details

property	value
status	complete
benchmark	p-22.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n149.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.414438962936 seconds
cpu usage	0.435850287
max memory	9912320.0

stage attributes

key	value
output-size	2376
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = f6#(x1, x2) -> f5#(x1, x2) f5#(I0, I1) -> f4#(I0, I1) f4#(I4, I5) -> f1#(I4, -1 + I5) [0 <= -1 + I5] f2#(I9, I10) -> f1#(I9, I10) f1#(I11, I12) -> f2#(I11, -1 + I12) [0 <= -1 + I12] R = f6(x1, x2) -> f5(x1, x2) f5(I0, I1) -> f4(I0, I1) f4(I2, I3) -> f3(rnd1, I3) [I3 <= 0 /\ y1 = y1 /\ rnd1 = rnd1] f4(I4, I5) -> f1(I4, -1 + I5) [0 <= -1 + I5] f1(I6, I7) -> f3(I8, I7) [I7 <= 0 /\ B0 = B0 /\ I8 = I8] f2(I9, I10) -> f1(I9, I10) f1(I11, I12) -> f2(I11, -1 + I12) [0 <= -1 + I12] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 4 3 -> 4 4 -> 3 Where: 0) f6#(x1, x2) -> f5#(x1, x2) 1) f5#(I0, I1) -> f4#(I0, I1) 2) f4#(I4, I5) -> f1#(I4, -1 + I5) [0 <= -1 + I5] 3) f2#(I9, I10) -> f1#(I9, I10) 4) f1#(I11, I12) -> f2#(I11, -1 + I12) [0 <= -1 + I12] We have the following SCCs. { 3, 4 } DP problem for innermost termination. P = f2#(I9, I10) -> f1#(I9, I10) f1#(I11, I12) -> f2#(I11, -1 + I12) [0 <= -1 + I12] R = f6(x1, x2) -> f5(x1, x2) f5(I0, I1) -> f4(I0, I1) f4(I2, I3) -> f3(rnd1, I3) [I3 <= 0 /\ y1 = y1 /\ rnd1 = rnd1] f4(I4, I5) -> f1(I4, -1 + I5) [0 <= -1 + I5] f1(I6, I7) -> f3(I8, I7) [I7 <= 0 /\ B0 = B0 /\ I8 = I8] f2(I9, I10) -> f1(I9, I10) f1(I11, I12) -> f2(I11, -1 + I12) [0 <= -1 + I12] We use the basic value criterion with the projection function NU: NU[f1#(z1,z2)] = z2 NU[f2#(z1,z2)] = z2 This gives the following inequalities: ==> I10 (>! \union =) I10 0 <= -1 + I12 ==> I12 >! -1 + I12 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I9, I10) -> f1#(I9, I10) R = f6(x1, x2) -> f5(x1, x2) f5(I0, I1) -> f4(I0, I1) f4(I2, I3) -> f3(rnd1, I3) [I3 <= 0 /\ y1 = y1 /\ rnd1 = rnd1] f4(I4, I5) -> f1(I4, -1 + I5) [0 <= -1 + I5] f1(I6, I7) -> f3(I8, I7) [I7 <= 0 /\ B0 = B0 /\ I8 = I8] f2(I9, I10) -> f1(I9, I10) f1(I11, I12) -> f2(I11, -1 + I12) [0 <= -1 + I12] The dependency graph for this problem is: 3 -> Where: 3) f2#(I9, I10) -> f1#(I9, I10) We have the following SCCs. popout

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