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	n146.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	4.11069 seconds
cpu usage	4.16792
user time	2.24094
system time	1.92698
max virtual memory	693348.0
max residence set size	8768.0

stage attributes

key	value
starexec-result	YES

output

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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job log

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actions

all output return to ITS