Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
Integ Trans Syste 27634 pair #381738236
details
property
value
status
complete
benchmark
GCD2.jar-obl-8.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n073.star.cs.uiowa.edu
space
From_AProVE_2014
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.564914941788 seconds
cpu usage
0.590475626
max memory
1.110016E7
stage attributes
key
value
output-size
4645
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, I0) [I0 <= I1 - 1] f2#(I2, I3) -> f2#(I2, 0) [I2 = I3 /\ 0 <= I2 - 1] f3#(I4, I5) -> f3#(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1] f2#(I6, I7) -> f3#(I6, I7) [0 <= I7 - 1 /\ -1 <= I6 - 1 /\ I7 <= I6 - 1] f2#(I8, I9) -> f3#(I8, I9) [0 <= I9 - 1 /\ -1 <= I8 - 1 /\ I8 <= I9 - 1] f1#(I10, I11) -> f2#(I12, I13) [0 <= I10 - 1 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1] f2(I2, I3) -> f2(I2, 0) [I2 = I3 /\ 0 <= I2 - 1] f3(I4, I5) -> f3(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1] f2(I6, I7) -> f3(I6, I7) [0 <= I7 - 1 /\ -1 <= I6 - 1 /\ I7 <= I6 - 1] f2(I8, I9) -> f3(I8, I9) [0 <= I9 - 1 /\ -1 <= I8 - 1 /\ I8 <= I9 - 1] f1(I10, I11) -> f2(I12, I13) [0 <= I10 - 1 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1] The dependency graph for this problem is: 0 -> 6 1 -> 4 2 -> 3 -> 1, 3 4 -> 3 5 -> 1 6 -> 2, 4, 5 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1] 2) f2#(I2, I3) -> f2#(I2, 0) [I2 = I3 /\ 0 <= I2 - 1] 3) f3#(I4, I5) -> f3#(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1] 4) f2#(I6, I7) -> f3#(I6, I7) [0 <= I7 - 1 /\ -1 <= I6 - 1 /\ I7 <= I6 - 1] 5) f2#(I8, I9) -> f3#(I8, I9) [0 <= I9 - 1 /\ -1 <= I8 - 1 /\ I8 <= I9 - 1] 6) f1#(I10, I11) -> f2#(I12, I13) [0 <= I10 - 1 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1] We have the following SCCs. { 1, 3, 4 } DP problem for innermost termination. P = f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1] f3#(I4, I5) -> f3#(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1] f2#(I6, I7) -> f3#(I6, I7) [0 <= I7 - 1 /\ -1 <= I6 - 1 /\ I7 <= I6 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1] f2(I2, I3) -> f2(I2, 0) [I2 = I3 /\ 0 <= I2 - 1] f3(I4, I5) -> f3(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1] f2(I6, I7) -> f3(I6, I7) [0 <= I7 - 1 /\ -1 <= I6 - 1 /\ I7 <= I6 - 1] f2(I8, I9) -> f3(I8, I9) [0 <= I9 - 1 /\ -1 <= I8 - 1 /\ I8 <= I9 - 1] f1(I10, I11) -> f2(I12, I13) [0 <= I10 - 1 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 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 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1 ==> I5 (>! \union =) I5 0 <= I7 - 1 /\ -1 <= I6 - 1 /\ I7 <= I6 - 1 ==> I6 >! I7 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1] f3#(I4, I5) -> f3#(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1] f2(I2, I3) -> f2(I2, 0) [I2 = I3 /\ 0 <= I2 - 1] f3(I4, I5) -> f3(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1] f2(I6, I7) -> f3(I6, I7) [0 <= I7 - 1 /\ -1 <= I6 - 1 /\ I7 <= I6 - 1] f2(I8, I9) -> f3(I8, I9) [0 <= I9 - 1 /\ -1 <= I8 - 1 /\ I8 <= I9 - 1] f1(I10, I11) -> f2(I12, I13) [0 <= I10 - 1 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1] The dependency graph for this problem is: 1 -> 3 -> 1, 3 Where: 1) f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1] 3) f3#(I4, I5) -> f3#(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1] We have the following SCCs. { 3 } DP problem for innermost termination. P = f3#(I4, I5) -> f3#(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1]
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to Integ Trans Syste 27634