details

property	value
status	complete
benchmark	byron-2.t2_fixed.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n108.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	29.0419747829 seconds
cpu usage	28.290376024
max memory	2.5649152E7

stage attributes

key	value
output-size	11776
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 = f8#(x1, x2, x3, x4, x5, x6, x7) -> f1#(x1, x2, x3, x4, x5, x6, x7) f7#(I0, I1, I2, I3, I4, I5, I6) -> f2#(I0, I1, I2, I3, I4, I5, I6) f6#(I7, I8, I9, I10, I11, I12, I13) -> f7#(rnd1, I8, I9, I10, I11, -1 + I12, rnd7) [y1 = y1 /\ rnd7 = y1 /\ rnd1 = rnd1] f5#(I14, I15, I16, I17, I18, I19, I20) -> f6#(I14, I15, I16, I17, I18, I19, I20) [1 + I16 <= 0] f5#(I21, I22, I23, I24, I25, I26, I27) -> f6#(I21, I22, I23, I24, I25, I26, I27) [1 <= I23] f2#(I28, I29, I30, I31, I32, I33, I34) -> f5#(I35, I29, rnd3, I31, I32, I33, I34) [1 <= I33 /\ I36 = I36 /\ rnd3 = I36 /\ I35 = I35] f4#(I37, I38, I39, I40, I41, I42, I43) -> f2#(I37, I38, I39, I40, I41, I42, I43) f2#(I44, I45, I46, I47, I48, I49, I50) -> f4#(I51, I45, I52, I47, rnd5, I49, -1 + I50) [1 <= I49 /\ I53 = I53 /\ I52 = I53 /\ I51 = I51 /\ 0 <= I52 /\ I52 <= 0 /\ rnd5 = rnd5 /\ 2 <= -1 + I50] f1#(I61, I62, I63, I64, I65, I66, I67) -> f2#(I61, I62, I63, I64, I65, I66, I67) R = f8(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) f7(I0, I1, I2, I3, I4, I5, I6) -> f2(I0, I1, I2, I3, I4, I5, I6) f6(I7, I8, I9, I10, I11, I12, I13) -> f7(rnd1, I8, I9, I10, I11, -1 + I12, rnd7) [y1 = y1 /\ rnd7 = y1 /\ rnd1 = rnd1] f5(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, I15, I16, I17, I18, I19, I20) [1 + I16 <= 0] f5(I21, I22, I23, I24, I25, I26, I27) -> f6(I21, I22, I23, I24, I25, I26, I27) [1 <= I23] f2(I28, I29, I30, I31, I32, I33, I34) -> f5(I35, I29, rnd3, I31, I32, I33, I34) [1 <= I33 /\ I36 = I36 /\ rnd3 = I36 /\ I35 = I35] f4(I37, I38, I39, I40, I41, I42, I43) -> f2(I37, I38, I39, I40, I41, I42, I43) f2(I44, I45, I46, I47, I48, I49, I50) -> f4(I51, I45, I52, I47, rnd5, I49, -1 + I50) [1 <= I49 /\ I53 = I53 /\ I52 = I53 /\ I51 = I51 /\ 0 <= I52 /\ I52 <= 0 /\ rnd5 = rnd5 /\ 2 <= -1 + I50] f2(I54, I55, I56, I57, I58, I59, I60) -> f3(I54, I57, I56, I57, I58, I59, I60) [I59 <= 0] f1(I61, I62, I63, I64, I65, I66, I67) -> f2(I61, I62, I63, I64, I65, I66, I67) The dependency graph for this problem is: 0 -> 8 1 -> 5, 7 2 -> 1 3 -> 2 4 -> 2 5 -> 3, 4 6 -> 5, 7 7 -> 6 8 -> 5, 7 Where: 0) f8#(x1, x2, x3, x4, x5, x6, x7) -> f1#(x1, x2, x3, x4, x5, x6, x7) 1) f7#(I0, I1, I2, I3, I4, I5, I6) -> f2#(I0, I1, I2, I3, I4, I5, I6) 2) f6#(I7, I8, I9, I10, I11, I12, I13) -> f7#(rnd1, I8, I9, I10, I11, -1 + I12, rnd7) [y1 = y1 /\ rnd7 = y1 /\ rnd1 = rnd1] 3) f5#(I14, I15, I16, I17, I18, I19, I20) -> f6#(I14, I15, I16, I17, I18, I19, I20) [1 + I16 <= 0] 4) f5#(I21, I22, I23, I24, I25, I26, I27) -> f6#(I21, I22, I23, I24, I25, I26, I27) [1 <= I23] 5) f2#(I28, I29, I30, I31, I32, I33, I34) -> f5#(I35, I29, rnd3, I31, I32, I33, I34) [1 <= I33 /\ I36 = I36 /\ rnd3 = I36 /\ I35 = I35] 6) f4#(I37, I38, I39, I40, I41, I42, I43) -> f2#(I37, I38, I39, I40, I41, I42, I43) 7) f2#(I44, I45, I46, I47, I48, I49, I50) -> f4#(I51, I45, I52, I47, rnd5, I49, -1 + I50) [1 <= I49 /\ I53 = I53 /\ I52 = I53 /\ I51 = I51 /\ 0 <= I52 /\ I52 <= 0 /\ rnd5 = rnd5 /\ 2 <= -1 + I50] 8) f1#(I61, I62, I63, I64, I65, I66, I67) -> f2#(I61, I62, I63, I64, I65, I66, I67) We have the following SCCs. { 1, 2, 3, 4, 5, 6, 7 } DP problem for innermost termination. P = f7#(I0, I1, I2, I3, I4, I5, I6) -> f2#(I0, I1, I2, I3, I4, I5, I6) f6#(I7, I8, I9, I10, I11, I12, I13) -> f7#(rnd1, I8, I9, I10, I11, -1 + I12, rnd7) [y1 = y1 /\ rnd7 = y1 /\ rnd1 = rnd1] f5#(I14, I15, I16, I17, I18, I19, I20) -> f6#(I14, I15, I16, I17, I18, I19, I20) [1 + I16 <= 0] f5#(I21, I22, I23, I24, I25, I26, I27) -> f6#(I21, I22, I23, I24, I25, I26, I27) [1 <= I23] f2#(I28, I29, I30, I31, I32, I33, I34) -> f5#(I35, I29, rnd3, I31, I32, I33, I34) [1 <= I33 /\ I36 = I36 /\ rnd3 = I36 /\ I35 = I35] f4#(I37, I38, I39, I40, I41, I42, I43) -> f2#(I37, I38, I39, I40, I41, I42, I43) f2#(I44, I45, I46, I47, I48, I49, I50) -> f4#(I51, I45, I52, I47, rnd5, I49, -1 + I50) [1 <= I49 /\ I53 = I53 /\ I52 = I53 /\ I51 = I51 /\ 0 <= I52 /\ I52 <= 0 /\ rnd5 = rnd5 /\ 2 <= -1 + I50] R = f8(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) f7(I0, I1, I2, I3, I4, I5, I6) -> f2(I0, I1, I2, I3, I4, I5, I6) f6(I7, I8, I9, I10, I11, I12, I13) -> f7(rnd1, I8, I9, I10, I11, -1 + I12, rnd7) [y1 = y1 /\ rnd7 = y1 /\ rnd1 = rnd1] f5(I14, I15, I16, I17, I18, I19, I20) -> f6(I14, I15, I16, I17, I18, I19, I20) [1 + I16 <= 0] f5(I21, I22, I23, I24, I25, I26, I27) -> f6(I21, I22, I23, I24, I25, I26, I27) [1 <= I23] f2(I28, I29, I30, I31, I32, I33, I34) -> f5(I35, I29, rnd3, I31, I32, I33, I34) [1 <= I33 /\ I36 = I36 /\ rnd3 = I36 /\ I35 = I35] f4(I37, I38, I39, I40, I41, I42, I43) -> f2(I37, I38, I39, I40, I41, I42, I43) f2(I44, I45, I46, I47, I48, I49, I50) -> f4(I51, I45, I52, I47, rnd5, I49, -1 + I50) [1 <= I49 /\ I53 = I53 /\ I52 = I53 /\ I51 = I51 /\ 0 <= I52 /\ I52 <= 0 /\ rnd5 = rnd5 /\ 2 <= -1 + I50] f2(I54, I55, I56, I57, I58, I59, I60) -> f3(I54, I57, I56, I57, I58, I59, I60) [I59 <= 0] f1(I61, I62, I63, I64, I65, I66, I67) -> f2(I61, I62, I63, I64, I65, I66, I67) We use the extended value criterion with the projection function NU: NU[f4#(x0,x1,x2,x3,x4,x5,x6)] = x5 - 1 NU[f5#(x0,x1,x2,x3,x4,x5,x6)] = x5 - 2 NU[f6#(x0,x1,x2,x3,x4,x5,x6)] = x5 - 2 NU[f2#(x0,x1,x2,x3,x4,x5,x6)] = x5 - 1 NU[f7#(x0,x1,x2,x3,x4,x5,x6)] = x5 - 1 This gives the following inequalities: ==> I5 - 1 >= I5 - 1 y1 = y1 /\ rnd7 = y1 /\ rnd1 = rnd1 ==> I12 - 2 >= (-1 + I12) - 1 1 + I16 <= 0 ==> I19 - 2 >= I19 - 2 1 <= I23 ==> I26 - 2 >= I26 - 2 1 <= I33 /\ I36 = I36 /\ rnd3 = I36 /\ I35 = I35 ==> I33 - 1 > I33 - 2 with I33 - 1 >= 0 ==> I42 - 1 >= I42 - 1 1 <= I49 /\ I53 = I53 /\ I52 = I53 /\ I51 = I51 /\ 0 <= I52 /\ I52 <= 0 /\ rnd5 = rnd5 /\ 2 <= -1 + I50 ==> I49 - 1 >= I49 - 1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f7#(I0, I1, I2, I3, I4, I5, I6) -> f2#(I0, I1, I2, I3, I4, I5, I6) f6#(I7, I8, I9, I10, I11, I12, I13) -> f7#(rnd1, I8, I9, I10, I11, -1 + I12, rnd7) [y1 = y1 /\ rnd7 = y1 /\ rnd1 = rnd1] f5#(I14, I15, I16, I17, I18, I19, I20) -> f6#(I14, I15, I16, I17, I18, I19, I20) [1 + I16 <= 0] f5#(I21, I22, I23, I24, I25, I26, I27) -> f6#(I21, I22, I23, I24, I25, I26, I27) [1 <= I23] popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to Integ Trans Syste 27634