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Integer_Transition_Systems 2019-03-29 01.54 pair #432274131
details
property
value
status
complete
benchmark
byron-2.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n065.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
37.8083 seconds
cpu usage
38.0584
user time
19.5946
system time
18.4638
max virtual memory
725200.0
max residence set size
17016.0
stage attributes
key
value
starexec-result
YES
output
38.02/37.80 YES 38.02/37.80 38.02/37.80 DP problem for innermost termination. 38.02/37.80 P = 38.02/37.80 f8#(x1, x2, x3, x4, x5, x6, x7, x8) -> f1#(x1, x2, x3, x4, x5, x6, x7, x8) 38.02/37.80 f7#(I0, I1, I2, I3, I4, I5, I6, I7) -> f2#(I0, I1, I2, I3, I4, I5, I6, I7) 38.02/37.80 f6#(I8, I9, I10, I11, I12, I13, I14, I15) -> f7#(I8, rnd2, I10, I11, I12, I13, -1 + I14, rnd8) [y1 = y1 /\ rnd8 = y1 /\ rnd2 = rnd2] 38.02/37.80 f5#(I16, I17, I18, I19, I20, I21, I22, I23) -> f6#(I16, I17, I18, I19, I20, I21, I22, I23) [I16 = I16] 38.02/37.80 f2#(I24, I25, I26, I27, I28, I29, I30, I31) -> f5#(I24, I32, I26, rnd4, I28, I29, I30, I31) [1 <= I30 /\ I33 = I33 /\ rnd4 = I33 /\ I32 = I32] 38.02/37.80 f4#(I34, I35, I36, I37, I38, I39, I40, I41) -> f2#(I34, I35, I36, I37, I38, I39, I40, I41) 38.02/37.80 f2#(I42, I43, I44, I45, I46, I47, I48, I49) -> f4#(I42, I50, I44, I51, I46, rnd6, I48, -1 + I49) [1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49] 38.02/37.80 f1#(I61, I62, I63, I64, I65, I66, I67, I68) -> f2#(I61, I62, I63, I64, I65, I66, I67, I68) 38.02/37.80 R = 38.02/37.80 f8(x1, x2, x3, x4, x5, x6, x7, x8) -> f1(x1, x2, x3, x4, x5, x6, x7, x8) 38.02/37.80 f7(I0, I1, I2, I3, I4, I5, I6, I7) -> f2(I0, I1, I2, I3, I4, I5, I6, I7) 38.02/37.80 f6(I8, I9, I10, I11, I12, I13, I14, I15) -> f7(I8, rnd2, I10, I11, I12, I13, -1 + I14, rnd8) [y1 = y1 /\ rnd8 = y1 /\ rnd2 = rnd2] 38.02/37.80 f5(I16, I17, I18, I19, I20, I21, I22, I23) -> f6(I16, I17, I18, I19, I20, I21, I22, I23) [I16 = I16] 38.02/37.80 f2(I24, I25, I26, I27, I28, I29, I30, I31) -> f5(I24, I32, I26, rnd4, I28, I29, I30, I31) [1 <= I30 /\ I33 = I33 /\ rnd4 = I33 /\ I32 = I32] 38.02/37.80 f4(I34, I35, I36, I37, I38, I39, I40, I41) -> f2(I34, I35, I36, I37, I38, I39, I40, I41) 38.02/37.80 f2(I42, I43, I44, I45, I46, I47, I48, I49) -> f4(I42, I50, I44, I51, I46, rnd6, I48, -1 + I49) [1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49] 38.02/37.80 f2(I53, I54, I55, I56, I57, I58, I59, I60) -> f3(I53, I54, I57, I56, I57, I58, I59, I60) [I59 <= 0] 38.02/37.80 f1(I61, I62, I63, I64, I65, I66, I67, I68) -> f2(I61, I62, I63, I64, I65, I66, I67, I68) 38.02/37.80 38.02/37.80 The dependency graph for this problem is: 38.02/37.80 0 -> 7 38.02/37.80 1 -> 4, 6 38.02/37.80 2 -> 1 38.02/37.80 3 -> 2 38.02/37.80 4 -> 3 38.02/37.80 5 -> 4, 6 38.02/37.80 6 -> 5 38.02/37.80 7 -> 4, 6 38.02/37.80 Where: 38.02/37.80 