details

property	value
status	complete
benchmark	PastaA10.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n140.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.478422 seconds
cpu usage	0.486463
user time	0.261931
system time	0.224532
max virtual memory	113188.0
max residence set size	8492.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = init#(x1, x2) -> f1#(rnd1, rnd2) f2#(I0, I1) -> f2#(I0, I1 + 1) [I1 <= I0 - 1] f2#(I2, I3) -> f2#(I2 + 1, I3) [I2 <= I3 - 1] f1#(I4, I5) -> f2#(I6, I7) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ -1 <= I5 - 1 /\ -1 <= I7 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(I0, I1 + 1) [I1 <= I0 - 1] f2(I2, I3) -> f2(I2 + 1, I3) [I2 <= I3 - 1] f1(I4, I5) -> f2(I6, I7) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ -1 <= I5 - 1 /\ -1 <= I7 - 1] The dependency graph for this problem is: 0 -> 3 1 -> 1 2 -> 2 3 -> 1, 2 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f2#(I0, I1) -> f2#(I0, I1 + 1) [I1 <= I0 - 1] 2) f2#(I2, I3) -> f2#(I2 + 1, I3) [I2 <= I3 - 1] 3) f1#(I4, I5) -> f2#(I6, I7) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ -1 <= I5 - 1 /\ -1 <= I7 - 1] We have the following SCCs. { 2 } { 1 } DP problem for innermost termination. P = f2#(I0, I1) -> f2#(I0, I1 + 1) [I1 <= I0 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(I0, I1 + 1) [I1 <= I0 - 1] f2(I2, I3) -> f2(I2 + 1, I3) [I2 <= I3 - 1] f1(I4, I5) -> f2(I6, I7) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ -1 <= I5 - 1 /\ -1 <= I7 - 1] We use the reverse value criterion with the projection function NU: NU[f2#(z1,z2)] = z1 - 1 + -1 * z2 This gives the following inequalities: I1 <= I0 - 1 ==> I0 - 1 + -1 * I1 > I0 - 1 + -1 * (I1 + 1) with I0 - 1 + -1 * I1 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f2#(I2, I3) -> f2#(I2 + 1, I3) [I2 <= I3 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(I0, I1 + 1) [I1 <= I0 - 1] f2(I2, I3) -> f2(I2 + 1, I3) [I2 <= I3 - 1] f1(I4, I5) -> f2(I6, I7) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ -1 <= I5 - 1 /\ -1 <= I7 - 1] We use the reverse value criterion with the projection function NU: NU[f2#(z1,z2)] = z2 - 1 + -1 * z1 This gives the following inequalities: I2 <= I3 - 1 ==> I3 - 1 + -1 * I2 > I3 - 1 + -1 * (I2 + 1) with I3 - 1 + -1 * I2 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. popout

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