details

property	value
status	complete
benchmark	Velroyen08-moduloLower.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n054.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	2.38451004028 seconds
cpu usage	2.549140457
max memory	1.7666048E7

stage attributes

key	value
output-size	3792
starexec-result	MAYBE

output

/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = init#(x1, x2) -> f1#(rnd1, rnd2) f3#(I0, I1) -> f2#(I0 - 1, I2) [0 <= I0 - 5 * y1 - 1 /\ I0 - 5 * y1 <= 4 /\ 2 <= I0 - 1] f2#(I3, I4) -> f3#(I3, I5) [0 <= I3 - 5 * I6 - 1 /\ 2 <= I3 - 1] f3#(I7, I8) -> f2#(I7, I9) [2 <= I7 - 1 /\ I7 - 5 * I10 = 0 /\ I7 - 5 * I10 <= 4 /\ 0 <= I7 - 5 * I10] f2#(I11, I12) -> f3#(I11, I13) [I11 - 5 * I14 = 0 /\ 2 <= I11 - 1] f1#(I15, I16) -> f2#(I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I0 - 1, I2) [0 <= I0 - 5 * y1 - 1 /\ I0 - 5 * y1 <= 4 /\ 2 <= I0 - 1] f2(I3, I4) -> f3(I3, I5) [0 <= I3 - 5 * I6 - 1 /\ 2 <= I3 - 1] f3(I7, I8) -> f2(I7, I9) [2 <= I7 - 1 /\ I7 - 5 * I10 = 0 /\ I7 - 5 * I10 <= 4 /\ 0 <= I7 - 5 * I10] f2(I11, I12) -> f3(I11, I13) [I11 - 5 * I14 = 0 /\ 2 <= I11 - 1] f1(I15, I16) -> f2(I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1] The dependency graph for this problem is: 0 -> 5 1 -> 2, 4 2 -> 1, 3 3 -> 2, 4 4 -> 3 5 -> 2, 4 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f3#(I0, I1) -> f2#(I0 - 1, I2) [0 <= I0 - 5 * y1 - 1 /\ I0 - 5 * y1 <= 4 /\ 2 <= I0 - 1] 2) f2#(I3, I4) -> f3#(I3, I5) [0 <= I3 - 5 * I6 - 1 /\ 2 <= I3 - 1] 3) f3#(I7, I8) -> f2#(I7, I9) [2 <= I7 - 1 /\ I7 - 5 * I10 = 0 /\ I7 - 5 * I10 <= 4 /\ 0 <= I7 - 5 * I10] 4) f2#(I11, I12) -> f3#(I11, I13) [I11 - 5 * I14 = 0 /\ 2 <= I11 - 1] 5) f1#(I15, I16) -> f2#(I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1] We have the following SCCs. { 1, 2, 3, 4 } DP problem for innermost termination. P = f3#(I0, I1) -> f2#(I0 - 1, I2) [0 <= I0 - 5 * y1 - 1 /\ I0 - 5 * y1 <= 4 /\ 2 <= I0 - 1] f2#(I3, I4) -> f3#(I3, I5) [0 <= I3 - 5 * I6 - 1 /\ 2 <= I3 - 1] f3#(I7, I8) -> f2#(I7, I9) [2 <= I7 - 1 /\ I7 - 5 * I10 = 0 /\ I7 - 5 * I10 <= 4 /\ 0 <= I7 - 5 * I10] f2#(I11, I12) -> f3#(I11, I13) [I11 - 5 * I14 = 0 /\ 2 <= I11 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I0 - 1, I2) [0 <= I0 - 5 * y1 - 1 /\ I0 - 5 * y1 <= 4 /\ 2 <= I0 - 1] f2(I3, I4) -> f3(I3, I5) [0 <= I3 - 5 * I6 - 1 /\ 2 <= I3 - 1] f3(I7, I8) -> f2(I7, I9) [2 <= I7 - 1 /\ I7 - 5 * I10 = 0 /\ I7 - 5 * I10 <= 4 /\ 0 <= I7 - 5 * I10] f2(I11, I12) -> f3(I11, I13) [I11 - 5 * I14 = 0 /\ 2 <= I11 - 1] f1(I15, I16) -> f2(I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1] We use the basic value criterion with the projection function NU: NU[f2#(z1,z2)] = z1 NU[f3#(z1,z2)] = z1 This gives the following inequalities: 0 <= I0 - 5 * y1 - 1 /\ I0 - 5 * y1 <= 4 /\ 2 <= I0 - 1 ==> I0 >! I0 - 1 0 <= I3 - 5 * I6 - 1 /\ 2 <= I3 - 1 ==> I3 (>! \union =) I3 2 <= I7 - 1 /\ I7 - 5 * I10 = 0 /\ I7 - 5 * I10 <= 4 /\ 0 <= I7 - 5 * I10 ==> I7 (>! \union =) I7 I11 - 5 * I14 = 0 /\ 2 <= I11 - 1 ==> I11 (>! \union =) I11 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I3, I4) -> f3#(I3, I5) [0 <= I3 - 5 * I6 - 1 /\ 2 <= I3 - 1] f3#(I7, I8) -> f2#(I7, I9) [2 <= I7 - 1 /\ I7 - 5 * I10 = 0 /\ I7 - 5 * I10 <= 4 /\ 0 <= I7 - 5 * I10] f2#(I11, I12) -> f3#(I11, I13) [I11 - 5 * I14 = 0 /\ 2 <= I11 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I0 - 1, I2) [0 <= I0 - 5 * y1 - 1 /\ I0 - 5 * y1 <= 4 /\ 2 <= I0 - 1] f2(I3, I4) -> f3(I3, I5) [0 <= I3 - 5 * I6 - 1 /\ 2 <= I3 - 1] f3(I7, I8) -> f2(I7, I9) [2 <= I7 - 1 /\ I7 - 5 * I10 = 0 /\ I7 - 5 * I10 <= 4 /\ 0 <= I7 - 5 * I10] f2(I11, I12) -> f3(I11, I13) [I11 - 5 * I14 = 0 /\ 2 <= I11 - 1] f1(I15, I16) -> f2(I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1] popout

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