MAYBE

The TRS could not be proven terminating. The proof attempt took 612 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (0ms).
 | – Problem 2 was processed with processor PolynomialLinearRange4iUR (0ms).
 |    | – Problem 5 was processed with processor PolynomialLinearRange4iUR (0ms).
 | – Problem 3 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (14ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (31ms), DependencyGraph (0ms), PolynomialLinearRange4iUR (13ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (29ms), DependencyGraph (1ms), ReductionPairSAT (19ms), DependencyGraph (1ms), SizeChangePrinciple (7ms)].
 | – Problem 4 was processed with processor SubtermCriterion (0ms).

The following open problems remain:



Open Dependency Pair Problem 3

Dependency Pairs

nats#(N)nats#(s(N))

Rewrite Rules

filter(cons(X, Y), 0, M)cons(0, filter(Y, M, M))filter(cons(X, Y), s(N), M)cons(X, filter(Y, N, M))
sieve(cons(0, Y))cons(0, sieve(Y))sieve(cons(s(N), Y))cons(s(N), sieve(filter(Y, N, N)))
nats(N)cons(N, nats(s(N)))zprimessieve(nats(s(s(0))))

Original Signature

Termination of terms over the following signature is verified: nats, 0, sieve, s, zprimes, filter, cons


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

filter#(cons(X, Y), 0, M)filter#(Y, M, M)sieve#(cons(s(N), Y))filter#(Y, N, N)
sieve#(cons(0, Y))sieve#(Y)filter#(cons(X, Y), s(N), M)filter#(Y, N, M)
sieve#(cons(s(N), Y))sieve#(filter(Y, N, N))zprimes#nats#(s(s(0)))
zprimes#sieve#(nats(s(s(0))))nats#(N)nats#(s(N))

Rewrite Rules

filter(cons(X, Y), 0, M)cons(0, filter(Y, M, M))filter(cons(X, Y), s(N), M)cons(X, filter(Y, N, M))
sieve(cons(0, Y))cons(0, sieve(Y))sieve(cons(s(N), Y))cons(s(N), sieve(filter(Y, N, N)))
nats(N)cons(N, nats(s(N)))zprimessieve(nats(s(s(0))))

Original Signature

Termination of terms over the following signature is verified: nats, 0, s, sieve, zprimes, filter, cons

Strategy


The following SCCs where found

filter#(cons(X, Y), 0, M) → filter#(Y, M, M)filter#(cons(X, Y), s(N), M) → filter#(Y, N, M)

nats#(N) → nats#(s(N))

sieve#(cons(0, Y)) → sieve#(Y)sieve#(cons(s(N), Y)) → sieve#(filter(Y, N, N))

Problem 2: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

sieve#(cons(0, Y))sieve#(Y)sieve#(cons(s(N), Y))sieve#(filter(Y, N, N))

Rewrite Rules

filter(cons(X, Y), 0, M)cons(0, filter(Y, M, M))filter(cons(X, Y), s(N), M)cons(X, filter(Y, N, M))
sieve(cons(0, Y))cons(0, sieve(Y))sieve(cons(s(N), Y))cons(s(N), sieve(filter(Y, N, N)))
nats(N)cons(N, nats(s(N)))zprimessieve(nats(s(s(0))))

Original Signature

Termination of terms over the following signature is verified: nats, 0, s, sieve, zprimes, filter, cons

Strategy


Polynomial Interpretation

Improved Usable rules

filter(cons(X, Y), 0, M)cons(0, filter(Y, M, M))filter(cons(X, Y), s(N), M)cons(X, filter(Y, N, M))

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

sieve#(cons(0, Y))sieve#(Y)

Problem 5: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

sieve#(cons(s(N), Y))sieve#(filter(Y, N, N))

Rewrite Rules

filter(cons(X, Y), 0, M)cons(0, filter(Y, M, M))filter(cons(X, Y), s(N), M)cons(X, filter(Y, N, M))
sieve(cons(0, Y))cons(0, sieve(Y))sieve(cons(s(N), Y))cons(s(N), sieve(filter(Y, N, N)))
nats(N)cons(N, nats(s(N)))zprimessieve(nats(s(s(0))))

Original Signature

Termination of terms over the following signature is verified: nats, 0, sieve, s, zprimes, filter, cons

Strategy


Polynomial Interpretation

Improved Usable rules

filter(cons(X, Y), 0, M)cons(0, filter(Y, M, M))filter(cons(X, Y), s(N), M)cons(X, filter(Y, N, M))

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

sieve#(cons(s(N), Y))sieve#(filter(Y, N, N))

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

filter#(cons(X, Y), 0, M)filter#(Y, M, M)filter#(cons(X, Y), s(N), M)filter#(Y, N, M)

Rewrite Rules

filter(cons(X, Y), 0, M)cons(0, filter(Y, M, M))filter(cons(X, Y), s(N), M)cons(X, filter(Y, N, M))
sieve(cons(0, Y))cons(0, sieve(Y))sieve(cons(s(N), Y))cons(s(N), sieve(filter(Y, N, N)))
nats(N)cons(N, nats(s(N)))zprimessieve(nats(s(s(0))))

Original Signature

Termination of terms over the following signature is verified: nats, 0, s, sieve, zprimes, filter, cons

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

filter#(cons(X, Y), 0, M)filter#(Y, M, M)filter#(cons(X, Y), s(N), M)filter#(Y, N, M)