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	n105.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	6.55464410782 seconds
cpu usage	7.004805071
max memory	2.1000192E7

stage attributes

key	value
output-size	4281
starexec-result	MAYBE

output

/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- 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) popout

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