details

property	value
status	complete
benchmark	RunningPointers.jar-obl-9.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n008.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.345847129822 seconds
cpu usage	0.360187046
max memory	9895936.0

stage attributes

key	value
output-size	4008
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = init#(x1, x2) -> f3#(rnd1, rnd2) f5#(I0, I1) -> f5#(I0 - 1, I2) [0 <= I0 - 1] f3#(I3, I4) -> f5#(I5, I6) [0 <= I3 - 1 /\ -1 <= I5 - 1 /\ -1 <= I4 - 1] f4#(I7, I8) -> f4#(I9, I10) [-1 <= I10 - 1 /\ -1 <= I9 - 1 /\ 2 <= I8 - 1 /\ 0 <= I7 - 1 /\ I10 + 3 <= I8 /\ I9 + 1 <= I7] f2#(I11, I12) -> f4#(I13, I14) [-1 <= I14 - 1 /\ 0 <= I13 - 1 /\ 0 <= I11 - 1 /\ I14 + 1 <= I11 /\ I13 <= I11] f3#(I15, I16) -> f2#(I17, I18) [-1 <= I17 - 1 /\ 0 <= I15 - 1] f1#(I19, I20) -> f2#(I21, I22) [-1 <= I21 - 1 /\ -1 <= I19 - 1 /\ I21 <= I19] R = init(x1, x2) -> f3(rnd1, rnd2) f5(I0, I1) -> f5(I0 - 1, I2) [0 <= I0 - 1] f3(I3, I4) -> f5(I5, I6) [0 <= I3 - 1 /\ -1 <= I5 - 1 /\ -1 <= I4 - 1] f4(I7, I8) -> f4(I9, I10) [-1 <= I10 - 1 /\ -1 <= I9 - 1 /\ 2 <= I8 - 1 /\ 0 <= I7 - 1 /\ I10 + 3 <= I8 /\ I9 + 1 <= I7] f2(I11, I12) -> f4(I13, I14) [-1 <= I14 - 1 /\ 0 <= I13 - 1 /\ 0 <= I11 - 1 /\ I14 + 1 <= I11 /\ I13 <= I11] f3(I15, I16) -> f2(I17, I18) [-1 <= I17 - 1 /\ 0 <= I15 - 1] f1(I19, I20) -> f2(I21, I22) [-1 <= I21 - 1 /\ -1 <= I19 - 1 /\ I21 <= I19] The dependency graph for this problem is: 0 -> 2, 5 1 -> 1 2 -> 1 3 -> 3 4 -> 3 5 -> 4 6 -> 4 Where: 0) init#(x1, x2) -> f3#(rnd1, rnd2) 1) f5#(I0, I1) -> f5#(I0 - 1, I2) [0 <= I0 - 1] 2) f3#(I3, I4) -> f5#(I5, I6) [0 <= I3 - 1 /\ -1 <= I5 - 1 /\ -1 <= I4 - 1] 3) f4#(I7, I8) -> f4#(I9, I10) [-1 <= I10 - 1 /\ -1 <= I9 - 1 /\ 2 <= I8 - 1 /\ 0 <= I7 - 1 /\ I10 + 3 <= I8 /\ I9 + 1 <= I7] 4) f2#(I11, I12) -> f4#(I13, I14) [-1 <= I14 - 1 /\ 0 <= I13 - 1 /\ 0 <= I11 - 1 /\ I14 + 1 <= I11 /\ I13 <= I11] 5) f3#(I15, I16) -> f2#(I17, I18) [-1 <= I17 - 1 /\ 0 <= I15 - 1] 6) f1#(I19, I20) -> f2#(I21, I22) [-1 <= I21 - 1 /\ -1 <= I19 - 1 /\ I21 <= I19] We have the following SCCs. { 1 } { 3 } DP problem for innermost termination. P = f4#(I7, I8) -> f4#(I9, I10) [-1 <= I10 - 1 /\ -1 <= I9 - 1 /\ 2 <= I8 - 1 /\ 0 <= I7 - 1 /\ I10 + 3 <= I8 /\ I9 + 1 <= I7] R = init(x1, x2) -> f3(rnd1, rnd2) f5(I0, I1) -> f5(I0 - 1, I2) [0 <= I0 - 1] f3(I3, I4) -> f5(I5, I6) [0 <= I3 - 1 /\ -1 <= I5 - 1 /\ -1 <= I4 - 1] f4(I7, I8) -> f4(I9, I10) [-1 <= I10 - 1 /\ -1 <= I9 - 1 /\ 2 <= I8 - 1 /\ 0 <= I7 - 1 /\ I10 + 3 <= I8 /\ I9 + 1 <= I7] f2(I11, I12) -> f4(I13, I14) [-1 <= I14 - 1 /\ 0 <= I13 - 1 /\ 0 <= I11 - 1 /\ I14 + 1 <= I11 /\ I13 <= I11] f3(I15, I16) -> f2(I17, I18) [-1 <= I17 - 1 /\ 0 <= I15 - 1] f1(I19, I20) -> f2(I21, I22) [-1 <= I21 - 1 /\ -1 <= I19 - 1 /\ I21 <= I19] We use the basic value criterion with the projection function NU: NU[f4#(z1,z2)] = z2 This gives the following inequalities: -1 <= I10 - 1 /\ -1 <= I9 - 1 /\ 2 <= I8 - 1 /\ 0 <= I7 - 1 /\ I10 + 3 <= I8 /\ I9 + 1 <= I7 ==> I8 >! I10 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f5#(I0, I1) -> f5#(I0 - 1, I2) [0 <= I0 - 1] R = init(x1, x2) -> f3(rnd1, rnd2) f5(I0, I1) -> f5(I0 - 1, I2) [0 <= I0 - 1] f3(I3, I4) -> f5(I5, I6) [0 <= I3 - 1 /\ -1 <= I5 - 1 /\ -1 <= I4 - 1] f4(I7, I8) -> f4(I9, I10) [-1 <= I10 - 1 /\ -1 <= I9 - 1 /\ 2 <= I8 - 1 /\ 0 <= I7 - 1 /\ I10 + 3 <= I8 /\ I9 + 1 <= I7] f2(I11, I12) -> f4(I13, I14) [-1 <= I14 - 1 /\ 0 <= I13 - 1 /\ 0 <= I11 - 1 /\ I14 + 1 <= I11 /\ I13 <= I11] f3(I15, I16) -> f2(I17, I18) [-1 <= I17 - 1 /\ 0 <= I15 - 1] f1(I19, I20) -> f2(I21, I22) [-1 <= I21 - 1 /\ -1 <= I19 - 1 /\ I21 <= I19] We use the basic value criterion with the projection function NU: NU[f5#(z1,z2)] = z1 This gives the following inequalities: 0 <= I0 - 1 ==> I0 >! I0 - 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