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Integ Trans Syste 27634 pair #381739046
details
property
value
status
complete
benchmark
GCD5.jar-obl-8.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n109.star.cs.uiowa.edu
space
From_AProVE_2014
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.408791065216 seconds
cpu usage
0.42685464
max memory
1.0088448E7
stage attributes
key
value
output-size
2829
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = init#(x1, x2) -> f1#(rnd1, rnd2) f3#(I0, I1) -> f2#(I1, I2) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2] f2#(I3, I4) -> f3#(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1] f1#(I5, I6) -> f2#(I7, I8) [0 <= I5 - 1 /\ -1 <= I7 - 1 /\ -1 <= I6 - 1 /\ -1 <= I8 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I1, I2) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2] f2(I3, I4) -> f3(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1] f1(I5, I6) -> f2(I7, I8) [0 <= I5 - 1 /\ -1 <= I7 - 1 /\ -1 <= I6 - 1 /\ -1 <= I8 - 1] The dependency graph for this problem is: 0 -> 3 1 -> 2 2 -> 1 3 -> 2 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f3#(I0, I1) -> f2#(I1, I2) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2] 2) f2#(I3, I4) -> f3#(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1] 3) f1#(I5, I6) -> f2#(I7, I8) [0 <= I5 - 1 /\ -1 <= I7 - 1 /\ -1 <= I6 - 1 /\ -1 <= I8 - 1] We have the following SCCs. { 1, 2 } DP problem for innermost termination. P = f3#(I0, I1) -> f2#(I1, I2) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2] f2#(I3, I4) -> f3#(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I1, I2) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2] f2(I3, I4) -> f3(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1] f1(I5, I6) -> f2(I7, I8) [0 <= I5 - 1 /\ -1 <= I7 - 1 /\ -1 <= I6 - 1 /\ -1 <= I8 - 1] We use the basic value criterion with the projection function NU: NU[f2#(z1,z2)] = z2 NU[f3#(z1,z2)] = z2 This gives the following inequalities: 0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2 ==> I1 >! I2 0 <= I3 - 1 /\ 0 <= I4 - 1 ==> I4 (>! \union =) I4 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I3, I4) -> f3#(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I1, I2) [0 <= I0 - 1 /\ 0 <= I1 - 1 /\ I0 - I1 * y1 <= I1 - 1 /\ 0 <= I0 - I1 * y1 /\ I0 - I1 * y1 = I2] f2(I3, I4) -> f3(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1] f1(I5, I6) -> f2(I7, I8) [0 <= I5 - 1 /\ -1 <= I7 - 1 /\ -1 <= I6 - 1 /\ -1 <= I8 - 1] The dependency graph for this problem is: 2 -> Where: 2) f2#(I3, I4) -> f3#(I3, I4) [0 <= I3 - 1 /\ 0 <= I4 - 1] We have the following SCCs.
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