0) f8#(x1, x2, x3, x4, x5, x6, x7, x8) -> f1#(x1, x2, x3, x4, x5, x6, x7, x8) 38.02/37.80 1) f7#(I0, I1, I2, I3, I4, I5, I6, I7) -> f2#(I0, I1, I2, I3, I4, I5, I6, I7) 38.02/37.80 2) f6#(I8, I9, I10, I11, I12, I13, I14, I15) -> f7#(I8, rnd2, I10, I11, I12, I13, -1 + I14, rnd8) [y1 = y1 /\ rnd8 = y1 /\ rnd2 = rnd2] 38.02/37.80 3) f5#(I16, I17, I18, I19, I20, I21, I22, I23) -> f6#(I16, I17, I18, I19, I20, I21, I22, I23) [I16 = I16] 38.02/37.80 4) f2#(I24, I25, I26, I27, I28, I29, I30, I31) -> f5#(I24, I32, I26, rnd4, I28, I29, I30, I31) [1 <= I30 /\ I33 = I33 /\ rnd4 = I33 /\ I32 = I32] 38.02/37.80 5) f4#(I34, I35, I36, I37, I38, I39, I40, I41) -> f2#(I34, I35, I36, I37, I38, I39, I40, I41) 38.02/37.80 6) f2#(I42, I43, I44, I45, I46, I47, I48, I49) -> f4#(I42, I50, I44, I51, I46, rnd6, I48, -1 + I49) [1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49] 38.02/37.80 7) f1#(I61, I62, I63, I64, I65, I66, I67, I68) -> f2#(I61, I62, I63, I64, I65, I66, I67, I68) 38.02/37.80 38.02/37.80 We have the following SCCs. 38.02/37.80 { 1, 2, 3, 4, 5, 6 } 38.02/37.80 38.02/37.80 DP problem for innermost termination. 38.02/37.80 P = 38.02/37.80 f7#(I0, I1, I2, I3, I4, I5, I6, I7) -> f2#(I0, I1, I2, I3, I4, I5, I6, I7) 38.02/37.80 f6#(I8, I9, I10, I11, I12, I13, I14, I15) -> f7#(I8, rnd2, I10, I11, I12, I13, -1 + I14, rnd8) [y1 = y1 /\ rnd8 = y1 /\ rnd2 = rnd2] 38.02/37.80 f5#(I16, I17, I18, I19, I20, I21, I22, I23) -> f6#(I16, I17, I18, I19, I20, I21, I22, I23) [I16 = I16] 38.02/37.80 f2#(I24, I25, I26, I27, I28, I29, I30, I31) -> f5#(I24, I32, I26, rnd4, I28, I29, I30, I31) [1 <= I30 /\ I33 = I33 /\ rnd4 = I33 /\ I32 = I32] 38.02/37.80 f4#(I34, I35, I36, I37, I38, I39, I40, I41) -> f2#(I34, I35, I36, I37, I38, I39, I40, I41) 38.02/37.80 f2#(I42, I43, I44, I45, I46, I47, I48, I49) -> f4#(I42, I50, I44, I51, I46, rnd6, I48, -1 + I49) [1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49] 38.02/37.80 R = 38.02/37.80 f8(x1, x2, x3, x4, x5, x6, x7, x8) -> f1(x1, x2, x3, x4, x5, x6, x7, x8) 38.02/37.80 f7(I0, I1, I2, I3, I4, I5, I6, I7) -> f2(I0, I1, I2, I3, I4, I5, I6, I7) 38.02/37.80 f6(I8, I9, I10, I11, I12, I13, I14, I15) -> f7(I8, rnd2, I10, I11, I12, I13, -1 + I14, rnd8) [y1 = y1 /\ rnd8 = y1 /\ rnd2 = rnd2] 38.02/37.80 f5(I16, I17, I18, I19, I20, I21, I22, I23) -> f6(I16, I17, I18, I19, I20, I21, I22, I23) [I16 = I16] 38.02/37.80 f2(I24, I25, I26, I27, I28, I29, I30, I31) -> f5(I24, I32, I26, rnd4, I28, I29, I30, I31) [1 <= I30 /\ I33 = I33 /\ rnd4 = I33 /\ I32 = I32] 38.02/37.80 f4(I34, I35, I36, I37, I38, I39, I40, I41) -> f2(I34, I35, I36, I37, I38, I39, I40, I41) 38.02/37.80 f2(I42, I43, I44, I45, I46, I47, I48, I49) -> f4(I42, I50, I44, I51, I46, rnd6, I48, -1 + I49) [1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49] 38.02/37.80 f2(I53, I54, I55, I56, I57, I58, I59, I60) -> f3(I53, I54, I57, I56, I57, I58, I59, I60) [I59 <= 0] 38.02/37.80 