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
↳Overlay and local confluence Check
The TRS is overlay and locally confluent (all critical pairs are trivially joinable).Hence, we can switch to innermost.
R
↳OC
→TRS2
↳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 four SCCs.
R
↳OC
→TRS2
↳DPs
→DP Problem 1
↳Usable Rules (Innermost)
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))
Strategy:
innermost
As we are in the innermost case, we can delete all 10 non-usable-rules.
R
↳OC
→TRS2
↳DPs
→DP Problem 1
↳UsableRules
...
→DP Problem 5
↳Non Termination
Dependency Pair:
FROM(X) -> FROM(s(X))
Rule:
none
Strategy:
innermost
Found an infinite P-chain over R:
P =
FROM(X) -> FROM(s(X))
R = none
s = FROM(X)
evaluates to t =FROM(s(X))
Thus, s starts an infinite chain as s matches t.
Non-Termination of R could be shown.
Duration:
0:01 minutes