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ITS pair #487097206
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
n148.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
34.5553 seconds
cpu usage
34.7234
user time
17.9439
system time
16.7795
max virtual memory
740200.0
max residence set size
15580.0
stage attributes
key
value
starexec-result
YES
output
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] 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]
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