details

property	value
status	complete
benchmark	java_EqUserDefRec.c.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n037.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	7.97187304497 seconds
cpu usage	7.75996103
max memory	2.7697152E7

stage attributes

key	value
output-size	8352
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 = f7#(x1, x2, x3, x4, x5, x6) -> f6#(x1, x2, x3, x4, x5, x6) f6#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) f6#(I6, I7, I8, I9, I10, I11) -> f5#(I6, I7, I8, I9, I10, I11) f6#(I12, I13, I14, I15, I16, I17) -> f4#(I12, I13, I14, I15, I16, I17) f6#(I18, I19, I20, I21, I22, I23) -> f1#(I18, I19, I20, I21, I22, I23) f3#(I30, I31, I32, I33, I34, I35) -> f5#(I34, I35, I32, I33, I34, I35) f5#(I36, I37, I38, I39, I40, I41) -> f4#(I40, I41, I38, I39, I40, I41) [I40 <= 0] f5#(I42, I43, I44, I45, I46, I47) -> f4#(I46, I47, I44, I45, I46, I47) [I47 <= 0] f5#(I48, I49, I50, I51, I52, I53) -> f1#(I52, I53, I50, I51, I52, I53) [1 <= I53 /\ 1 <= I52] f1#(I100, I101, I102, I103, I104, I105) -> f3#(I104, I105, I102, I103, -1 + I104, -1 + I105) R = f7(x1, x2, x3, x4, x5, x6) -> f6(x1, x2, x3, x4, x5, x6) f6(I0, I1, I2, I3, I4, I5) -> f3(I0, I1, I2, I3, I4, I5) f6(I6, I7, I8, I9, I10, I11) -> f5(I6, I7, I8, I9, I10, I11) f6(I12, I13, I14, I15, I16, I17) -> f4(I12, I13, I14, I15, I16, I17) f6(I18, I19, I20, I21, I22, I23) -> f1(I18, I19, I20, I21, I22, I23) f6(I24, I25, I26, I27, I28, I29) -> f2(I24, I25, I26, I27, I28, I29) f3(I30, I31, I32, I33, I34, I35) -> f5(I34, I35, I32, I33, I34, I35) f5(I36, I37, I38, I39, I40, I41) -> f4(I40, I41, I38, I39, I40, I41) [I40 <= 0] f5(I42, I43, I44, I45, I46, I47) -> f4(I46, I47, I44, I45, I46, I47) [I47 <= 0] f5(I48, I49, I50, I51, I52, I53) -> f1(I52, I53, I50, I51, I52, I53) [1 <= I53 /\ 1 <= I52] f4(I54, I55, I56, I57, I58, I59) -> f2(I58, I59, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ 1 + I58 <= 0 /\ rnd4 = rnd4 /\ rnd3 = rnd3] f4(I60, I61, I62, I63, I64, I65) -> f2(I64, I65, I66, I67, I68, I69) [I69 = I67 /\ I68 = I66 /\ 1 <= I64 /\ I67 = I67 /\ I66 = I66] f4(I70, I71, I72, I73, I74, I75) -> f2(I74, I75, I76, I77, I78, I79) [I79 = I77 /\ I78 = I76 /\ 1 + I75 <= 0 /\ I77 = I77 /\ I76 = I76] f4(I80, I81, I82, I83, I84, I85) -> f2(I84, I85, I86, I87, I88, I89) [I89 = I87 /\ I88 = I86 /\ 1 <= I85 /\ I87 = I87 /\ I86 = I86] f4(I90, I91, I92, I93, I94, I95) -> f2(I94, I95, I96, I97, I98, I99) [I99 = I97 /\ I98 = I96 /\ 0 <= I95 /\ I95 <= 0 /\ 0 <= I94 /\ I94 <= 0 /\ I97 = I97 /\ I96 = I96] f1(I100, I101, I102, I103, I104, I105) -> f3(I104, I105, I102, I103, -1 + I104, -1 + I105) f1(I106, I107, I108, I109, I110, I111) -> f2(I110, I111, I112, I113, I114, I115) [I115 = I113 /\ I114 = I112 /\ I113 = I113 /\ I112 = I112] The dependency graph for this problem is: 0 -> 1, 2, 3, 4 1 -> 5 2 -> 6, 7, 8 3 -> 4 -> 9 5 -> 6, 7, 8 6 -> 7 -> 8 -> 9 9 -> 5 Where: 0) f7#(x1, x2, x3, x4, x5, x6) -> f6#(x1, x2, x3, x4, x5, x6) 1) f6#(I0, I1, I2, I3, I4, I5) -> f3#(I0, I1, I2, I3, I4, I5) 2) f6#(I6, I7, I8, I9, I10, I11) -> f5#(I6, I7, I8, I9, I10, I11) 