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Integ Trans Syste 27634 pair #381738566
details
property
value
status
complete
benchmark
GCD.jar-obl-8.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n083.star.cs.uiowa.edu
space
From_AProVE_2014
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.604004859924 seconds
cpu usage
0.636358304
max memory
1.1235328E7
stage attributes
key
value
output-size
4994
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) -> f1#(rnd1, rnd2) f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1] f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1] f2#(I4, I5) -> f2#(I4, 0) [I4 = I5 /\ 0 <= I4 - 1] f2#(I6, I7) -> f2#(I7, 0) [0 = I6 /\ 0 <= I7 - 1] f2#(I8, I9) -> f3#(I8, I9) [0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1] f2#(I10, I11) -> f3#(I10, I11) [0 <= I10 - 1 /\ I10 <= I11 - 1 /\ 0 <= I11 - 1] f1#(I12, I13) -> f2#(I14, I15) [0 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ -1 <= I15 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1] f3(I2, I3) -> f3(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1] f2(I4, I5) -> f2(I4, 0) [I4 = I5 /\ 0 <= I4 - 1] f2(I6, I7) -> f2(I7, 0) [0 = I6 /\ 0 <= I7 - 1] f2(I8, I9) -> f3(I8, I9) [0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1] f2(I10, I11) -> f3(I10, I11) [0 <= I10 - 1 /\ I10 <= I11 - 1 /\ 0 <= I11 - 1] f1(I12, I13) -> f2(I14, I15) [0 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ -1 <= I15 - 1] The dependency graph for this problem is: 0 -> 7 1 -> 5 2 -> 1, 2 3 -> 4 -> 5 -> 2 6 -> 1 7 -> 3, 4, 5, 6 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1] 2) f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1] 3) f2#(I4, I5) -> f2#(I4, 0) [I4 = I5 /\ 0 <= I4 - 1] 4) f2#(I6, I7) -> f2#(I7, 0) [0 = I6 /\ 0 <= I7 - 1] 5) f2#(I8, I9) -> f3#(I8, I9) [0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1] 6) f2#(I10, I11) -> f3#(I10, I11) [0 <= I10 - 1 /\ I10 <= I11 - 1 /\ 0 <= I11 - 1] 7) f1#(I12, I13) -> f2#(I14, I15) [0 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ -1 <= I15 - 1] We have the following SCCs. { 1, 2, 5 } DP problem for innermost termination. P = f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1] f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1] f2#(I8, I9) -> f3#(I8, I9) [0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1] f3(I2, I3) -> f3(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1] f2(I4, I5) -> f2(I4, 0) [I4 = I5 /\ 0 <= I4 - 1] f2(I6, I7) -> f2(I7, 0) [0 = I6 /\ 0 <= I7 - 1] f2(I8, I9) -> f3(I8, I9) [0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1] f2(I10, I11) -> f3(I10, I11) [0 <= I10 - 1 /\ I10 <= I11 - 1 /\ 0 <= I11 - 1] f1(I12, I13) -> f2(I14, I15) [0 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ -1 <= I15 - 1] We use the basic value criterion with the projection function NU: NU[f2#(z1,z2)] = z1 NU[f3#(z1,z2)] = z2 This gives the following inequalities: I0 <= I1 - 1 ==> I1 (>! \union =) I1 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1 ==> I3 (>! \union =) I3 0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1 ==> I8 >! I9 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1] f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1] f3(I2, I3) -> f3(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1] f2(I4, I5) -> f2(I4, 0) [I4 = I5 /\ 0 <= I4 - 1] f2(I6, I7) -> f2(I7, 0) [0 = I6 /\ 0 <= I7 - 1] f2(I8, I9) -> f3(I8, I9) [0 <= I8 - 1 /\ I9 <= I8 - 1 /\ 0 <= I9 - 1] f2(I10, I11) -> f3(I10, I11) [0 <= I10 - 1 /\ I10 <= I11 - 1 /\ 0 <= I11 - 1] f1(I12, I13) -> f2(I14, I15) [0 <= I12 - 1 /\ -1 <= I14 - 1 /\ -1 <= I13 - 1 /\ -1 <= I15 - 1] The dependency graph for this problem is: 1 -> 2 -> 1, 2 Where: 1) f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1] 2) f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2 - 1] We have the following SCCs. { 2 }
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