding the implicit AFS) are oriented.
minus(X, 0) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
Used ordering:
Polynomial Order with Interpretation:
POL( QUOT(x1, x2) ) = x1
POL( s(x1) ) = 1
POL( minus(x1, x2) ) = 0
POL( 0 ) = 0
This results in one new DP problem.
R
↳DPs
→DP Problem 1
↳SCP
→DP Problem 2
↳Neg POLO
→DP Problem 5
↳Dependency Graph
→DP Problem 3
↳SCP
→DP Problem 4
↳SCP
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
↳SCP
→DP Problem 2
↳Neg POLO
→DP Problem 3
↳Size-Change Principle
→DP Problem 4
↳SCP
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
We number the DPs as follows:
- 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)
and get the following Size-Change Graph(s):
which lead(s) to this/these maximal multigraph(s):
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)
nzWquot(x1, x2) -> nzWquot(x1, x2)
We obtain no new DP problems.
R
↳DPs
→DP Problem 1
↳SCP
→DP Problem 2
↳Neg POLO
→DP Problem 3
↳SCP
→DP Problem 4
↳Size-Change Principle
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
We number the DPs as follows:
- SEL(s(N), cons(X, XS)) -> SEL(N, activate(XS))
and get the following Size-Change Graph(s):
which lead(s) to this/these maximal multigraph(s):
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.
Termination of R successfully shown.
Duration:
0:00 minutes