details

property	value
status	complete
benchmark	alternKonv_rec.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n154.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.690366983414 seconds
cpu usage	0.735835354
max memory	1.1206656E7

stage attributes

key	value
output-size	2290
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) -> f1#(rnd1, rnd2) f2#(I0, I1) -> f2#(-1 * I0 - 2, I2) [I0 <= 0 /\ 3 <= -1 * I0 /\ I0 <= -2 - 1 /\ I0 <= -1 /\ I0 <= -1 - 1] f2#(I3, I4) -> f2#(-1 * I3 + 2, I5) [-1 * I3 <= -1 - 1 /\ 2 <= I3 - 1] f2#(I6, I7) -> f2#(I6 - 2, I8) [I6 <= 2 /\ 0 <= I6 - 1] f2#(I9, I10) -> f2#(I9 + 2, I11) [I9 <= -1 /\ I9 <= 0 /\ -3 <= I9 - 1] f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(-1 * I0 - 2, I2) [I0 <= 0 /\ 3 <= -1 * I0 /\ I0 <= -2 - 1 /\ I0 <= -1 /\ I0 <= -1 - 1] f2(I3, I4) -> f2(-1 * I3 + 2, I5) [-1 * I3 <= -1 - 1 /\ 2 <= I3 - 1] f2(I6, I7) -> f2(I6 - 2, I8) [I6 <= 2 /\ 0 <= I6 - 1] f2(I9, I10) -> f2(I9 + 2, I11) [I9 <= -1 /\ I9 <= 0 /\ -3 <= I9 - 1] f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] The dependency graph for this problem is: 0 -> 5 1 -> 2, 3 2 -> 1, 4 3 -> 4 4 -> 3 5 -> 2, 3 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f2#(I0, I1) -> f2#(-1 * I0 - 2, I2) [I0 <= 0 /\ 3 <= -1 * I0 /\ I0 <= -2 - 1 /\ I0 <= -1 /\ I0 <= -1 - 1] 2) f2#(I3, I4) -> f2#(-1 * I3 + 2, I5) [-1 * I3 <= -1 - 1 /\ 2 <= I3 - 1] 3) f2#(I6, I7) -> f2#(I6 - 2, I8) [I6 <= 2 /\ 0 <= I6 - 1] 4) f2#(I9, I10) -> f2#(I9 + 2, I11) [I9 <= -1 /\ I9 <= 0 /\ -3 <= I9 - 1] 5) f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] We have the following SCCs. { 1, 2 } { 3, 4 } DP problem for innermost termination. P = f2#(I6, I7) -> f2#(I6 - 2, I8) [I6 <= 2 /\ 0 <= I6 - 1] f2#(I9, I10) -> f2#(I9 + 2, I11) [I9 <= -1 /\ I9 <= 0 /\ -3 <= I9 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(-1 * I0 - 2, I2) [I0 <= 0 /\ 3 <= -1 * I0 /\ I0 <= -2 - 1 /\ I0 <= -1 /\ I0 <= -1 - 1] f2(I3, I4) -> f2(-1 * I3 + 2, I5) [-1 * I3 <= -1 - 1 /\ 2 <= I3 - 1] f2(I6, I7) -> f2(I6 - 2, I8) [I6 <= 2 /\ 0 <= I6 - 1] f2(I9, I10) -> f2(I9 + 2, I11) [I9 <= -1 /\ I9 <= 0 /\ -3 <= I9 - 1] f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] popout

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