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Integer Transition Systems pair #487528689
details
property
value
status
complete
benchmark
LeUserDefRec.jar-obl-8.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n140.star.cs.uiowa.edu
space
From_AProVE_2014
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.127876996994 seconds
cpu usage
0.124471317
max memory
8507392.0
stage attributes
key
value
output-size
1804
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 - 1, I1 - 1) [I1 - 1 <= I1 - 1 /\ I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1 /\ 0 <= I1 - 1] f1#(I2, I3) -> f2#(I4, I5) [0 <= I2 - 1 /\ -1 <= I5 - 1 /\ 1 <= I3 - 1 /\ -1 <= I4 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(I0 - 1, I1 - 1) [I1 - 1 <= I1 - 1 /\ I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1 /\ 0 <= I1 - 1] f1(I2, I3) -> f2(I4, I5) [0 <= I2 - 1 /\ -1 <= I5 - 1 /\ 1 <= I3 - 1 /\ -1 <= I4 - 1] The dependency graph for this problem is: 0 -> 2 1 -> 1 2 -> 1 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f2#(I0, I1) -> f2#(I0 - 1, I1 - 1) [I1 - 1 <= I1 - 1 /\ I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1 /\ 0 <= I1 - 1] 2) f1#(I2, I3) -> f2#(I4, I5) [0 <= I2 - 1 /\ -1 <= I5 - 1 /\ 1 <= I3 - 1 /\ -1 <= I4 - 1] We have the following SCCs. { 1 } DP problem for innermost termination. P = f2#(I0, I1) -> f2#(I0 - 1, I1 - 1) [I1 - 1 <= I1 - 1 /\ I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1 /\ 0 <= I1 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f2(I0, I1) -> f2(I0 - 1, I1 - 1) [I1 - 1 <= I1 - 1 /\ I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1 /\ 0 <= I1 - 1] f1(I2, I3) -> f2(I4, I5) [0 <= I2 - 1 /\ -1 <= I5 - 1 /\ 1 <= I3 - 1 /\ -1 <= I4 - 1] We use the basic value criterion with the projection function NU: NU[f2#(z1,z2)] = z2 This gives the following inequalities: I1 - 1 <= I1 - 1 /\ I0 - 1 <= I0 - 1 /\ 0 <= I0 - 1 /\ 0 <= I1 - 1 ==> I1 >! I1 - 1 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
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