details

property	value
status	complete
benchmark	SharingPair.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n143.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.691368 seconds
cpu usage	0.597733
user time	0.333995
system time	0.263738
max virtual memory	123600.0
max residence set size	8564.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) f2#(I0, I1, I2) -> f2#(I3, I4, 0) [0 = I2 /\ 0 <= I4 - 1 /\ 3 <= I3 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I4 <= I1 /\ I3 - 3 <= I1 /\ I3 - 3 <= I0] f2#(I5, I6, I7) -> f2#(I8, I9, I7) [3 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ I9 - 3 <= I6 /\ I9 - 3 <= I5 /\ 0 <= I7 - 1 /\ I8 <= I5] f1#(I10, I11, I12) -> f2#(I13, I14, 0) [1 <= I14 - 1 /\ 3 <= I13 - 1 /\ 0 <= I10 - 1 /\ I14 - 1 <= I10 /\ -1 <= I11 - 1 /\ I13 - 3 <= I10] f1#(I15, I16, I17) -> f2#(I18, I19, I20) [3 <= I19 - 1 /\ 1 <= I18 - 1 /\ 0 <= I15 - 1 /\ I19 - 3 <= I15 /\ I18 - 1 <= I15 /\ -1 <= I16 - 1 /\ 0 <= I20 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f2(I0, I1, I2) -> f2(I3, I4, 0) [0 = I2 /\ 0 <= I4 - 1 /\ 3 <= I3 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I4 <= I1 /\ I3 - 3 <= I1 /\ I3 - 3 <= I0] f2(I5, I6, I7) -> f2(I8, I9, I7) [3 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ I9 - 3 <= I6 /\ I9 - 3 <= I5 /\ 0 <= I7 - 1 /\ I8 <= I5] f1(I10, I11, I12) -> f2(I13, I14, 0) [1 <= I14 - 1 /\ 3 <= I13 - 1 /\ 0 <= I10 - 1 /\ I14 - 1 <= I10 /\ -1 <= I11 - 1 /\ I13 - 3 <= I10] f1(I15, I16, I17) -> f2(I18, I19, I20) [3 <= I19 - 1 /\ 1 <= I18 - 1 /\ 0 <= I15 - 1 /\ I19 - 3 <= I15 /\ I18 - 1 <= I15 /\ -1 <= I16 - 1 /\ 0 <= I20 - 1] The dependency graph for this problem is: 0 -> 3, 4 1 -> 1 2 -> 2 3 -> 1 4 -> 2 Where: 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 1) f2#(I0, I1, I2) -> f2#(I3, I4, 0) [0 = I2 /\ 0 <= I4 - 1 /\ 3 <= I3 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I4 <= I1 /\ I3 - 3 <= I1 /\ I3 - 3 <= I0] 2) f2#(I5, I6, I7) -> f2#(I8, I9, I7) [3 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ I9 - 3 <= I6 /\ I9 - 3 <= I5 /\ 0 <= I7 - 1 /\ I8 <= I5] 3) f1#(I10, I11, I12) -> f2#(I13, I14, 0) [1 <= I14 - 1 /\ 3 <= I13 - 1 /\ 0 <= I10 - 1 /\ I14 - 1 <= I10 /\ -1 <= I11 - 1 /\ I13 - 3 <= I10] 4) f1#(I15, I16, I17) -> f2#(I18, I19, I20) [3 <= I19 - 1 /\ 1 <= I18 - 1 /\ 0 <= I15 - 1 /\ I19 - 3 <= I15 /\ I18 - 1 <= I15 /\ -1 <= I16 - 1 /\ 0 <= I20 - 1] We have the following SCCs. { 1 } { 2 } DP problem for innermost termination. P = f2#(I5, I6, I7) -> f2#(I8, I9, I7) [3 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ I9 - 3 <= I6 /\ I9 - 3 <= I5 /\ 0 <= I7 - 1 /\ I8 <= I5] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f2(I0, I1, I2) -> f2(I3, I4, 0) [0 = I2 /\ 0 <= I4 - 1 /\ 3 <= I3 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I4 <= I1 /\ I3 - 3 <= I1 /\ I3 - 3 <= I0] f2(I5, I6, I7) -> f2(I8, I9, I7) [3 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ I9 - 3 <= I6 /\ I9 - 3 <= I5 /\ 0 <= I7 - 1 /\ I8 <= I5] f1(I10, I11, I12) -> f2(I13, I14, 0) [1 <= I14 - 1 /\ 3 <= I13 - 1 /\ 0 <= I10 - 1 /\ I14 - 1 <= I10 /\ -1 <= I11 - 1 /\ I13 - 3 <= I10] f1(I15, I16, I17) -> f2(I18, I19, I20) [3 <= I19 - 1 /\ 1 <= I18 - 1 /\ 0 <= I15 - 1 /\ I19 - 3 <= I15 /\ I18 - 1 <= I15 /\ -1 <= I16 - 1 /\ 0 <= I20 - 1] popout

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