Term Rewriting System R:
[X, M, N]
filter(cons(X), 0, M) -> cons(0)
filter(cons(X), s(N), M) -> cons(X)
sieve(cons(0)) -> cons(0)
sieve(cons(s(N))) -> cons(s(N))
nats(N) -> cons(N)
zprimes -> sieve(nats(s(s(0))))
Termination of R to be shown.
R
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
filter(cons(X), 0, M) -> cons(0)
filter(cons(X), s(N), M) -> cons(X)
where the Polynomial interpretation:
POL(filter(x1, x2, x3)) | = 1 + x1 + x2 + x3 |
POL(sieve(x1)) | = x1 |
POL(0) | = 0 |
POL(cons(x1)) | = x1 |
POL(nats(x1)) | = x1 |
POL(s(x1)) | = x1 |
POL(zprimes) | = 0 |
was used.
Not all Rules of R can be deleted, so we still have to regard a part of R.
R
↳RRRPolo
→TRS2
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
sieve(cons(0)) -> cons(0)
sieve(cons(s(N))) -> cons(s(N))
where the Polynomial interpretation:
POL(sieve(x1)) | = 2·x1 |
POL(0) | = 0 |
POL(cons(x1)) | = 1 + x1 |
POL(nats(x1)) | = 1 + x1 |
POL(s(x1)) | = x1 |
POL(zprimes) | = 2 |
was used.
Not all Rules of R can be deleted, so we still have to regard a part of R.
R
↳RRRPolo
→TRS2
↳RRRPolo
→TRS3
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
zprimes -> sieve(nats(s(s(0))))
where the Polynomial interpretation:
POL(sieve(x1)) | = x1 |
POL(0) | = 0 |
POL(cons(x1)) | = x1 |
POL(nats(x1)) | = x1 |
POL(s(x1)) | = x1 |
POL(zprimes) | = 1 |
was used.
Not all Rules of R can be deleted, so we still have to regard a part of R.
R
↳RRRPolo
→TRS2
↳RRRPolo
→TRS3
↳RRRPolo
...
→TRS4
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
nats(N) -> cons(N)
where the Polynomial interpretation:
POL(cons(x1)) | = x1 |
POL(nats(x1)) | = 1 + x1 |
was used.
All Rules of R can be deleted.
R
↳RRRPolo
→TRS2
↳RRRPolo
→TRS3
↳RRRPolo
...
→TRS5
↳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
↳RRRPolo
→TRS2
↳RRRPolo
→TRS3
↳RRRPolo
...
→TRS6
↳Dependency Pair Analysis
R contains no Dependency Pairs and therefore no SCCs.
Termination of R successfully shown.
Duration:
0:00 minutes