details

property	value
status	complete
benchmark	acqrel-succeed2.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n148.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	6.55183 seconds
cpu usage	6.64932
user time	3.25876
system time	3.39055
max virtual memory	723276.0
max residence set size	8612.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = f5#(x1, x2, x3, x4) -> f4#(x1, x2, x3, x4) f4#(I0, I1, I2, I3) -> f2#(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3] f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1] f3#(I8, I9, I10, I11) -> f1#(I8, I9, I10, I11) f1#(I12, I13, I14, I15) -> f3#(I12, I13, I14, -1 + I15) [1 <= I15] f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20] R = f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) f4(I0, I1, I2, I3) -> f2(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3] f2(I4, I5, I6, I7) -> f1(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1] f3(I8, I9, I10, I11) -> f1(I8, I9, I10, I11) f1(I12, I13, I14, I15) -> f3(I12, I13, I14, -1 + I15) [1 <= I15] f1(I16, I17, I18, I19) -> f2(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 4, 5 3 -> 4, 5 4 -> 3 5 -> 2 Where: 0) f5#(x1, x2, x3, x4) -> f4#(x1, x2, x3, x4) 1) f4#(I0, I1, I2, I3) -> f2#(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3] 2) f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1] 3) f3#(I8, I9, I10, I11) -> f1#(I8, I9, I10, I11) 4) f1#(I12, I13, I14, I15) -> f3#(I12, I13, I14, -1 + I15) [1 <= I15] 5) f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20] We have the following SCCs. { 2, 3, 4, 5 } DP problem for innermost termination. P = f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1] f3#(I8, I9, I10, I11) -> f1#(I8, I9, I10, I11) f1#(I12, I13, I14, I15) -> f3#(I12, I13, I14, -1 + I15) [1 <= I15] f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20] R = f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) f4(I0, I1, I2, I3) -> f2(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3] f2(I4, I5, I6, I7) -> f1(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1] f3(I8, I9, I10, I11) -> f1(I8, I9, I10, I11) f1(I12, I13, I14, I15) -> f3(I12, I13, I14, -1 + I15) [1 <= I15] f1(I16, I17, I18, I19) -> f2(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20] We use the basic value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4)] = z4 NU[f1#(z1,z2,z3,z4)] = z4 NU[f2#(z1,z2,z3,z4)] = z4 This gives the following inequalities: I6 <= 0 /\ y1 = 1 ==> I7 (>! \union =) I7 ==> I11 (>! \union =) I11 1 <= I15 ==> I15 >! -1 + I15 I19 <= 0 /\ I21 = 0 /\ I20 = I20 ==> I19 (>! \union =) I19 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1] f3#(I8, I9, I10, I11) -> f1#(I8, I9, I10, I11) f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20] R = f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) f4(I0, I1, I2, I3) -> f2(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3] f2(I4, I5, I6, I7) -> f1(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1] f3(I8, I9, I10, I11) -> f1(I8, I9, I10, I11) f1(I12, I13, I14, I15) -> f3(I12, I13, I14, -1 + I15) [1 <= I15] f1(I16, I17, I18, I19) -> f2(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20] The dependency graph for this problem is: 2 -> 5 3 -> 5 5 -> 2 Where: 2) f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1] 3) f3#(I8, I9, I10, I11) -> f1#(I8, I9, I10, I11) 5) f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20] We have the following SCCs. { 2, 5 } DP problem for innermost termination. P = f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1] f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20] R = f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) f4(I0, I1, I2, I3) -> f2(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3] f2(I4, I5, I6, I7) -> f1(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1] f3(I8, I9, I10, I11) -> f1(I8, I9, I10, I11) f1(I12, I13, I14, I15) -> f3(I12, I13, I14, -1 + I15) [1 <= I15] f1(I16, I17, I18, I19) -> f2(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20] popout

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job log

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