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Integ Trans Syste 27634 pair #381738722
details
property
value
status
complete
benchmark
Velroyen08-sunset.jar-obl-8.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n113.star.cs.uiowa.edu
space
From_AProVE_2014
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.628289937973 seconds
cpu usage
0.660070901
max memory
1.1677696E7
stage attributes
key
value
output-size
2592
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 = init#(x1, x2) -> f1#(rnd1, rnd2) f2#(I0, I1) -> f2#(20, I2) [30 <= I0 - 1] f2#(I3, I4) -> f2#(29, I5) [25 = I3] f2#(I6, I7) -> f2#(I6 - 1, I8) [25 <= I6 - 1 /\ I6 <= 30] f2#(I9, I10) -> f2#(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30] f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(20, I2) [30 <= I0 - 1] f2(I3, I4) -> f2(29, I5) [25 = I3] f2(I6, I7) -> f2(I6 - 1, I8) [25 <= I6 - 1 /\ I6 <= 30] f2(I9, I10) -> f2(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30] f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] The dependency graph for this problem is: 0 -> 5 1 -> 4 2 -> 3 3 -> 2, 3 4 -> 4 5 -> 1, 2, 3, 4 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f2#(I0, I1) -> f2#(20, I2) [30 <= I0 - 1] 2) f2#(I3, I4) -> f2#(29, I5) [25 = I3] 3) f2#(I6, I7) -> f2#(I6 - 1, I8) [25 <= I6 - 1 /\ I6 <= 30] 4) f2#(I9, I10) -> f2#(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30] 5) f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] We have the following SCCs. { 2, 3 } { 4 } DP problem for innermost termination. P = f2#(I9, I10) -> f2#(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(20, I2) [30 <= I0 - 1] f2(I3, I4) -> f2(29, I5) [25 = I3] f2(I6, I7) -> f2(I6 - 1, I8) [25 <= I6 - 1 /\ I6 <= 30] f2(I9, I10) -> f2(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30] f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] We use the basic value criterion with the projection function NU: NU[f2#(z1,z2)] = z1 This gives the following inequalities: 10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30 ==> I9 >! I9 - 1 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f2#(I3, I4) -> f2#(29, I5) [25 = I3] f2#(I6, I7) -> f2#(I6 - 1, I8) [25 <= I6 - 1 /\ I6 <= 30] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(20, I2) [30 <= I0 - 1] f2(I3, I4) -> f2(29, I5) [25 = I3] f2(I6, I7) -> f2(I6 - 1, I8) [25 <= I6 - 1 /\ I6 <= 30] f2(I9, I10) -> f2(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30] f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
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