details

property	value
status	complete
benchmark	ex2.t2.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_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	3.48261904716 seconds
cpu usage	3.726460094
max memory	2.1798912E7

stage attributes

key	value
output-size	3339
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 = f8#(x1, x2, x3, x4, x5) -> f7#(x1, x2, x3, x4, x5) f7#(I0, I1, I2, I3, I4) -> f4#(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4] f5#(I5, I6, I7, I8, I9) -> f3#(I5, 1 + I6, I7, I8, I9) f3#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, I19) f4#(I20, I21, I22, I23, I24) -> f2#(I20, I21, I22, I23, I24) [I22 <= 0] f4#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) [1 <= I27] f2#(I30, I31, I32, I33, I34) -> f3#(I30, 0, I32, I33, I34) f1#(I35, I36, I37, I38, I39) -> f2#(I35, I36, I37, I38, I39) [I35 <= I37] f1#(I40, I41, I42, I43, I44) -> f2#(I40, I41, I42, I43, rnd5) [rnd5 = rnd5 /\ 1 + I42 <= I40] R = f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) f7(I0, I1, I2, I3, I4) -> f4(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4] f5(I5, I6, I7, I8, I9) -> f3(I5, 1 + I6, I7, I8, I9) f5(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, I13, I14) f3(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) f4(I20, I21, I22, I23, I24) -> f2(I20, I21, I22, I23, I24) [I22 <= 0] f4(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) [1 <= I27] f2(I30, I31, I32, I33, I34) -> f3(I30, 0, I32, I33, I34) f1(I35, I36, I37, I38, I39) -> f2(I35, I36, I37, I38, I39) [I35 <= I37] f1(I40, I41, I42, I43, I44) -> f2(I40, I41, I42, I43, rnd5) [rnd5 = rnd5 /\ 1 + I42 <= I40] The dependency graph for this problem is: 0 -> 1 1 -> 4, 5 2 -> 3 3 -> 2 4 -> 6 5 -> 7, 8 6 -> 3 7 -> 6 8 -> 6 Where: 0) f8#(x1, x2, x3, x4, x5) -> f7#(x1, x2, x3, x4, x5) 1) f7#(I0, I1, I2, I3, I4) -> f4#(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4] 2) f5#(I5, I6, I7, I8, I9) -> f3#(I5, 1 + I6, I7, I8, I9) 3) f3#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, I19) 4) f4#(I20, I21, I22, I23, I24) -> f2#(I20, I21, I22, I23, I24) [I22 <= 0] 5) f4#(I25, I26, I27, I28, I29) -> f1#(I25, I26, I27, I28, I29) [1 <= I27] 6) f2#(I30, I31, I32, I33, I34) -> f3#(I30, 0, I32, I33, I34) 7) f1#(I35, I36, I37, I38, I39) -> f2#(I35, I36, I37, I38, I39) [I35 <= I37] 8) f1#(I40, I41, I42, I43, I44) -> f2#(I40, I41, I42, I43, rnd5) [rnd5 = rnd5 /\ 1 + I42 <= I40] We have the following SCCs. { 2, 3 } DP problem for innermost termination. P = f5#(I5, I6, I7, I8, I9) -> f3#(I5, 1 + I6, I7, I8, I9) f3#(I15, I16, I17, I18, I19) -> f5#(I15, I16, I17, I18, I19) R = f8(x1, x2, x3, x4, x5) -> f7(x1, x2, x3, x4, x5) f7(I0, I1, I2, I3, I4) -> f4(I0, I1, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4] f5(I5, I6, I7, I8, I9) -> f3(I5, 1 + I6, I7, I8, I9) f5(I10, I11, I12, I13, I14) -> f6(I10, I11, I12, I13, I14) f3(I15, I16, I17, I18, I19) -> f5(I15, I16, I17, I18, I19) f4(I20, I21, I22, I23, I24) -> f2(I20, I21, I22, I23, I24) [I22 <= 0] f4(I25, I26, I27, I28, I29) -> f1(I25, I26, I27, I28, I29) [1 <= I27] f2(I30, I31, I32, I33, I34) -> f3(I30, 0, I32, I33, I34) f1(I35, I36, I37, I38, I39) -> f2(I35, I36, I37, I38, I39) [I35 <= I37] f1(I40, I41, I42, I43, I44) -> f2(I40, I41, I42, I43, rnd5) [rnd5 = rnd5 /\ 1 + I42 <= I40] popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to Integer Transition Systems