details

property	value
status	complete
benchmark	w1.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n145.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.255237 seconds
cpu usage	0.26008
user time	0.148302
system time	0.111778
max virtual memory	113188.0
max residence set size	8400.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = f4#(x1) -> f3#(x1) f3#(I0) -> f1#(rnd1) [2 <= rnd1 /\ rnd1 = rnd1] f2#(I1) -> f1#(I1) f1#(I2) -> f2#(I3) [y1 = y1 /\ 1 <= y1 /\ 2 <= y1 /\ I3 = I3 /\ -1 <= I3 /\ 1 + I3 <= 2] R = f4(x1) -> f3(x1) f3(I0) -> f1(rnd1) [2 <= rnd1 /\ rnd1 = rnd1] f2(I1) -> f1(I1) f1(I2) -> f2(I3) [y1 = y1 /\ 1 <= y1 /\ 2 <= y1 /\ I3 = I3 /\ -1 <= I3 /\ 1 + I3 <= 2] The dependency graph for this problem is: 0 -> 1 1 -> 3 2 -> 3 3 -> 2 Where: 0) f4#(x1) -> f3#(x1) 1) f3#(I0) -> f1#(rnd1) [2 <= rnd1 /\ rnd1 = rnd1] 2) f2#(I1) -> f1#(I1) 3) f1#(I2) -> f2#(I3) [y1 = y1 /\ 1 <= y1 /\ 2 <= y1 /\ I3 = I3 /\ -1 <= I3 /\ 1 + I3 <= 2] We have the following SCCs. { 2, 3 } DP problem for innermost termination. P = f2#(I1) -> f1#(I1) f1#(I2) -> f2#(I3) [y1 = y1 /\ 1 <= y1 /\ 2 <= y1 /\ I3 = I3 /\ -1 <= I3 /\ 1 + I3 <= 2] R = f4(x1) -> f3(x1) f3(I0) -> f1(rnd1) [2 <= rnd1 /\ rnd1 = rnd1] f2(I1) -> f1(I1) f1(I2) -> f2(I3) [y1 = y1 /\ 1 <= y1 /\ 2 <= y1 /\ I3 = I3 /\ -1 <= I3 /\ 1 + I3 <= 2] popout

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