details

property	value
status	complete
benchmark	slayer-2-filtered.t2_fixed.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n191.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	3.62869000435 seconds
cpu usage	3.819761033
max memory	2.3007232E7

stage attributes

key	value
output-size	4823
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 = f7#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) f2#(I5, I6, I7, I8, I9) -> f6#(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3] f4#(I15, I16, I17, I18, I19) -> f2#(I15, 1 + I16, I17, I18, I19) [1 + I16 <= 1 /\ 1 <= 1 + I16 /\ 1 + I16 <= 10] f3#(I20, I21, I22, I23, I24) -> f2#(rnd1, rnd2, I22, I23, I24) [1 + I21 <= 10 /\ y1 = 1 + I21 /\ rnd2 = rnd2 /\ 2 <= rnd2 /\ rnd2 <= 2 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1] f1#(I25, I26, I27, I28, I29) -> f2#(I25, 0, I27, I28, I29) [0 <= 0 /\ 0 <= 0] R = f7(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) f6(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) f2(I5, I6, I7, I8, I9) -> f6(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3] f2(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I14, I14) [10 <= I11] f4(I15, I16, I17, I18, I19) -> f2(I15, 1 + I16, I17, I18, I19) [1 + I16 <= 1 /\ 1 <= 1 + I16 /\ 1 + I16 <= 10] f3(I20, I21, I22, I23, I24) -> f2(rnd1, rnd2, I22, I23, I24) [1 + I21 <= 10 /\ y1 = 1 + I21 /\ rnd2 = rnd2 /\ 2 <= rnd2 /\ rnd2 <= 2 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1] f1(I25, I26, I27, I28, I29) -> f2(I25, 0, I27, I28, I29) [0 <= 0 /\ 0 <= 0] The dependency graph for this problem is: 0 -> 5 1 -> 2 2 -> 1 3 -> 2 4 -> 2 5 -> 2 Where: 0) f7#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) 1) f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 2) f2#(I5, I6, I7, I8, I9) -> f6#(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3] 3) f4#(I15, I16, I17, I18, I19) -> f2#(I15, 1 + I16, I17, I18, I19) [1 + I16 <= 1 /\ 1 <= 1 + I16 /\ 1 + I16 <= 10] 4) f3#(I20, I21, I22, I23, I24) -> f2#(rnd1, rnd2, I22, I23, I24) [1 + I21 <= 10 /\ y1 = 1 + I21 /\ rnd2 = rnd2 /\ 2 <= rnd2 /\ rnd2 <= 2 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1] 5) f1#(I25, I26, I27, I28, I29) -> f2#(I25, 0, I27, I28, I29) [0 <= 0 /\ 0 <= 0] We have the following SCCs. { 1, 2 } DP problem for innermost termination. P = f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) f2#(I5, I6, I7, I8, I9) -> f6#(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3] R = f7(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) f6(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) f2(I5, I6, I7, I8, I9) -> f6(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3] f2(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I14, I14) [10 <= I11] f4(I15, I16, I17, I18, I19) -> f2(I15, 1 + I16, I17, I18, I19) [1 + I16 <= 1 /\ 1 <= 1 + I16 /\ 1 + I16 <= 10] f3(I20, I21, I22, I23, I24) -> f2(rnd1, rnd2, I22, I23, I24) [1 + I21 <= 10 /\ y1 = 1 + I21 /\ rnd2 = rnd2 /\ 2 <= rnd2 /\ rnd2 <= 2 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1] f1(I25, I26, I27, I28, I29) -> f2(I25, 0, I27, I28, I29) [0 <= 0 /\ 0 <= 0] We use the reverse value criterion with the projection function NU: NU[f2#(z1,z2,z3,z4,z5)] = 10 + -1 * (1 + z2) NU[f6#(z1,z2,z3,z4,z5)] = 10 + -1 * (1 + z2) This gives the following inequalities: ==> 10 + -1 * (1 + I1) >= 10 + -1 * (1 + I1) 1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3 ==> 10 + -1 * (1 + I6) > 10 + -1 * (1 + (1 + I6)) with 10 + -1 * (1 + I6) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) R = f7(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) f6(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) f2(I5, I6, I7, I8, I9) -> f6(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3] f2(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I14, I14) [10 <= I11] f4(I15, I16, I17, I18, I19) -> f2(I15, 1 + I16, I17, I18, I19) [1 + I16 <= 1 /\ 1 <= 1 + I16 /\ 1 + I16 <= 10] f3(I20, I21, I22, I23, I24) -> f2(rnd1, rnd2, I22, I23, I24) [1 + I21 <= 10 /\ y1 = 1 + I21 /\ rnd2 = rnd2 /\ 2 <= rnd2 /\ rnd2 <= 2 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1] f1(I25, I26, I27, I28, I29) -> f2(I25, 0, I27, I28, I29) [0 <= 0 /\ 0 <= 0] The dependency graph for this problem is: 1 -> Where: 1) f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 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