details

property	value
status	complete
benchmark	ex7.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n070.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	5.46944499016 seconds
cpu usage	5.481158005
max memory	2.2892544E7

stage attributes

key	value
output-size	3251
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 = f5#(x1, x2, x3, x4, x5, x6) -> f4#(x1, x2, x3, x4, x5, x6) f4#(I0, I1, I2, I3, I4, I5) -> f3#(I0, 0, I0, rnd4, rnd5, I5) [rnd4 = rnd5 /\ rnd5 = rnd5] f3#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) f1#(I12, I13, I14, I15, I16, I17) -> f3#(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14] R = f5(x1, x2, x3, x4, x5, x6) -> f4(x1, x2, x3, x4, x5, x6) f4(I0, I1, I2, I3, I4, I5) -> f3(I0, 0, I0, rnd4, rnd5, I5) [rnd4 = rnd5 /\ rnd5 = rnd5] f3(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) f1(I12, I13, I14, I15, I16, I17) -> f3(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14] f1(I18, I19, I20, I21, I22, I23) -> f2(I18, I19, I20, I21, I22, I23) [I20 <= I19] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 3 3 -> 2 Where: 0) f5#(x1, x2, x3, x4, x5, x6) -> f4#(x1, x2, x3, x4, x5, x6) 1) f4#(I0, I1, I2, I3, I4, I5) -> f3#(I0, 0, I0, rnd4, rnd5, I5) [rnd4 = rnd5 /\ rnd5 = rnd5] 2) f3#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) 3) f1#(I12, I13, I14, I15, I16, I17) -> f3#(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14] We have the following SCCs. { 2, 3 } DP problem for innermost termination. P = f3#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) f1#(I12, I13, I14, I15, I16, I17) -> f3#(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14] R = f5(x1, x2, x3, x4, x5, x6) -> f4(x1, x2, x3, x4, x5, x6) f4(I0, I1, I2, I3, I4, I5) -> f3(I0, 0, I0, rnd4, rnd5, I5) [rnd4 = rnd5 /\ rnd5 = rnd5] f3(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) f1(I12, I13, I14, I15, I16, I17) -> f3(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14] f1(I18, I19, I20, I21, I22, I23) -> f2(I18, I19, I20, I21, I22, I23) [I20 <= I19] We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5,z6)] = z3 + -1 * (1 + z2) NU[f3#(z1,z2,z3,z4,z5,z6)] = z3 + -1 * (1 + z2) This gives the following inequalities: ==> I8 + -1 * (1 + I7) >= I8 + -1 * (1 + I7) rnd6 = rnd6 /\ 1 + I13 <= I14 ==> I14 + -1 * (1 + I13) > I14 + -1 * (1 + (1 + I13)) with I14 + -1 * (1 + I13) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) R = f5(x1, x2, x3, x4, x5, x6) -> f4(x1, x2, x3, x4, x5, x6) f4(I0, I1, I2, I3, I4, I5) -> f3(I0, 0, I0, rnd4, rnd5, I5) [rnd4 = rnd5 /\ rnd5 = rnd5] f3(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) f1(I12, I13, I14, I15, I16, I17) -> f3(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14] f1(I18, I19, I20, I21, I22, I23) -> f2(I18, I19, I20, I21, I22, I23) [I20 <= I19] The dependency graph for this problem is: 2 -> Where: 2) f3#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) We have the following SCCs. popout

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job log

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actions

all output return to Integ Trans Syste 27634