details

property	value
status	complete
benchmark	PastaB15.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n110.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.451148986816 seconds
cpu usage	0.470834767
max memory	1.0194944E7

stage attributes

key	value
output-size	2597
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, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) f3#(I0, I1, I2, I3) -> f3#(I0 - 1, I1 - 1, I1 - 1, I3) [I1 = I2 /\ I3 <= I1 - 1] f3#(I4, I5, I6, I7) -> f2#(I7, I4, I5, I8) [I5 = I6 /\ I5 <= I7] f2#(I9, I10, I11, I12) -> f3#(I10, I10, I10, I9) [I10 = I11 /\ I9 <= I10 - 1] f1#(I13, I14, I15, I16) -> f2#(I17, I18, I19, I20) [0 <= I13 - 1 /\ -1 <= I19 - 1 /\ -1 <= I17 - 1 /\ -1 <= I14 - 1 /\ -1 <= I18 - 1] R = init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) f3(I0, I1, I2, I3) -> f3(I0 - 1, I1 - 1, I1 - 1, I3) [I1 = I2 /\ I3 <= I1 - 1] f3(I4, I5, I6, I7) -> f2(I7, I4, I5, I8) [I5 = I6 /\ I5 <= I7] f2(I9, I10, I11, I12) -> f3(I10, I10, I10, I9) [I10 = I11 /\ I9 <= I10 - 1] f1(I13, I14, I15, I16) -> f2(I17, I18, I19, I20) [0 <= I13 - 1 /\ -1 <= I19 - 1 /\ -1 <= I17 - 1 /\ -1 <= I14 - 1 /\ -1 <= I18 - 1] The dependency graph for this problem is: 0 -> 4 1 -> 1, 2 2 -> 3 -> 1 4 -> 3 Where: 0) init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 1) f3#(I0, I1, I2, I3) -> f3#(I0 - 1, I1 - 1, I1 - 1, I3) [I1 = I2 /\ I3 <= I1 - 1] 2) f3#(I4, I5, I6, I7) -> f2#(I7, I4, I5, I8) [I5 = I6 /\ I5 <= I7] 3) f2#(I9, I10, I11, I12) -> f3#(I10, I10, I10, I9) [I10 = I11 /\ I9 <= I10 - 1] 4) f1#(I13, I14, I15, I16) -> f2#(I17, I18, I19, I20) [0 <= I13 - 1 /\ -1 <= I19 - 1 /\ -1 <= I17 - 1 /\ -1 <= I14 - 1 /\ -1 <= I18 - 1] We have the following SCCs. { 1 } DP problem for innermost termination. P = f3#(I0, I1, I2, I3) -> f3#(I0 - 1, I1 - 1, I1 - 1, I3) [I1 = I2 /\ I3 <= I1 - 1] R = init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) f3(I0, I1, I2, I3) -> f3(I0 - 1, I1 - 1, I1 - 1, I3) [I1 = I2 /\ I3 <= I1 - 1] f3(I4, I5, I6, I7) -> f2(I7, I4, I5, I8) [I5 = I6 /\ I5 <= I7] f2(I9, I10, I11, I12) -> f3(I10, I10, I10, I9) [I10 = I11 /\ I9 <= I10 - 1] f1(I13, I14, I15, I16) -> f2(I17, I18, I19, I20) [0 <= I13 - 1 /\ -1 <= I19 - 1 /\ -1 <= I17 - 1 /\ -1 <= I14 - 1 /\ -1 <= I18 - 1] We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4)] = z2 - 1 + -1 * z4 This gives the following inequalities: I1 = I2 /\ I3 <= I1 - 1 ==> I1 - 1 + -1 * I3 > I1 - 1 - 1 + -1 * I3 with I1 - 1 + -1 * I3 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. popout

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job log

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actions

all output return to Integ Trans Syste 27634