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Integ Trans Syste 27634 pair #381737102
details
property
value
status
complete
benchmark
SharingPair.jar-obl-8.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n037.star.cs.uiowa.edu
space
From_AProVE_2014
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.473881959915 seconds
cpu usage
0.490771055
max memory
1.0653696E7
stage attributes
key
value
output-size
3242
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, x3) -> f1#(rnd1, rnd2, rnd3) f2#(I0, I1, I2) -> f2#(I3, I4, 0) [0 = I2 /\ 0 <= I4 - 1 /\ 3 <= I3 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I4 <= I1 /\ I3 - 3 <= I1 /\ I3 - 3 <= I0] f2#(I5, I6, I7) -> f2#(I8, I9, I7) [3 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ I9 - 3 <= I6 /\ I9 - 3 <= I5 /\ 0 <= I7 - 1 /\ I8 <= I5] f1#(I10, I11, I12) -> f2#(I13, I14, 0) [1 <= I14 - 1 /\ 3 <= I13 - 1 /\ 0 <= I10 - 1 /\ I14 - 1 <= I10 /\ -1 <= I11 - 1 /\ I13 - 3 <= I10] f1#(I15, I16, I17) -> f2#(I18, I19, I20) [3 <= I19 - 1 /\ 1 <= I18 - 1 /\ 0 <= I15 - 1 /\ I19 - 3 <= I15 /\ I18 - 1 <= I15 /\ -1 <= I16 - 1 /\ 0 <= I20 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f2(I0, I1, I2) -> f2(I3, I4, 0) [0 = I2 /\ 0 <= I4 - 1 /\ 3 <= I3 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I4 <= I1 /\ I3 - 3 <= I1 /\ I3 - 3 <= I0] f2(I5, I6, I7) -> f2(I8, I9, I7) [3 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ I9 - 3 <= I6 /\ I9 - 3 <= I5 /\ 0 <= I7 - 1 /\ I8 <= I5] f1(I10, I11, I12) -> f2(I13, I14, 0) [1 <= I14 - 1 /\ 3 <= I13 - 1 /\ 0 <= I10 - 1 /\ I14 - 1 <= I10 /\ -1 <= I11 - 1 /\ I13 - 3 <= I10] f1(I15, I16, I17) -> f2(I18, I19, I20) [3 <= I19 - 1 /\ 1 <= I18 - 1 /\ 0 <= I15 - 1 /\ I19 - 3 <= I15 /\ I18 - 1 <= I15 /\ -1 <= I16 - 1 /\ 0 <= I20 - 1] The dependency graph for this problem is: 0 -> 3, 4 1 -> 1 2 -> 2 3 -> 1 4 -> 2 Where: 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 1) f2#(I0, I1, I2) -> f2#(I3, I4, 0) [0 = I2 /\ 0 <= I4 - 1 /\ 3 <= I3 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I4 <= I1 /\ I3 - 3 <= I1 /\ I3 - 3 <= I0] 2) f2#(I5, I6, I7) -> f2#(I8, I9, I7) [3 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ I9 - 3 <= I6 /\ I9 - 3 <= I5 /\ 0 <= I7 - 1 /\ I8 <= I5] 3) f1#(I10, I11, I12) -> f2#(I13, I14, 0) [1 <= I14 - 1 /\ 3 <= I13 - 1 /\ 0 <= I10 - 1 /\ I14 - 1 <= I10 /\ -1 <= I11 - 1 /\ I13 - 3 <= I10] 4) f1#(I15, I16, I17) -> f2#(I18, I19, I20) [3 <= I19 - 1 /\ 1 <= I18 - 1 /\ 0 <= I15 - 1 /\ I19 - 3 <= I15 /\ I18 - 1 <= I15 /\ -1 <= I16 - 1 /\ 0 <= I20 - 1] We have the following SCCs. { 1 } { 2 } DP problem for innermost termination. P = f2#(I5, I6, I7) -> f2#(I8, I9, I7) [3 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ I9 - 3 <= I6 /\ I9 - 3 <= I5 /\ 0 <= I7 - 1 /\ I8 <= I5] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f2(I0, I1, I2) -> f2(I3, I4, 0) [0 = I2 /\ 0 <= I4 - 1 /\ 3 <= I3 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I4 <= I1 /\ I3 - 3 <= I1 /\ I3 - 3 <= I0] f2(I5, I6, I7) -> f2(I8, I9, I7) [3 <= I9 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ I9 - 3 <= I6 /\ I9 - 3 <= I5 /\ 0 <= I7 - 1 /\ I8 <= I5] f1(I10, I11, I12) -> f2(I13, I14, 0) [1 <= I14 - 1 /\ 3 <= I13 - 1 /\ 0 <= I10 - 1 /\ I14 - 1 <= I10 /\ -1 <= I11 - 1 /\ I13 - 3 <= I10] f1(I15, I16, I17) -> f2(I18, I19, I20) [3 <= I19 - 1 /\ 1 <= I18 - 1 /\ 0 <= I15 - 1 /\ I19 - 3 <= I15 /\ I18 - 1 <= I15 /\ -1 <= I16 - 1 /\ 0 <= I20 - 1]
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