3) f6#(I12, I13, I14, I15, I16, I17) -> f4#(I12, I13, I14, I15, I16, I17) 4) f6#(I18, I19, I20, I21, I22, I23) -> f1#(I18, I19, I20, I21, I22, I23) 5) f3#(I30, I31, I32, I33, I34, I35) -> f5#(I34, I35, I32, I33, I34, I35) 6) f5#(I36, I37, I38, I39, I40, I41) -> f4#(I40, I41, I38, I39, I40, I41) [I40 <= 0] 7) f5#(I42, I43, I44, I45, I46, I47) -> f4#(I46, I47, I44, I45, I46, I47) [I47 <= 0] 8) f5#(I48, I49, I50, I51, I52, I53) -> f1#(I52, I53, I50, I51, I52, I53) [1 <= I53 /\ 1 <= I52] 9) f1#(I100, I101, I102, I103, I104, I105) -> f3#(I104, I105, I102, I103, -1 + I104, -1 + I105) We have the following SCCs. { 5, 8, 9 } DP problem for innermost termination. P = f3#(I30, I31, I32, I33, I34, I35) -> f5#(I34, I35, I32, I33, I34, I35) f5#(I48, I49, I50, I51, I52, I53) -> f1#(I52, I53, I50, I51, I52, I53) [1 <= I53 /\ 1 <= I52] f1#(I100, I101, I102, I103, I104, I105) -> f3#(I104, I105, I102, I103, -1 + I104, -1 + I105) R = f7(x1, x2, x3, x4, x5, x6) -> f6(x1, x2, x3, x4, x5, x6) f6(I0, I1, I2, I3, I4, I5) -> f3(I0, I1, I2, I3, I4, I5) f6(I6, I7, I8, I9, I10, I11) -> f5(I6, I7, I8, I9, I10, I11) f6(I12, I13, I14, I15, I16, I17) -> f4(I12, I13, I14, I15, I16, I17) f6(I18, I19, I20, I21, I22, I23) -> f1(I18, I19, I20, I21, I22, I23) f6(I24, I25, I26, I27, I28, I29) -> f2(I24, I25, I26, I27, I28, I29) f3(I30, I31, I32, I33, I34, I35) -> f5(I34, I35, I32, I33, I34, I35) f5(I36, I37, I38, I39, I40, I41) -> f4(I40, I41, I38, I39, I40, I41) [I40 <= 0] f5(I42, I43, I44, I45, I46, I47) -> f4(I46, I47, I44, I45, I46, I47) [I47 <= 0] f5(I48, I49, I50, I51, I52, I53) -> f1(I52, I53, I50, I51, I52, I53) [1 <= I53 /\ 1 <= I52] f4(I54, I55, I56, I57, I58, I59) -> f2(I58, I59, rnd3, rnd4, rnd5, rnd6) [rnd6 = rnd4 /\ rnd5 = rnd3 /\ 1 + I58 <= 0 /\ rnd4 = rnd4 /\ rnd3 = rnd3] f4(I60, I61, I62, I63, I64, I65) -> f2(I64, I65, I66, I67, I68, I69) [I69 = I67 /\ I68 = I66 /\ 1 <= I64 /\ I67 = I67 /\ I66 = I66] f4(I70, I71, I72, I73, I74, I75) -> f2(I74, I75, I76, I77, I78, I79) [I79 = I77 /\ I78 = I76 /\ 1 + I75 <= 0 /\ I77 = I77 /\ I76 = I76] f4(I80, I81, I82, I83, I84, I85) -> f2(I84, I85, I86, I87, I88, I89) [I89 = I87 /\ I88 = I86 /\ 1 <= I85 /\ I87 = I87 /\ I86 = I86] f4(I90, I91, I92, I93, I94, I95) -> f2(I94, I95, I96, I97, I98, I99) [I99 = I97 /\ I98 = I96 /\ 0 <= I95 /\ I95 <= 0 /\ 0 <= I94 /\ I94 <= 0 /\ I97 = I97 /\ I96 = I96] f1(I100, I101, I102, I103, I104, I105) -> f3(I104, I105, I102, I103, -1 + I104, -1 + I105) f1(I106, I107, I108, I109, I110, I111) -> f2(I110, I111, I112, I113, I114, I115) [I115 = I113 /\ I114 = I112 /\ I113 = I113 /\ I112 = I112] We use the extended value criterion with the projection function NU: NU[f1#(x0,x1,x2,x3,x4,x5)] = x5 NU[f5#(x0,x1,x2,x3,x4,x5)] = x5 + 1 NU[f3#(x0,x1,x2,x3,x4,x5)] = x5 + 1 This gives the following inequalities: ==> I35 + 1 >= I35 + 1 1 <= I53 /\ 1 <= I52 ==> I53 + 1 > I53 with I53 + 1 >= 0 ==> I105 >= (-1 + I105) + 1 We remove all the strictly oriented dependency pairs. popout

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job log

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actions

all output return to Integ Trans Syste 27634