details

property	value
status	complete
benchmark	ListReverseAcyclicList.jar-obl-9.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n100.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.391896009445 seconds
cpu usage	0.398784129
max memory	9785344.0

stage attributes

key	value
output-size	3388
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) -> f1#(rnd1, rnd2) f3#(I0, I1) -> f3#(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0] f2#(I4, I5) -> f2#(I6, I5 - 1) [0 <= I6 - 1 /\ -1 <= I4 - 1 /\ 0 <= I5 - 1 /\ I6 - 2 <= I4] f2#(I7, I8) -> f3#(I9, I10) [-1 <= I9 - 1 /\ -1 <= I7 - 1 /\ I8 <= 0 /\ I9 <= I7] f1#(I11, I12) -> f2#(I13, I14) [0 <= I12 - 1 /\ -1 <= y1 - 1 /\ I13 + 1 <= I11 /\ 0 <= I11 - 1 /\ -1 <= I13 - 1 /\ y1 - 1 = I14] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f3(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0] f2(I4, I5) -> f2(I6, I5 - 1) [0 <= I6 - 1 /\ -1 <= I4 - 1 /\ 0 <= I5 - 1 /\ I6 - 2 <= I4] f2(I7, I8) -> f3(I9, I10) [-1 <= I9 - 1 /\ -1 <= I7 - 1 /\ I8 <= 0 /\ I9 <= I7] f1(I11, I12) -> f2(I13, I14) [0 <= I12 - 1 /\ -1 <= y1 - 1 /\ I13 + 1 <= I11 /\ 0 <= I11 - 1 /\ -1 <= I13 - 1 /\ y1 - 1 = I14] The dependency graph for this problem is: 0 -> 4 1 -> 1 2 -> 2, 3 3 -> 1 4 -> 2, 3 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f3#(I0, I1) -> f3#(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0] 2) f2#(I4, I5) -> f2#(I6, I5 - 1) [0 <= I6 - 1 /\ -1 <= I4 - 1 /\ 0 <= I5 - 1 /\ I6 - 2 <= I4] 3) f2#(I7, I8) -> f3#(I9, I10) [-1 <= I9 - 1 /\ -1 <= I7 - 1 /\ I8 <= 0 /\ I9 <= I7] 4) f1#(I11, I12) -> f2#(I13, I14) [0 <= I12 - 1 /\ -1 <= y1 - 1 /\ I13 + 1 <= I11 /\ 0 <= I11 - 1 /\ -1 <= I13 - 1 /\ y1 - 1 = I14] We have the following SCCs. { 2 } { 1 } DP problem for innermost termination. P = f3#(I0, I1) -> f3#(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f3(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0] f2(I4, I5) -> f2(I6, I5 - 1) [0 <= I6 - 1 /\ -1 <= I4 - 1 /\ 0 <= I5 - 1 /\ I6 - 2 <= I4] f2(I7, I8) -> f3(I9, I10) [-1 <= I9 - 1 /\ -1 <= I7 - 1 /\ I8 <= 0 /\ I9 <= I7] f1(I11, I12) -> f2(I13, I14) [0 <= I12 - 1 /\ -1 <= y1 - 1 /\ I13 + 1 <= I11 /\ 0 <= I11 - 1 /\ -1 <= I13 - 1 /\ y1 - 1 = I14] We use the basic value criterion with the projection function NU: NU[f3#(z1,z2)] = z1 This gives the following inequalities: -1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0 ==> I0 >! I2 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f2#(I4, I5) -> f2#(I6, I5 - 1) [0 <= I6 - 1 /\ -1 <= I4 - 1 /\ 0 <= I5 - 1 /\ I6 - 2 <= I4] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f3(I2, I3) [-1 <= I2 - 1 /\ 0 <= I0 - 1 /\ I2 + 1 <= I0] f2(I4, I5) -> f2(I6, I5 - 1) [0 <= I6 - 1 /\ -1 <= I4 - 1 /\ 0 <= I5 - 1 /\ I6 - 2 <= I4] f2(I7, I8) -> f3(I9, I10) [-1 <= I9 - 1 /\ -1 <= I7 - 1 /\ I8 <= 0 /\ I9 <= I7] f1(I11, I12) -> f2(I13, I14) [0 <= I12 - 1 /\ -1 <= y1 - 1 /\ I13 + 1 <= I11 /\ 0 <= I11 - 1 /\ -1 <= I13 - 1 /\ y1 - 1 = I14] We use the basic value criterion with the projection function NU: NU[f2#(z1,z2)] = z2 This gives the following inequalities: 0 <= I6 - 1 /\ -1 <= I4 - 1 /\ 0 <= I5 - 1 /\ I6 - 2 <= I4 ==> I5 >! I5 - 1 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. popout

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