details

property	value
status	complete
benchmark	costa09-example_4.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n020.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.211405038834 seconds
cpu usage	0.2174349
max memory	9170944.0

stage attributes

key	value
output-size	1850
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) f2#(I0, I1, I2) -> f2#(I3, I1 + 10, I2 - 1) [I2 + 4 <= I0 /\ 2 <= I3 - 1 /\ 2 <= I0 - 1 /\ I3 <= I0 /\ 0 <= I2 - 1 /\ -1 <= I1 - 1] f1#(I4, I5, I6) -> f2#(I7, 0, 10) [13 <= I7 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f2(I0, I1, I2) -> f2(I3, I1 + 10, I2 - 1) [I2 + 4 <= I0 /\ 2 <= I3 - 1 /\ 2 <= I0 - 1 /\ I3 <= I0 /\ 0 <= I2 - 1 /\ -1 <= I1 - 1] f1(I4, I5, I6) -> f2(I7, 0, 10) [13 <= I7 - 1] The dependency graph for this problem is: 0 -> 2 1 -> 1 2 -> 1 Where: 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 1) f2#(I0, I1, I2) -> f2#(I3, I1 + 10, I2 - 1) [I2 + 4 <= I0 /\ 2 <= I3 - 1 /\ 2 <= I0 - 1 /\ I3 <= I0 /\ 0 <= I2 - 1 /\ -1 <= I1 - 1] 2) f1#(I4, I5, I6) -> f2#(I7, 0, 10) [13 <= I7 - 1] We have the following SCCs. { 1 } DP problem for innermost termination. P = f2#(I0, I1, I2) -> f2#(I3, I1 + 10, I2 - 1) [I2 + 4 <= I0 /\ 2 <= I3 - 1 /\ 2 <= I0 - 1 /\ I3 <= I0 /\ 0 <= I2 - 1 /\ -1 <= I1 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f2(I0, I1, I2) -> f2(I3, I1 + 10, I2 - 1) [I2 + 4 <= I0 /\ 2 <= I3 - 1 /\ 2 <= I0 - 1 /\ I3 <= I0 /\ 0 <= I2 - 1 /\ -1 <= I1 - 1] f1(I4, I5, I6) -> f2(I7, 0, 10) [13 <= I7 - 1] We use the basic value criterion with the projection function NU: NU[f2#(z1,z2,z3)] = z3 This gives the following inequalities: I2 + 4 <= I0 /\ 2 <= I3 - 1 /\ 2 <= I0 - 1 /\ I3 <= I0 /\ 0 <= I2 - 1 /\ -1 <= I1 - 1 ==> I2 >! I2 - 1 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