details

property	value
status	complete
benchmark	mc91test.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n145.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	78.1016 seconds
cpu usage	78.7069
user time	43.4114
system time	35.2955
max virtual memory	784368.0
max residence set size	24232.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = f8#(x1, x2, x3, x4, x5, x6, x7, x8) -> f7#(x1, x2, x3, x4, x5, x6, x7, x8) f7#(I0, I1, I2, I3, I4, I5, I6, I7) -> f1#(I0, I1, I2, 0, 1, rnd6, I6, I7) [rnd6 = rnd6] f6#(I8, I9, I10, I11, I12, I13, I14, I15) -> f1#(I8, I9, I10, I11, I12, I13, I14, I15) f1#(I16, I17, I18, I19, I20, I21, I22, I23) -> f6#(I16, I17, I18, I19, -1 + I20, -1 * I17 + I21, I22, I23) [1 + I16 <= I21 /\ 1 <= I20] f5#(I24, I25, I26, I27, I28, I29, I30, I31) -> f1#(I24, I25, I26, I27, I28, I29, I30, I31) f1#(I32, I33, I34, I35, I36, I37, I38, I39) -> f5#(I32, I33, I34, I35, 1 + I36, I34 + I37, I38, I39) [I37 <= I32 /\ 1 <= I36] f4#(I40, I41, I42, I43, I44, I45, I46, I47) -> f1#(I40, I41, I42, I43, I44, I45, I46, I47) f1#(I48, I49, I50, I51, I52, I53, I54, I55) -> f4#(I48, I49, I50, 1, -1 + I52, -1 * I49 + I53, I52, I53) [1 + I48 <= I53 /\ 1 <= I52 /\ I51 <= 0] f3#(I56, I57, I58, I59, I60, I61, I62, I63) -> f1#(I56, I57, I58, I59, I60, I61, I62, I63) f1#(I64, I65, I66, I67, I68, I69, I70, I71) -> f3#(I64, I65, I66, 1, 1 + I68, I66 + I69, I68, I69) [I69 <= I64 /\ 1 <= I68 /\ I67 <= 0] R = f8(x1, x2, x3, x4, x5, x6, x7, x8) -> f7(x1, x2, x3, x4, x5, x6, x7, x8) f7(I0, I1, I2, I3, I4, I5, I6, I7) -> f1(I0, I1, I2, 0, 1, rnd6, I6, I7) [rnd6 = rnd6] f6(I8, I9, I10, I11, I12, I13, I14, I15) -> f1(I8, I9, I10, I11, I12, I13, I14, I15) f1(I16, I17, I18, I19, I20, I21, I22, I23) -> f6(I16, I17, I18, I19, -1 + I20, -1 * I17 + I21, I22, I23) [1 + I16 <= I21 /\ 1 <= I20] f5(I24, I25, I26, I27, I28, I29, I30, I31) -> f1(I24, I25, I26, I27, I28, I29, I30, I31) f1(I32, I33, I34, I35, I36, I37, I38, I39) -> f5(I32, I33, I34, I35, 1 + I36, I34 + I37, I38, I39) [I37 <= I32 /\ 1 <= I36] f4(I40, I41, I42, I43, I44, I45, I46, I47) -> f1(I40, I41, I42, I43, I44, I45, I46, I47) f1(I48, I49, I50, I51, I52, I53, I54, I55) -> f4(I48, I49, I50, 1, -1 + I52, -1 * I49 + I53, I52, I53) [1 + I48 <= I53 /\ 1 <= I52 /\ I51 <= 0] f3(I56, I57, I58, I59, I60, I61, I62, I63) -> f1(I56, I57, I58, I59, I60, I61, I62, I63) f1(I64, I65, I66, I67, I68, I69, I70, I71) -> f3(I64, I65, I66, 1, 1 + I68, I66 + I69, I68, I69) [I69 <= I64 /\ 1 <= I68 /\ I67 <= 0] f1(I72, I73, I74, I75, I76, I77, I78, I79) -> f2(I72, I73, I74, I75, I76, I77, I78, I79) [I77 <= I79 /\ I78 <= I76 /\ 1 <= I75] The dependency graph for this problem is: 0 -> 1 1 -> 3, 5, 7, 9 2 -> 3, 5, 7, 9 3 -> 2 4 -> 3, 5, 7, 9 5 -> 4 6 -> 3, 5, 7, 9 7 -> 6 8 -> 3, 5, 7, 9 9 -> 8 Where: 0) f8#(x1, x2, x3, x4, x5, x6, x7, x8) -> f7#(x1, x2, x3, x4, x5, x6, x7, x8) 1) f7#(I0, I1, I2, I3, I4, I5, I6, I7) -> f1#(I0, I1, I2, 0, 1, rnd6, I6, I7) [rnd6 = rnd6] 2) f6#(I8, I9, I10, I11, I12, I13, I14, I15) -> f1#(I8, I9, I10, I11, I12, I13, I14, I15) 3) f1#(I16, I17, I18, I19, I20, I21, I22, I23) -> f6#(I16, I17, I18, I19, -1 + I20, -1 * I17 + I21, I22, I23) [1 + I16 <= I21 /\ 1 <= I20] 4) f5#(I24, I25, I26, I27, I28, I29, I30, I31) -> f1#(I24, I25, I26, I27, I28, I29, I30, I31) 5) f1#(I32, I33, I34, I35, I36, I37, I38, I39) -> f5#(I32, I33, I34, I35, 1 + I36, I34 + I37, I38, I39) [I37 <= I32 /\ 1 <= I36] 6) f4#(I40, I41, I42, I43, I44, I45, I46, I47) -> f1#(I40, I41, I42, I43, I44, I45, I46, I47) 7) f1#(I48, I49, I50, I51, I52, I53, I54, I55) -> f4#(I48, I49, I50, 1, -1 + I52, -1 * I49 + I53, I52, I53) [1 + I48 <= I53 /\ 1 <= I52 /\ I51 <= 0] 8) f3#(I56, I57, I58, I59, I60, I61, I62, I63) -> f1#(I56, I57, I58, I59, I60, I61, I62, I63) 9) f1#(I64, I65, I66, I67, I68, I69, I70, I71) -> f3#(I64, I65, I66, 1, 1 + I68, I66 + I69, I68, I69) [I69 <= I64 /\ 1 <= I68 /\ I67 <= 0] We have the following SCCs. { 2, 3, 4, 5, 6, 7, 8, 9 } DP problem for innermost termination. P = f6#(I8, I9, I10, I11, I12, I13, I14, I15) -> f1#(I8, I9, I10, I11, I12, I13, I14, I15) f1#(I16, I17, I18, I19, I20, I21, I22, I23) -> f6#(I16, I17, I18, I19, -1 + I20, -1 * I17 + I21, I22, I23) [1 + I16 <= I21 /\ 1 <= I20] f5#(I24, I25, I26, I27, I28, I29, I30, I31) -> f1#(I24, I25, I26, I27, I28, I29, I30, I31) f1#(I32, I33, I34, I35, I36, I37, I38, I39) -> f5#(I32, I33, I34, I35, 1 + I36, I34 + I37, I38, I39) [I37 <= I32 /\ 1 <= I36] f4#(I40, I41, I42, I43, I44, I45, I46, I47) -> f1#(I40, I41, I42, I43, I44, I45, I46, I47) f1#(I48, I49, I50, I51, I52, I53, I54, I55) -> f4#(I48, I49, I50, 1, -1 + I52, -1 * I49 + I53, I52, I53) [1 + I48 <= I53 /\ 1 <= I52 /\ I51 <= 0] f3#(I56, I57, I58, I59, I60, I61, I62, I63) -> f1#(I56, I57, I58, I59, I60, I61, I62, I63) f1#(I64, I65, I66, I67, I68, I69, I70, I71) -> f3#(I64, I65, I66, 1, 1 + I68, I66 + I69, I68, I69) [I69 <= I64 /\ 1 <= I68 /\ I67 <= 0] R = f8(x1, x2, x3, x4, x5, x6, x7, x8) -> f7(x1, x2, x3, x4, x5, x6, x7, x8) f7(I0, I1, I2, I3, I4, I5, I6, I7) -> f1(I0, I1, I2, 0, 1, rnd6, I6, I7) [rnd6 = rnd6] f6(I8, I9, I10, I11, I12, I13, I14, I15) -> f1(I8, I9, I10, I11, I12, I13, I14, I15) f1(I16, I17, I18, I19, I20, I21, I22, I23) -> f6(I16, I17, I18, I19, -1 + I20, -1 * I17 + I21, I22, I23) [1 + I16 <= I21 /\ 1 <= I20] f5(I24, I25, I26, I27, I28, I29, I30, I31) -> f1(I24, I25, I26, I27, I28, I29, I30, I31) f1(I32, I33, I34, I35, I36, I37, I38, I39) -> f5(I32, I33, I34, I35, 1 + I36, I34 + I37, I38, I39) [I37 <= I32 /\ 1 <= I36] f4(I40, I41, I42, I43, I44, I45, I46, I47) -> f1(I40, I41, I42, I43, I44, I45, I46, I47) f1(I48, I49, I50, I51, I52, I53, I54, I55) -> f4(I48, I49, I50, 1, -1 + I52, -1 * I49 + I53, I52, I53) [1 + I48 <= I53 /\ 1 <= I52 /\ I51 <= 0] f3(I56, I57, I58, I59, I60, I61, I62, I63) -> f1(I56, I57, I58, I59, I60, I61, I62, I63) f1(I64, I65, I66, I67, I68, I69, I70, I71) -> f3(I64, I65, I66, 1, 1 + I68, I66 + I69, I68, I69) [I69 <= I64 /\ 1 <= I68 /\ I67 <= 0] f1(I72, I73, I74, I75, I76, I77, I78, I79) -> f2(I72, I73, I74, I75, I76, I77, I78, I79) [I77 <= I79 /\ I78 <= I76 /\ 1 <= I75] We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4,z5,z6,z7,z8)] = 0 + -1 * z4 NU[f4#(z1,z2,z3,z4,z5,z6,z7,z8)] = 0 + -1 * z4 NU[f5#(z1,z2,z3,z4,z5,z6,z7,z8)] = 0 + -1 * z4 NU[f1#(z1,z2,z3,z4,z5,z6,z7,z8)] = 0 + -1 * z4 NU[f6#(z1,z2,z3,z4,z5,z6,z7,z8)] = 0 + -1 * z4 This gives the following inequalities: ==> 0 + -1 * I11 >= 0 + -1 * I11 1 + I16 <= I21 /\ 1 <= I20 ==> 0 + -1 * I19 >= 0 + -1 * I19 ==> 0 + -1 * I27 >= 0 + -1 * I27 I37 <= I32 /\ 1 <= I36 ==> 0 + -1 * I35 >= 0 + -1 * I35 ==> 0 + -1 * I43 >= 0 + -1 * I43 1 + I48 <= I53 /\ 1 <= I52 /\ I51 <= 0 ==> 0 + -1 * I51 > 0 + -1 * 1 with 0 + -1 * I51 >= 0 ==> 0 + -1 * I59 >= 0 + -1 * I59 I69 <= I64 /\ 1 <= I68 /\ I67 <= 0 ==> 0 + -1 * I67 > 0 + -1 * 1 with 0 + -1 * I67 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f6#(I8, I9, I10, I11, I12, I13, I14, I15) -> f1#(I8, I9, I10, I11, I12, I13, I14, I15) f1#(I16, I17, I18, I19, I20, I21, I22, I23) -> f6#(I16, I17, I18, I19, -1 + I20, -1 * I17 + I21, I22, I23) [1 + I16 <= I21 /\ 1 <= I20] f5#(I24, I25, I26, I27, I28, I29, I30, I31) -> f1#(I24, I25, I26, I27, I28, I29, I30, I31) popout

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

job log

popout

actions

all output return to ITS