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Integ Trans Syste 27634 pair #381736970
details
property
value
status
complete
benchmark
EvenOdd.jar-obl-8.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n079.star.cs.uiowa.edu
space
From_AProVE_2014
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
4.06409406662 seconds
cpu usage
4.313418689
max memory
2.4506368E7
stage attributes
key
value
output-size
3376
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) f3#(I0, I1) -> f2#(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 1 <= I0 - 1] f3#(I3, I4) -> f2#(I3 - 1, I5) [I3 <= 1 /\ I3 - 1 <= I3 - 1 /\ I3 <= -1 /\ I3 <= 0] f2#(I6, I7) -> f3#(I6 - 1, I8) [I6 - 1 <= I6 - 1 /\ 1 <= I6 - 1] f2#(I9, I10) -> f3#(I9 - 1, I11) [I9 <= 1 /\ I9 - 1 <= I9 - 1 /\ I9 <= -1 /\ I9 <= 0] f1#(I12, I13) -> f2#(I13 - 5, I14) [0 <= I12 - 1 /\ -1 <= I13 - 1 /\ I13 - 5 <= I13 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 1 <= I0 - 1] f3(I3, I4) -> f2(I3 - 1, I5) [I3 <= 1 /\ I3 - 1 <= I3 - 1 /\ I3 <= -1 /\ I3 <= 0] f2(I6, I7) -> f3(I6 - 1, I8) [I6 - 1 <= I6 - 1 /\ 1 <= I6 - 1] f2(I9, I10) -> f3(I9 - 1, I11) [I9 <= 1 /\ I9 - 1 <= I9 - 1 /\ I9 <= -1 /\ I9 <= 0] f1(I12, I13) -> f2(I13 - 5, I14) [0 <= I12 - 1 /\ -1 <= I13 - 1 /\ I13 - 5 <= I13 - 1] The dependency graph for this problem is: 0 -> 5 1 -> 3 2 -> 4 3 -> 1 4 -> 2 5 -> 3, 4 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f3#(I0, I1) -> f2#(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 1 <= I0 - 1] 2) f3#(I3, I4) -> f2#(I3 - 1, I5) [I3 <= 1 /\ I3 - 1 <= I3 - 1 /\ I3 <= -1 /\ I3 <= 0] 3) f2#(I6, I7) -> f3#(I6 - 1, I8) [I6 - 1 <= I6 - 1 /\ 1 <= I6 - 1] 4) f2#(I9, I10) -> f3#(I9 - 1, I11) [I9 <= 1 /\ I9 - 1 <= I9 - 1 /\ I9 <= -1 /\ I9 <= 0] 5) f1#(I12, I13) -> f2#(I13 - 5, I14) [0 <= I12 - 1 /\ -1 <= I13 - 1 /\ I13 - 5 <= I13 - 1] We have the following SCCs. { 2, 4 } { 1, 3 } DP problem for innermost termination. P = f3#(I0, I1) -> f2#(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 1 <= I0 - 1] f2#(I6, I7) -> f3#(I6 - 1, I8) [I6 - 1 <= I6 - 1 /\ 1 <= I6 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 1 <= I0 - 1] f3(I3, I4) -> f2(I3 - 1, I5) [I3 <= 1 /\ I3 - 1 <= I3 - 1 /\ I3 <= -1 /\ I3 <= 0] f2(I6, I7) -> f3(I6 - 1, I8) [I6 - 1 <= I6 - 1 /\ 1 <= I6 - 1] f2(I9, I10) -> f3(I9 - 1, I11) [I9 <= 1 /\ I9 - 1 <= I9 - 1 /\ I9 <= -1 /\ I9 <= 0] f1(I12, I13) -> f2(I13 - 5, I14) [0 <= I12 - 1 /\ -1 <= I13 - 1 /\ I13 - 5 <= I13 - 1] We use the basic value criterion with the projection function NU: NU[f2#(z1,z2)] = z1 NU[f3#(z1,z2)] = z1 This gives the following inequalities: I0 - 1 <= I0 - 1 /\ 1 <= I0 - 1 ==> I0 >! I0 - 1 I6 - 1 <= I6 - 1 /\ 1 <= I6 - 1 ==> I6 >! I6 - 1 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f3#(I3, I4) -> f2#(I3 - 1, I5) [I3 <= 1 /\ I3 - 1 <= I3 - 1 /\ I3 <= -1 /\ I3 <= 0] f2#(I9, I10) -> f3#(I9 - 1, I11) [I9 <= 1 /\ I9 - 1 <= I9 - 1 /\ I9 <= -1 /\ I9 <= 0] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I0 - 1, I2) [I0 - 1 <= I0 - 1 /\ 1 <= I0 - 1] f3(I3, I4) -> f2(I3 - 1, I5) [I3 <= 1 /\ I3 - 1 <= I3 - 1 /\ I3 <= -1 /\ I3 <= 0] f2(I6, I7) -> f3(I6 - 1, I8) [I6 - 1 <= I6 - 1 /\ 1 <= I6 - 1] f2(I9, I10) -> f3(I9 - 1, I11) [I9 <= 1 /\ I9 - 1 <= I9 - 1 /\ I9 <= -1 /\ I9 <= 0] f1(I12, I13) -> f2(I13 - 5, I14) [0 <= I12 - 1 /\ -1 <= I13 - 1 /\ I13 - 5 <= I13 - 1]
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