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Integ Trans Syste 27634 pair #381738871
details
property
value
status
complete
benchmark
ex7.t2_fixed.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n012.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
3.59999585152 seconds
cpu usage
3.780998714
max memory
2.3437312E7
stage attributes
key
value
output-size
2979
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 = f5#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) f4#(I0, I1, I2, I3, I4) -> f3#(0, 10, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4] f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f1#(I10, I11, I12, I13, I14) -> f3#(1 + I10, I11, I12, I13, rnd5) [rnd5 = rnd5 /\ 1 + I10 <= I11] R = f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) f4(I0, I1, I2, I3, I4) -> f3(0, 10, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4] f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f1(I10, I11, I12, I13, I14) -> f3(1 + I10, I11, I12, I13, rnd5) [rnd5 = rnd5 /\ 1 + I10 <= I11] f1(I15, I16, I17, I18, I19) -> f2(I15, I16, I17, I18, I19) [I16 <= I15] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 3 3 -> 2 Where: 0) f5#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 1) f4#(I0, I1, I2, I3, I4) -> f3#(0, 10, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4] 2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 3) f1#(I10, I11, I12, I13, I14) -> f3#(1 + I10, I11, I12, I13, rnd5) [rnd5 = rnd5 /\ 1 + I10 <= I11] We have the following SCCs. { 2, 3 } DP problem for innermost termination. P = f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f1#(I10, I11, I12, I13, I14) -> f3#(1 + I10, I11, I12, I13, rnd5) [rnd5 = rnd5 /\ 1 + I10 <= I11] R = f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) f4(I0, I1, I2, I3, I4) -> f3(0, 10, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4] f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f1(I10, I11, I12, I13, I14) -> f3(1 + I10, I11, I12, I13, rnd5) [rnd5 = rnd5 /\ 1 + I10 <= I11] f1(I15, I16, I17, I18, I19) -> f2(I15, I16, I17, I18, I19) [I16 <= I15] We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5)] = z2 + -1 * (1 + z1) NU[f3#(z1,z2,z3,z4,z5)] = z2 + -1 * (1 + z1) This gives the following inequalities: ==> I6 + -1 * (1 + I5) >= I6 + -1 * (1 + I5) rnd5 = rnd5 /\ 1 + I10 <= I11 ==> I11 + -1 * (1 + I10) > I11 + -1 * (1 + (1 + I10)) with I11 + -1 * (1 + I10) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) R = f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) f4(I0, I1, I2, I3, I4) -> f3(0, 10, rnd3, rnd4, I4) [rnd3 = rnd4 /\ rnd4 = rnd4] f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f1(I10, I11, I12, I13, I14) -> f3(1 + I10, I11, I12, I13, rnd5) [rnd5 = rnd5 /\ 1 + I10 <= I11] f1(I15, I16, I17, I18, I19) -> f2(I15, I16, I17, I18, I19) [I16 <= I15] The dependency graph for this problem is: 2 -> Where: 2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) We have the following SCCs.
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