/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


--------------------------------------------------------------------------------


WORST_CASE(NON_POLY, ?)
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).

(0) CpxTRS
(1) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
(2) TRS for Loop Detection
(3) DecreasingLoopProof [LOWER BOUND(ID), 0 ms]
(4) BEST
    (5) proven lower bound
        (6) LowerBoundPropagationProof [FINISHED, 0 ms]
        (7) BOUNDS(n^1, INF)
    (8) TRS for Loop Detection
        (9) InfiniteLowerBoundProof [FINISHED, 0 ms]
        (10) BOUNDS(INF, INF)


----------------------------------------

(0)
Obligation:
The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).


The TRS R consists of the following 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)))
   zprimes -> sieve(nats(s(s(0))))

S is empty.
Rewrite Strategy: FULL
----------------------------------------

(1) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID))
Transformed a relative TRS into a decreasing-loop problem.
----------------------------------------

(2)
Obligation:
Analyzing the following TRS for decreasing loops:

The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).


The TRS R consists of the following 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)))
   zprimes -> sieve(nats(s(s(0))))

S is empty.
Rewrite Strategy: FULL
----------------------------------------

(3) DecreasingLoopProof (LOWER BOUND(ID))
The following loop(s) give(s) rise to the lower bound Omega(n^1):

The rewrite sequence

filter(cons(X, Y), s(N), M) ->^+ cons(X, filter(Y, N, M))

gives rise to a decreasing loop by considering the right hand sides subterm at position [1].

The pumping substitution is [Y / cons(X, Y), N / s(N)].

The result substitution is [ ].




----------------------------------------

(4)
Complex Obligation (BEST)

----------------------------------------

(5)
Obligation:
Proved the lower bound n^1 for the following obligation:

The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).


The TRS R consists of the following 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)))
   zprimes -> sieve(nats(s(s(0))))

S is empty.
Rewrite Strategy: FULL
----------------------------------------

(6) LowerBoundPropagationProof (FINISHED)
Propagated lower bound.
----------------------------------------

(7)
BOUNDS(n^1, INF)

----------------------------------------

(8)
Obligation:
Analyzing the following TRS for decreasing loops:

The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).


The TRS R consists of the following 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)))
   zprimes -> sieve(nats(s(s(0))))

S is empty.
Rewrite Strategy: FULL
----------------------------------------

(9) InfiniteLowerBoundProof (FINISHED)
The following loop proves infinite runtime complexity:

The rewrite sequence

nats(N) ->^+ cons(N, nats(s(N)))

gives rise to a decreasing loop by considering the right hand sides subterm at position [1].

The pumping substitution is [ ].

The result substitution is [N / s(N)].




----------------------------------------

(10)
BOUNDS(INF, INF)