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Integ Trans Syste 27634 pair #381739340
details
property
value
status
complete
benchmark
PastaB10.jar-obl-8.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n090.star.cs.uiowa.edu
space
From_AProVE_2014
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.348289012909 seconds
cpu usage
0.362995683
max memory
9994240.0
stage attributes
key
value
output-size
3373
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, x3) -> f1#(rnd1, rnd2, rnd3) f2#(I0, I1, I2) -> f2#(0, I1 - 1, I1 - 1) [0 = I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1] f2#(I3, I4, I5) -> f2#(0, 0, 0) [0 = I4 /\ 0 = I3 /\ 0 <= I5 - 1] f2#(I6, I7, I8) -> f2#(I6 - 1, I7, I6 - 1 + I7) [-1 <= I7 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1] f1#(I9, I10, I11) -> f2#(I12, I13, I14) [I12 + I13 = I14 /\ 0 <= I9 - 1 /\ -1 <= I13 - 1 /\ -1 <= I10 - 1 /\ -1 <= I12 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f2(I0, I1, I2) -> f2(0, I1 - 1, I1 - 1) [0 = I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1] f2(I3, I4, I5) -> f2(0, 0, 0) [0 = I4 /\ 0 = I3 /\ 0 <= I5 - 1] f2(I6, I7, I8) -> f2(I6 - 1, I7, I6 - 1 + I7) [-1 <= I7 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1] f1(I9, I10, I11) -> f2(I12, I13, I14) [I12 + I13 = I14 /\ 0 <= I9 - 1 /\ -1 <= I13 - 1 /\ -1 <= I10 - 1 /\ -1 <= I12 - 1] The dependency graph for this problem is: 0 -> 4 1 -> 1 2 -> 3 -> 1, 3 4 -> 1, 3 Where: 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 1) f2#(I0, I1, I2) -> f2#(0, I1 - 1, I1 - 1) [0 = I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1] 2) f2#(I3, I4, I5) -> f2#(0, 0, 0) [0 = I4 /\ 0 = I3 /\ 0 <= I5 - 1] 3) f2#(I6, I7, I8) -> f2#(I6 - 1, I7, I6 - 1 + I7) [-1 <= I7 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1] 4) f1#(I9, I10, I11) -> f2#(I12, I13, I14) [I12 + I13 = I14 /\ 0 <= I9 - 1 /\ -1 <= I13 - 1 /\ -1 <= I10 - 1 /\ -1 <= I12 - 1] We have the following SCCs. { 3 } { 1 } DP problem for innermost termination. P = f2#(I0, I1, I2) -> f2#(0, I1 - 1, I1 - 1) [0 = I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f2(I0, I1, I2) -> f2(0, I1 - 1, I1 - 1) [0 = I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1] f2(I3, I4, I5) -> f2(0, 0, 0) [0 = I4 /\ 0 = I3 /\ 0 <= I5 - 1] f2(I6, I7, I8) -> f2(I6 - 1, I7, I6 - 1 + I7) [-1 <= I7 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1] f1(I9, I10, I11) -> f2(I12, I13, I14) [I12 + I13 = I14 /\ 0 <= I9 - 1 /\ -1 <= I13 - 1 /\ -1 <= I10 - 1 /\ -1 <= I12 - 1] We use the basic value criterion with the projection function NU: NU[f2#(z1,z2,z3)] = z2 This gives the following inequalities: 0 = I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 ==> I1 >! I1 - 1 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f2#(I6, I7, I8) -> f2#(I6 - 1, I7, I6 - 1 + I7) [-1 <= I7 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1] R = init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) f2(I0, I1, I2) -> f2(0, I1 - 1, I1 - 1) [0 = I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1] f2(I3, I4, I5) -> f2(0, 0, 0) [0 = I4 /\ 0 = I3 /\ 0 <= I5 - 1] f2(I6, I7, I8) -> f2(I6 - 1, I7, I6 - 1 + I7) [-1 <= I7 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1] f1(I9, I10, I11) -> f2(I12, I13, I14) [I12 + I13 = I14 /\ 0 <= I9 - 1 /\ -1 <= I13 - 1 /\ -1 <= I10 - 1 /\ -1 <= I12 - 1] We use the basic value criterion with the projection function NU: NU[f2#(z1,z2,z3)] = z1 This gives the following inequalities: -1 <= I7 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1 ==> I6 >! I6 - 1 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
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