details

property	value
status	complete
benchmark	andrey.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n142.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.097443 seconds
cpu usage	0.099107
user time	0.049665
system time	0.049442
max virtual memory	113188.0
max residence set size	7760.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = f4#(x1) -> f3#(x1) f3#(I0) -> f1#(I0) f1#(I1) -> f3#(rnd1) [1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] R = f4(x1) -> f3(x1) f3(I0) -> f1(I0) f1(I1) -> f3(rnd1) [1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] f1(I2) -> f2(I3) [I2 <= 0 /\ I4 = I2 /\ y2 = I4 /\ y3 = y2 /\ I3 = y3] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 1 Where: 0) f4#(x1) -> f3#(x1) 1) f3#(I0) -> f1#(I0) 2) f1#(I1) -> f3#(rnd1) [1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] We have the following SCCs. { 1, 2 } DP problem for innermost termination. P = f3#(I0) -> f1#(I0) f1#(I1) -> f3#(rnd1) [1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] R = f4(x1) -> f3(x1) f3(I0) -> f1(I0) f1(I1) -> f3(rnd1) [1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] f1(I2) -> f2(I3) [I2 <= 0 /\ I4 = I2 /\ y2 = I4 /\ y3 = y2 /\ I3 = y3] We use the basic value criterion with the projection function NU: NU[f1#(z1)] = z1 NU[f3#(z1)] = z1 This gives the following inequalities: ==> I0 (>! \union =) I0 1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1 ==> I1 >! rnd1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I0) -> f1#(I0) R = f4(x1) -> f3(x1) f3(I0) -> f1(I0) f1(I1) -> f3(rnd1) [1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] f1(I2) -> f2(I3) [I2 <= 0 /\ I4 = I2 /\ y2 = I4 /\ y3 = y2 /\ I3 = y3] The dependency graph for this problem is: 1 -> Where: 1) f3#(I0) -> f1#(I0) We have the following SCCs. popout

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