details

property	value
status	complete
benchmark	Virtual.jar-obl-9.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n078.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.706601142883 seconds
cpu usage	0.74386989
max memory	1.124352E7

stage attributes

key	value
output-size	2977
starexec-result	MAYBE

output

/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) f3#(I0, I1, I2) -> f3#(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0] f2#(I6, I7, I8) -> f3#(I9, I10, I11) [-1 <= I9 - 1 /\ 0 <= I6 - 1 /\ I8 <= 0 /\ I9 + 1 <= I6] f2#(I12, I13, I14) -> f2#(I15, I13 - 1, I13) [3 <= I15 - 1 /\ 0 <= I12 - 1 /\ 0 <= I14 - 1 /\ I15 - 3 <= I12] f1#(I16, I17, I18) -> f2#(I19, 27, 28) [0 <= I19 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f3(I0, I1, I2) -> f3(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0] f2(I6, I7, I8) -> f3(I9, I10, I11) [-1 <= I9 - 1 /\ 0 <= I6 - 1 /\ I8 <= 0 /\ I9 + 1 <= I6] f2(I12, I13, I14) -> f2(I15, I13 - 1, I13) [3 <= I15 - 1 /\ 0 <= I12 - 1 /\ 0 <= I14 - 1 /\ I15 - 3 <= I12] f1(I16, I17, I18) -> f2(I19, 27, 28) [0 <= I19 - 1] The dependency graph for this problem is: 0 -> 4 1 -> 1 2 -> 1 3 -> 2, 3 4 -> 3 Where: 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 1) f3#(I0, I1, I2) -> f3#(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0] 2) f2#(I6, I7, I8) -> f3#(I9, I10, I11) [-1 <= I9 - 1 /\ 0 <= I6 - 1 /\ I8 <= 0 /\ I9 + 1 <= I6] 3) f2#(I12, I13, I14) -> f2#(I15, I13 - 1, I13) [3 <= I15 - 1 /\ 0 <= I12 - 1 /\ 0 <= I14 - 1 /\ I15 - 3 <= I12] 4) f1#(I16, I17, I18) -> f2#(I19, 27, 28) [0 <= I19 - 1] We have the following SCCs. { 3 } { 1 } DP problem for innermost termination. P = f3#(I0, I1, I2) -> f3#(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f3(I0, I1, I2) -> f3(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0] f2(I6, I7, I8) -> f3(I9, I10, I11) [-1 <= I9 - 1 /\ 0 <= I6 - 1 /\ I8 <= 0 /\ I9 + 1 <= I6] f2(I12, I13, I14) -> f2(I15, I13 - 1, I13) [3 <= I15 - 1 /\ 0 <= I12 - 1 /\ 0 <= I14 - 1 /\ I15 - 3 <= I12] f1(I16, I17, I18) -> f2(I19, 27, 28) [0 <= I19 - 1] We use the basic value criterion with the projection function NU: NU[f3#(z1,z2,z3)] = z1 This gives the following inequalities: -1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0 ==> I0 >! I3 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f2#(I12, I13, I14) -> f2#(I15, I13 - 1, I13) [3 <= I15 - 1 /\ 0 <= I12 - 1 /\ 0 <= I14 - 1 /\ I15 - 3 <= I12] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f3(I0, I1, I2) -> f3(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0] f2(I6, I7, I8) -> f3(I9, I10, I11) [-1 <= I9 - 1 /\ 0 <= I6 - 1 /\ I8 <= 0 /\ I9 + 1 <= I6] f2(I12, I13, I14) -> f2(I15, I13 - 1, I13) [3 <= I15 - 1 /\ 0 <= I12 - 1 /\ 0 <= I14 - 1 /\ I15 - 3 <= I12] f1(I16, I17, I18) -> f2(I19, 27, 28) [0 <= I19 - 1] popout

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