details

property	value
status	complete
benchmark	ex23.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n149.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	4.20987391472 seconds
cpu usage	4.488510521
max memory	2.1700608E7

stage attributes

key	value
output-size	3429
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 = f6#(x1, x2, x3, x4, x5) -> f5#(x1, x2, x3, x4, x5) f5#(I0, I1, I2, I3, I4) -> f4#(I0, I1, 0, I3, I4) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f4#(I15, I16, I17, I18, I19) -> f3#(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ I18 <= I15] f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, 1 + I22, I23, 1 + I24) [1 + I22 <= I21] R = f6(x1, x2, x3, x4, x5) -> f5(x1, x2, x3, x4, x5) f5(I0, I1, I2, I3, I4) -> f4(I0, I1, 0, I3, I4) f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f4(I10, I11, I12, I13, I14) -> f2(I10, I11, I12, I13, I14) [1 + I10 <= I13] f4(I15, I16, I17, I18, I19) -> f3(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ I18 <= I15] f1(I20, I21, I22, I23, I24) -> f3(I20, I21, 1 + I22, I23, 1 + I24) [1 + I22 <= I21] f1(I25, I26, I27, I28, I29) -> f2(I25, I26, I27, I28, I29) [I26 <= I27] The dependency graph for this problem is: 0 -> 1 1 -> 3 2 -> 4 3 -> 2 4 -> 2 Where: 0) f6#(x1, x2, x3, x4, x5) -> f5#(x1, x2, x3, x4, x5) 1) f5#(I0, I1, I2, I3, I4) -> f4#(I0, I1, 0, I3, I4) 2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 3) f4#(I15, I16, I17, I18, I19) -> f3#(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ I18 <= I15] 4) f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, 1 + I22, I23, 1 + I24) [1 + I22 <= I21] We have the following SCCs. { 2, 4 } DP problem for innermost termination. P = f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, 1 + I22, I23, 1 + I24) [1 + I22 <= I21] R = f6(x1, x2, x3, x4, x5) -> f5(x1, x2, x3, x4, x5) f5(I0, I1, I2, I3, I4) -> f4(I0, I1, 0, I3, I4) f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f4(I10, I11, I12, I13, I14) -> f2(I10, I11, I12, I13, I14) [1 + I10 <= I13] f4(I15, I16, I17, I18, I19) -> f3(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ I18 <= I15] f1(I20, I21, I22, I23, I24) -> f3(I20, I21, 1 + I22, I23, 1 + I24) [1 + I22 <= I21] f1(I25, I26, I27, I28, I29) -> f2(I25, I26, I27, I28, I29) [I26 <= I27] We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5)] = z2 + -1 * (1 + z3) NU[f3#(z1,z2,z3,z4,z5)] = z2 + -1 * (1 + z3) This gives the following inequalities: ==> I6 + -1 * (1 + I7) >= I6 + -1 * (1 + I7) 1 + I22 <= I21 ==> I21 + -1 * (1 + I22) > I21 + -1 * (1 + (1 + I22)) with I21 + -1 * (1 + I22) >= 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 = f6(x1, x2, x3, x4, x5) -> f5(x1, x2, x3, x4, x5) f5(I0, I1, I2, I3, I4) -> f4(I0, I1, 0, I3, I4) f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f4(I10, I11, I12, I13, I14) -> f2(I10, I11, I12, I13, I14) [1 + I10 <= I13] f4(I15, I16, I17, I18, I19) -> f3(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ I18 <= I15] f1(I20, I21, I22, I23, I24) -> f3(I20, I21, 1 + I22, I23, 1 + I24) [1 + I22 <= I21] f1(I25, I26, I27, I28, I29) -> f2(I25, I26, I27, I28, I29) [I26 <= I27] 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. popout

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