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Integ Trans Syste 27634 pair #381739226
details
property
value
status
complete
benchmark
Take.jar-obl-8.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n097.star.cs.uiowa.edu
space
From_AProVE_2014
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.572771072388 seconds
cpu usage
0.596121248
max memory
1.0842112E7
stage attributes
key
value
output-size
2899
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 = init#(x1, x2, x3, x4, x5) -> f1#(rnd1, rnd2, rnd3, rnd4, rnd5) f2#(I0, I1, I2, I3, I4) -> f2#(I5, I6, I2 + 1, I3 + 1, I4) [I3 + 2 <= I0 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I6 <= I1 /\ I6 <= I0 /\ I5 - 1 <= I0 /\ I2 <= I4 - 1 /\ -1 <= I3 - 1] f1#(I7, I8, I9, I10, I11) -> f2#(I12, I13, 0, I14, I15) [0 <= I13 - 1 /\ 0 <= I12 - 1 /\ 0 <= I7 - 1 /\ I13 <= I7 /\ -1 <= I14 - 1 /\ 1 <= I8 - 1 /\ -1 <= I15 - 1] R = init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) f2(I0, I1, I2, I3, I4) -> f2(I5, I6, I2 + 1, I3 + 1, I4) [I3 + 2 <= I0 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I6 <= I1 /\ I6 <= I0 /\ I5 - 1 <= I0 /\ I2 <= I4 - 1 /\ -1 <= I3 - 1] f1(I7, I8, I9, I10, I11) -> f2(I12, I13, 0, I14, I15) [0 <= I13 - 1 /\ 0 <= I12 - 1 /\ 0 <= I7 - 1 /\ I13 <= I7 /\ -1 <= I14 - 1 /\ 1 <= I8 - 1 /\ -1 <= I15 - 1] The dependency graph for this problem is: 0 -> 2 1 -> 1 2 -> 1 Where: 0) init#(x1, x2, x3, x4, x5) -> f1#(rnd1, rnd2, rnd3, rnd4, rnd5) 1) f2#(I0, I1, I2, I3, I4) -> f2#(I5, I6, I2 + 1, I3 + 1, I4) [I3 + 2 <= I0 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I6 <= I1 /\ I6 <= I0 /\ I5 - 1 <= I0 /\ I2 <= I4 - 1 /\ -1 <= I3 - 1] 2) f1#(I7, I8, I9, I10, I11) -> f2#(I12, I13, 0, I14, I15) [0 <= I13 - 1 /\ 0 <= I12 - 1 /\ 0 <= I7 - 1 /\ I13 <= I7 /\ -1 <= I14 - 1 /\ 1 <= I8 - 1 /\ -1 <= I15 - 1] We have the following SCCs. { 1 } DP problem for innermost termination. P = f2#(I0, I1, I2, I3, I4) -> f2#(I5, I6, I2 + 1, I3 + 1, I4) [I3 + 2 <= I0 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I6 <= I1 /\ I6 <= I0 /\ I5 - 1 <= I0 /\ I2 <= I4 - 1 /\ -1 <= I3 - 1] R = init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) f2(I0, I1, I2, I3, I4) -> f2(I5, I6, I2 + 1, I3 + 1, I4) [I3 + 2 <= I0 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I6 <= I1 /\ I6 <= I0 /\ I5 - 1 <= I0 /\ I2 <= I4 - 1 /\ -1 <= I3 - 1] f1(I7, I8, I9, I10, I11) -> f2(I12, I13, 0, I14, I15) [0 <= I13 - 1 /\ 0 <= I12 - 1 /\ 0 <= I7 - 1 /\ I13 <= I7 /\ -1 <= I14 - 1 /\ 1 <= I8 - 1 /\ -1 <= I15 - 1] We use the reverse value criterion with the projection function NU: NU[f2#(z1,z2,z3,z4,z5)] = z5 - 1 + -1 * z3 This gives the following inequalities: I3 + 2 <= I0 /\ 0 <= I6 - 1 /\ 0 <= I5 - 1 /\ 0 <= I1 - 1 /\ 0 <= I0 - 1 /\ I6 <= I1 /\ I6 <= I0 /\ I5 - 1 <= I0 /\ I2 <= I4 - 1 /\ -1 <= I3 - 1 ==> I4 - 1 + -1 * I2 > I4 - 1 + -1 * (I2 + 1) with I4 - 1 + -1 * I2 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
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