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Integ Trans Syste 27634 pair #381738728
details
property
value
status
complete
benchmark
PastaA10.jar-obl-8.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n026.star.cs.uiowa.edu
space
From_AProVE_2014
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.461975097656 seconds
cpu usage
0.482327848
max memory
1.0371072E7
stage attributes
key
value
output-size
2396
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) f2#(I0, I1) -> f2#(I0, I1 + 1) [I1 <= I0 - 1] f2#(I2, I3) -> f2#(I2 + 1, I3) [I2 <= I3 - 1] f1#(I4, I5) -> f2#(I6, I7) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ -1 <= I5 - 1 /\ -1 <= I7 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(I0, I1 + 1) [I1 <= I0 - 1] f2(I2, I3) -> f2(I2 + 1, I3) [I2 <= I3 - 1] f1(I4, I5) -> f2(I6, I7) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ -1 <= I5 - 1 /\ -1 <= I7 - 1] The dependency graph for this problem is: 0 -> 3 1 -> 1 2 -> 2 3 -> 1, 2 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f2#(I0, I1) -> f2#(I0, I1 + 1) [I1 <= I0 - 1] 2) f2#(I2, I3) -> f2#(I2 + 1, I3) [I2 <= I3 - 1] 3) f1#(I4, I5) -> f2#(I6, I7) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ -1 <= I5 - 1 /\ -1 <= I7 - 1] We have the following SCCs. { 2 } { 1 } DP problem for innermost termination. P = f2#(I0, I1) -> f2#(I0, I1 + 1) [I1 <= I0 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(I0, I1 + 1) [I1 <= I0 - 1] f2(I2, I3) -> f2(I2 + 1, I3) [I2 <= I3 - 1] f1(I4, I5) -> f2(I6, I7) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ -1 <= I5 - 1 /\ -1 <= I7 - 1] We use the reverse value criterion with the projection function NU: NU[f2#(z1,z2)] = z1 - 1 + -1 * z2 This gives the following inequalities: I1 <= I0 - 1 ==> I0 - 1 + -1 * I1 > I0 - 1 + -1 * (I1 + 1) with I0 - 1 + -1 * I1 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f2#(I2, I3) -> f2#(I2 + 1, I3) [I2 <= I3 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(I0, I1 + 1) [I1 <= I0 - 1] f2(I2, I3) -> f2(I2 + 1, I3) [I2 <= I3 - 1] f1(I4, I5) -> f2(I6, I7) [0 <= I4 - 1 /\ -1 <= I6 - 1 /\ -1 <= I5 - 1 /\ -1 <= I7 - 1] We use the reverse value criterion with the projection function NU: NU[f2#(z1,z2)] = z2 - 1 + -1 * z1 This gives the following inequalities: I2 <= I3 - 1 ==> I3 - 1 + -1 * I2 > I3 - 1 + -1 * (I2 + 1) with I3 - 1 + -1 * I2 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
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