details

property	value
status	complete
benchmark	PastaB12.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n079.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.242042064667 seconds
cpu usage	0.241356869
max memory	9236480.0

stage attributes

key	value
output-size	2304
starexec-result	YES

output

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

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all output return to Integ Trans Syste 27634