details

property	value
status	complete
benchmark	p-12.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n027.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.393192052841 seconds
cpu usage	0.370823264
max memory	9744384.0

stage attributes

key	value
output-size	1882
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) -> f1#(x1, x2) f3#(I2, I3) -> f2#(I2, I3) f2#(I4, I5) -> f3#(I4, -1 + I5) [-1 * I5 <= 0] f1#(I6, I7) -> f2#(I6, I7) R = f5(x1, x2) -> f1(x1, x2) f2(I0, I1) -> f4(rnd1, I1) [rnd1 = rnd1 /\ 0 <= -1 - I1] f3(I2, I3) -> f2(I2, I3) f2(I4, I5) -> f3(I4, -1 + I5) [-1 * I5 <= 0] f1(I6, I7) -> f2(I6, I7) The dependency graph for this problem is: 0 -> 3 1 -> 2 2 -> 1 3 -> 2 Where: 0) f5#(x1, x2) -> f1#(x1, x2) 1) f3#(I2, I3) -> f2#(I2, I3) 2) f2#(I4, I5) -> f3#(I4, -1 + I5) [-1 * I5 <= 0] 3) f1#(I6, I7) -> f2#(I6, I7) We have the following SCCs. { 1, 2 } DP problem for innermost termination. P = f3#(I2, I3) -> f2#(I2, I3) f2#(I4, I5) -> f3#(I4, -1 + I5) [-1 * I5 <= 0] R = f5(x1, x2) -> f1(x1, x2) f2(I0, I1) -> f4(rnd1, I1) [rnd1 = rnd1 /\ 0 <= -1 - I1] f3(I2, I3) -> f2(I2, I3) f2(I4, I5) -> f3(I4, -1 + I5) [-1 * I5 <= 0] f1(I6, I7) -> f2(I6, I7) We use the basic value criterion with the projection function NU: NU[f2#(z1,z2)] = z2 NU[f3#(z1,z2)] = z2 This gives the following inequalities: ==> I3 (>! \union =) I3 -1 * I5 <= 0 ==> I5 >! -1 + I5 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I2, I3) -> f2#(I2, I3) R = f5(x1, x2) -> f1(x1, x2) f2(I0, I1) -> f4(rnd1, I1) [rnd1 = rnd1 /\ 0 <= -1 - I1] f3(I2, I3) -> f2(I2, I3) f2(I4, I5) -> f3(I4, -1 + I5) [-1 * I5 <= 0] f1(I6, I7) -> f2(I6, I7) The dependency graph for this problem is: 1 -> Where: 1) f3#(I2, I3) -> f2#(I2, I3) We have the following SCCs. popout

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

job log

popout

actions

all output return to Integer Transition Systems