details

property	value
status	complete
benchmark	PastaC3.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)	1.01887106895 seconds
cpu usage	1.081355189
max memory	1.212416E7

stage attributes

key	value
output-size	2966
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, x3) -> f1#(rnd1, rnd2, rnd3) f2#(I0, I1, I2) -> f2#(I0, I1 + 1, I2) [I1 <= I0 - 1 /\ I1 <= I2 - 1] f2#(I3, I4, I5) -> f2#(I3 + 1, I4, I5) [I3 <= I4 /\ I4 <= I5 - 1] f1#(I6, I7, I8) -> f2#(I9, I10, I11) [0 <= I6 - 1 /\ -1 <= I11 - 1 /\ -1 <= I9 - 1 /\ -1 <= I7 - 1 /\ -1 <= I10 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f2(I0, I1, I2) -> f2(I0, I1 + 1, I2) [I1 <= I0 - 1 /\ I1 <= I2 - 1] f2(I3, I4, I5) -> f2(I3 + 1, I4, I5) [I3 <= I4 /\ I4 <= I5 - 1] f1(I6, I7, I8) -> f2(I9, I10, I11) [0 <= I6 - 1 /\ -1 <= I11 - 1 /\ -1 <= I9 - 1 /\ -1 <= I7 - 1 /\ -1 <= I10 - 1] The dependency graph for this problem is: 0 -> 3 1 -> 1, 2 2 -> 1, 2 3 -> 1, 2 Where: 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 1) f2#(I0, I1, I2) -> f2#(I0, I1 + 1, I2) [I1 <= I0 - 1 /\ I1 <= I2 - 1] 2) f2#(I3, I4, I5) -> f2#(I3 + 1, I4, I5) [I3 <= I4 /\ I4 <= I5 - 1] 3) f1#(I6, I7, I8) -> f2#(I9, I10, I11) [0 <= I6 - 1 /\ -1 <= I11 - 1 /\ -1 <= I9 - 1 /\ -1 <= I7 - 1 /\ -1 <= I10 - 1] We have the following SCCs. { 1, 2 } DP problem for innermost termination. P = f2#(I0, I1, I2) -> f2#(I0, I1 + 1, I2) [I1 <= I0 - 1 /\ I1 <= I2 - 1] f2#(I3, I4, I5) -> f2#(I3 + 1, I4, I5) [I3 <= I4 /\ I4 <= I5 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f2(I0, I1, I2) -> f2(I0, I1 + 1, I2) [I1 <= I0 - 1 /\ I1 <= I2 - 1] f2(I3, I4, I5) -> f2(I3 + 1, I4, I5) [I3 <= I4 /\ I4 <= I5 - 1] f1(I6, I7, I8) -> f2(I9, I10, I11) [0 <= I6 - 1 /\ -1 <= I11 - 1 /\ -1 <= I9 - 1 /\ -1 <= I7 - 1 /\ -1 <= I10 - 1] We use the reverse value criterion with the projection function NU: NU[f2#(z1,z2,z3)] = z3 - 1 + -1 * z2 This gives the following inequalities: I1 <= I0 - 1 /\ I1 <= I2 - 1 ==> I2 - 1 + -1 * I1 > I2 - 1 + -1 * (I1 + 1) with I2 - 1 + -1 * I1 >= 0 I3 <= I4 /\ I4 <= I5 - 1 ==> I5 - 1 + -1 * I4 >= I5 - 1 + -1 * I4 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I3, I4, I5) -> f2#(I3 + 1, I4, I5) [I3 <= I4 /\ I4 <= I5 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f2(I0, I1, I2) -> f2(I0, I1 + 1, I2) [I1 <= I0 - 1 /\ I1 <= I2 - 1] f2(I3, I4, I5) -> f2(I3 + 1, I4, I5) [I3 <= I4 /\ I4 <= I5 - 1] f1(I6, I7, I8) -> f2(I9, I10, I11) [0 <= I6 - 1 /\ -1 <= I11 - 1 /\ -1 <= I9 - 1 /\ -1 <= I7 - 1 /\ -1 <= I10 - 1] We use the reverse value criterion with the projection function NU: NU[f2#(z1,z2,z3)] = z2 + -1 * z1 This gives the following inequalities: I3 <= I4 /\ I4 <= I5 - 1 ==> I4 + -1 * I3 > I4 + -1 * (I3 + 1) with I4 + -1 * I3 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. popout

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