details

property	value
status	complete
benchmark	PastaA4.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n094.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.258382081985 seconds
cpu usage	0.266542851
max memory	9338880.0

stage attributes

key	value
output-size	1541
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = init#(x1, x2) -> f1#(rnd1, rnd2) f2#(I0, I1) -> f2#(I0, I1 + 1) [I1 <= I0 - 1] f1#(I2, I3) -> f2#(I4, I5) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1 /\ -1 <= I5 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(I0, I1 + 1) [I1 <= I0 - 1] f1(I2, I3) -> f2(I4, I5) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1 /\ -1 <= I5 - 1] The dependency graph for this problem is: 0 -> 2 1 -> 1 2 -> 1 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f2#(I0, I1) -> f2#(I0, I1 + 1) [I1 <= I0 - 1] 2) f1#(I2, I3) -> f2#(I4, I5) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1 /\ -1 <= I5 - 1] We have the following SCCs. { 1 } DP problem for innermost termination. P = f2#(I0, I1) -> f2#(I0, I1 + 1) [I1 <= I0 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(I0, I1 + 1) [I1 <= I0 - 1] f1(I2, I3) -> f2(I4, I5) [0 <= I2 - 1 /\ -1 <= I4 - 1 /\ -1 <= I3 - 1 /\ -1 <= I5 - 1] We use the reverse value criterion with the projection function NU: NU[f2#(z1,z2)] = z1 - 1 + -1 * z2 This gives the following inequalities: I1 <= I0 - 1 ==> I0 - 1 + -1 * I1 > I0 - 1 + -1 * (I1 + 1) with I0 - 1 + -1 * I1 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. popout

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all output return to Integ Trans Syste 27634