f1(I61, I62, I63, I64, I65, I66, I67, I68) -> f2(I61, I62, I63, I64, I65, I66, I67, I68) 38.02/37.80 38.02/37.80 We use the extended value criterion with the projection function NU: 38.02/37.80 NU[f4#(x0,x1,x2,x3,x4,x5,x6,x7)] = x6 - 1 38.02/37.80 NU[f5#(x0,x1,x2,x3,x4,x5,x6,x7)] = x6 - 2 38.02/37.80 NU[f6#(x0,x1,x2,x3,x4,x5,x6,x7)] = x6 - 2 38.02/37.80 NU[f2#(x0,x1,x2,x3,x4,x5,x6,x7)] = x6 - 1 38.02/37.80 NU[f7#(x0,x1,x2,x3,x4,x5,x6,x7)] = x6 - 1 38.02/37.80 38.02/37.80 This gives the following inequalities: 38.02/37.80 ==> I6 - 1 >= I6 - 1 38.02/37.80 y1 = y1 /\ rnd8 = y1 /\ rnd2 = rnd2 ==> I14 - 2 >= (-1 + I14) - 1 38.02/37.80 I16 = I16 ==> I22 - 2 >= I22 - 2 38.02/37.80 1 <= I30 /\ I33 = I33 /\ rnd4 = I33 /\ I32 = I32 ==> I30 - 1 > I30 - 2 with I30 - 1 >= 0 38.02/37.80 ==> I40 - 1 >= I40 - 1 38.02/37.80 1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49 ==> I48 - 1 >= I48 - 1 38.02/37.80 38.02/37.80 We remove all the strictly oriented dependency pairs. 38.02/37.80 38.02/37.80 DP problem for innermost termination. 38.02/37.80 P = 38.02/37.80 f7#(I0, I1, I2, I3, I4, I5, I6, I7) -> f2#(I0, I1, I2, I3, I4, I5, I6, I7) 38.02/37.80 f6#(I8, I9, I10, I11, I12, I13, I14, I15) -> f7#(I8, rnd2, I10, I11, I12, I13, -1 + I14, rnd8) [y1 = y1 /\ rnd8 = y1 /\ rnd2 = rnd2] 38.02/37.80 f5#(I16, I17, I18, I19, I20, I21, I22, I23) -> f6#(I16, I17, I18, I19, I20, I21, I22, I23) [I16 = I16] 38.02/37.80 f4#(I34, I35, I36, I37, I38, I39, I40, I41) -> f2#(I34, I35, I36, I37, I38, I39, I40, I41) 38.02/37.80 f2#(I42, I43, I44, I45, I46, I47, I48, I49) -> f4#(I42, I50, I44, I51, I46, rnd6, I48, -1 + I49) [1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49] 38.02/37.80 R = 38.02/37.80 f8(x1, x2, x3, x4, x5, x6, x7, x8) -> f1(x1, x2, x3, x4, x5, x6, x7, x8) 38.02/37.80 f7(I0, I1, I2, I3, I4, I5, I6, I7) -> f2(I0, I1, I2, I3, I4, I5, I6, I7) 38.02/37.80 f6(I8, I9, I10, I11, I12, I13, I14, I15) -> f7(I8, rnd2, I10, I11, I12, I13, -1 + I14, rnd8) [y1 = y1 /\ rnd8 = y1 /\ rnd2 = rnd2] 38.02/37.80 f5(I16, I17, I18, I19, I20, I21, I22, I23) -> f6(I16, I17, I18, I19, I20, I21, I22, I23) [I16 = I16] 38.02/37.80 f2(I24, I25, I26, I27, I28, I29, I30, I31) -> f5(I24, I32, I26, rnd4, I28, I29, I30, I31) [1 <= I30 /\ I33 = I33 /\ rnd4 = I33 /\ I32 = I32] 38.02/37.80 f4(I34, I35, I36, I37, I38, I39, I40, I41) -> f2(I34, I35, I36, I37, I38, I39, I40, I41) 38.02/37.80 f2(I42, I43, I44, I45, I46, I47, I48, I49) -> f4(I42, I50, I44, I51, I46, rnd6, I48, -1 + I49) [1 <= I48 /\ I52 = I52 /\ I51 = I52 /\ I50 = I50 /\ 0 <= I51 /\ I51 <= 0 /\ rnd6 = rnd6 /\ 2 <= -1 + I49] 38.02/37.80 f2(I53, I54, I55, I56, I57, I58, I59, I60) -> f3(I53, I54, I57, I56, I57, I58, I59, I60) [I59 <= 0] 38.02/37.80 f1(I61, I62, I63, I64, I65, I66, I67, I68) -> f2(I61, I62, I63, I64, I65, I66, I67, I68) 38.02/37.80 38.02/37.80 The dependency graph for this problem is:
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