details

property	value
status	complete
benchmark	slayer-n2.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n139.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	4.31061 seconds
cpu usage	4.37967
user time	2.2938
system time	2.08588
max virtual memory	723180.0
max residence set size	8636.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = f6#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) f5#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) f2#(I5, I6, I7, I8, I9) -> f5#(I5, -1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3] f3#(I15, I16, I17, I18, I19) -> f2#(rnd1, rnd2, I17, I18, I19) [1 + I16 <= 10 /\ y1 = -1 + I16 /\ rnd2 = rnd2 /\ -1 <= rnd2 /\ rnd2 <= -1 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1] f1#(I20, I21, I22, I23, I24) -> f2#(I20, 0, I22, I23, I24) [0 <= 0 /\ 0 <= 0] R = f6(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) f5(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) f2(I5, I6, I7, I8, I9) -> f5(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) -> f4(I10, I11, I12, I14, I14) [10 <= I11] f3(I15, I16, I17, I18, I19) -> f2(rnd1, rnd2, I17, I18, I19) [1 + I16 <= 10 /\ y1 = -1 + I16 /\ rnd2 = rnd2 /\ -1 <= rnd2 /\ rnd2 <= -1 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1] f1(I20, I21, I22, I23, I24) -> f2(I20, 0, I22, I23, I24) [0 <= 0 /\ 0 <= 0] The dependency graph for this problem is: 0 -> 4 1 -> 2 2 -> 1 3 -> 2 4 -> 2 Where: 0) f6#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) 1) f5#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 2) f2#(I5, I6, I7, I8, I9) -> f5#(I5, -1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3] 3) f3#(I15, I16, I17, I18, I19) -> f2#(rnd1, rnd2, I17, I18, I19) [1 + I16 <= 10 /\ y1 = -1 + I16 /\ rnd2 = rnd2 /\ -1 <= rnd2 /\ rnd2 <= -1 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1] 4) f1#(I20, I21, I22, I23, I24) -> f2#(I20, 0, I22, I23, I24) [0 <= 0 /\ 0 <= 0] We have the following SCCs. { 1, 2 } DP problem for innermost termination. P = f5#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) f2#(I5, I6, I7, I8, I9) -> f5#(I5, -1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ -1 + rnd3 <= -1 + I6 /\ -1 + I6 <= -1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3] R = f6(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) f5(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) f2(I5, I6, I7, I8, I9) -> f5(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) -> f4(I10, I11, I12, I14, I14) [10 <= I11] f3(I15, I16, I17, I18, I19) -> f2(rnd1, rnd2, I17, I18, I19) [1 + I16 <= 10 /\ y1 = -1 + I16 /\ rnd2 = rnd2 /\ -1 <= rnd2 /\ rnd2 <= -1 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1] f1(I20, I21, I22, I23, I24) -> f2(I20, 0, I22, I23, I24) [0 <= 0 /\ 0 <= 0] popout

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