details

property	value
status	complete
benchmark	BubbleSort.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n144.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	7.81887 seconds
cpu usage	7.95828
user time	4.20108
system time	3.7572
max virtual memory	714688.0
max residence set size	10336.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) f3#(I0, I1, I2) -> f3#(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1] f3#(I3, I4, I5) -> f3#(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1] f3#(I8, I9, I10) -> f2#(I8 + 1, I11, I12) [I10 <= I9] f2#(I13, I14, I15) -> f3#(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1] f1#(I16, I17, I18) -> f2#(1, I19, I20) R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f3(I0, I1, I2) -> f3(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1] f3(I3, I4, I5) -> f3(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1] f3(I8, I9, I10) -> f2(I8 + 1, I11, I12) [I10 <= I9] f2(I13, I14, I15) -> f3(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1] f1(I16, I17, I18) -> f2(1, I19, I20) The dependency graph for this problem is: 0 -> 5 1 -> 1, 2, 3 2 -> 1, 2, 3 3 -> 4 4 -> 1, 2 5 -> 4 Where: 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 1) f3#(I0, I1, I2) -> f3#(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1] 2) f3#(I3, I4, I5) -> f3#(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1] 3) f3#(I8, I9, I10) -> f2#(I8 + 1, I11, I12) [I10 <= I9] 4) f2#(I13, I14, I15) -> f3#(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1] 5) f1#(I16, I17, I18) -> f2#(1, I19, I20) We have the following SCCs. { 1, 2, 3, 4 } DP problem for innermost termination. P = f3#(I0, I1, I2) -> f3#(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1] f3#(I3, I4, I5) -> f3#(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1] f3#(I8, I9, I10) -> f2#(I8 + 1, I11, I12) [I10 <= I9] f2#(I13, I14, I15) -> f3#(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f3(I0, I1, I2) -> f3(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1] f3(I3, I4, I5) -> f3(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1] f3(I8, I9, I10) -> f2(I8 + 1, I11, I12) [I10 <= I9] f2(I13, I14, I15) -> f3(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1] f1(I16, I17, I18) -> f2(1, I19, I20) We use the reverse value criterion with the projection function NU: NU[f2#(z1,z2,z3)] = 99 + -1 * z1 NU[f3#(z1,z2,z3)] = 99 + -1 * (z1 + 1) This gives the following inequalities: I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1 ==> 99 + -1 * (I0 + 1) >= 99 + -1 * (I0 + 1) I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1 ==> 99 + -1 * (I3 + 1) >= 99 + -1 * (I3 + 1) I10 <= I9 ==> 99 + -1 * (I8 + 1) >= 99 + -1 * (I8 + 1) I13 <= 99 /\ 0 <= I13 - 1 ==> 99 + -1 * I13 > 99 + -1 * (I13 + 1) with 99 + -1 * I13 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I0, I1, I2) -> f3#(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1] f3#(I3, I4, I5) -> f3#(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1] f3#(I8, I9, I10) -> f2#(I8 + 1, I11, I12) [I10 <= I9] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f3(I0, I1, I2) -> f3(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1] f3(I3, I4, I5) -> f3(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1] f3(I8, I9, I10) -> f2(I8 + 1, I11, I12) [I10 <= I9] f2(I13, I14, I15) -> f3(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1] f1(I16, I17, I18) -> f2(1, I19, I20) The dependency graph for this problem is: 1 -> 1, 2, 3 2 -> 1, 2, 3 3 -> Where: 1) f3#(I0, I1, I2) -> f3#(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1] 2) f3#(I3, I4, I5) -> f3#(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1] 3) f3#(I8, I9, I10) -> f2#(I8 + 1, I11, I12) [I10 <= I9] We have the following SCCs. { 1, 2 } DP problem for innermost termination. P = f3#(I0, I1, I2) -> f3#(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1] f3#(I3, I4, I5) -> f3#(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f3(I0, I1, I2) -> f3(I0, I1 + 1, 100 - I0) [I1 <= 99 /\ I1 <= I2 - 1 /\ -1 <= I1 - 1 /\ I1 <= 98 /\ 0 <= I0 - 1 /\ y1 <= y2 - 1] f3(I3, I4, I5) -> f3(I3, I4 + 1, 100 - I3) [I4 <= 99 /\ I4 <= I5 - 1 /\ -1 <= I4 - 1 /\ I4 <= 98 /\ I6 <= I7 /\ 0 <= I3 - 1] f3(I8, I9, I10) -> f2(I8 + 1, I11, I12) [I10 <= I9] f2(I13, I14, I15) -> f3(I13, 0, 100 - I13) [I13 <= 99 /\ 0 <= I13 - 1] f1(I16, I17, I18) -> f2(1, I19, I20) We use the reverse value criterion with the projection function NU: popout

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job log

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actions

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