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ITS pair #487098553
details
property
value
status
complete
benchmark
java_MinusBuiltIn.c.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n138.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
13.0252 seconds
cpu usage
13.1595
user time
6.82929
system time
6.33019
max virtual memory
751772.0
max residence set size
11416.0
stage attributes
key
value
starexec-result
YES
output
YES DP problem for innermost termination. P = f7#(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f6#(x1, x2, x3, x4, x5, x6, x7, x8, x9) f6#(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f5#(I0, I1, I2, I3, I4, I5, I6, I7, I8) f6#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f4#(I9, I10, I11, I12, I13, I14, I15, I16, I17) f6#(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3#(I18, I19, I20, I21, I22, I23, I24, I25, I26) f6#(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f1#(I27, I28, I29, I30, I31, I32, I33, I34, I35) f6#(I45, I46, I47, I48, I49, I50, I51, I52, I53) -> f5#(I51, I52, I53, rnd4, rnd5, rnd6, rnd7, rnd8, rnd9) [rnd9 = rnd6 /\ rnd8 = rnd5 /\ rnd7 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5 /\ rnd4 = rnd4] f5#(I54, I55, I56, I57, I58, I59, I60, I61, I62) -> f4#(I60, I61, I62, I63, I64, I59, I65, I66, 0) [I66 = I64 /\ I65 = I63 /\ I64 = I64 /\ I63 = I63] f4#(I67, I68, I69, I70, I71, I72, I73, I74, I75) -> f3#(I73, I74, I75, I70, I71, I72, I73, I74, I75) [1 + I74 <= I73] f4#(I76, I77, I78, I79, I80, I81, I82, I83, I84) -> f1#(I82, I83, I84, I79, I80, I81, I82, I83, I84) [I82 <= I83] f3#(I85, I86, I87, I88, I89, I90, I91, I92, I93) -> f4#(I91, I92, I93, I88, I89, I90, I91, 1 + I92, 1 + I93) R = f7(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f6(x1, x2, x3, x4, x5, x6, x7, x8, x9) f6(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f5(I0, I1, I2, I3, I4, I5, I6, I7, I8) f6(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f4(I9, I10, I11, I12, I13, I14, I15, I16, I17) f6(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3(I18, I19, I20, I21, I22, I23, I24, I25, I26) f6(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f1(I27, I28, I29, I30, I31, I32, I33, I34, I35) f6(I36, I37, I38, I39, I40, I41, I42, I43, I44) -> f2(I36, I37, I38, I39, I40, I41, I42, I43, I44) f6(I45, I46, I47, I48, I49, I50, I51, I52, I53) -> f5(I51, I52, I53, rnd4, rnd5, rnd6, rnd7, rnd8, rnd9) [rnd9 = rnd6 /\ rnd8 = rnd5 /\ rnd7 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5 /\ rnd4 = rnd4] f5(I54, I55, I56, I57, I58, I59, I60, I61, I62) -> f4(I60, I61, I62, I63, I64, I59, I65, I66, 0) [I66 = I64 /\ I65 = I63 /\ I64 = I64 /\ I63 = I63] f4(I67, I68, I69, I70, I71, I72, I73, I74, I75) -> f3(I73, I74, I75, I70, I71, I72, I73, I74, I75) [1 + I74 <= I73] f4(I76, I77, I78, I79, I80, I81, I82, I83, I84) -> f1(I82, I83, I84, I79, I80, I81, I82, I83, I84) [I82 <= I83] f3(I85, I86, I87, I88, I89, I90, I91, I92, I93) -> f4(I91, I92, I93, I88, I89, I90, I91, 1 + I92, 1 + I93) f1(I94, I95, I96, I97, I98, I99, I100, I101, I102) -> f2(I100, I101, I102, I103, I104, I105, I106, I107, I108) [I108 = I105 /\ I107 = I104 /\ I106 = I103 /\ I105 = I105 /\ I104 = I104 /\ I103 = I103] The dependency graph for this problem is: 0 -> 1, 2, 3, 4, 5 1 -> 6 2 -> 7, 8 3 -> 9 4 -> 5 -> 6 6 -> 7, 8 7 -> 9 8 -> 9 -> 7, 8 Where: 0) f7#(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f6#(x1, x2, x3, x4, x5, x6, x7, x8, x9) 1) f6#(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f5#(I0, I1, I2, I3, I4, I5, I6, I7, I8) 2) f6#(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f4#(I9, I10, I11, I12, I13, I14, I15, I16, I17) 3) f6#(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3#(I18, I19, I20, I21, I22, I23, I24, I25, I26) 4) f6#(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f1#(I27, I28, I29, I30, I31, I32, I33, I34, I35) 5) f6#(I45, I46, I47, I48, I49, I50, I51, I52, I53) -> f5#(I51, I52, I53, rnd4, rnd5, rnd6, rnd7, rnd8, rnd9) [rnd9 = rnd6 /\ rnd8 = rnd5 /\ rnd7 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5 /\ rnd4 = rnd4] 6) f5#(I54, I55, I56, I57, I58, I59, I60, I61, I62) -> f4#(I60, I61, I62, I63, I64, I59, I65, I66, 0) [I66 = I64 /\ I65 = I63 /\ I64 = I64 /\ I63 = I63] 7) f4#(I67, I68, I69, I70, I71, I72, I73, I74, I75) -> f3#(I73, I74, I75, I70, I71, I72, I73, I74, I75) [1 + I74 <= I73] 8) f4#(I76, I77, I78, I79, I80, I81, I82, I83, I84) -> f1#(I82, I83, I84, I79, I80, I81, I82, I83, I84) [I82 <= I83] 9) f3#(I85, I86, I87, I88, I89, I90, I91, I92, I93) -> f4#(I91, I92, I93, I88, I89, I90, I91, 1 + I92, 1 + I93) We have the following SCCs. { 7, 9 } DP problem for innermost termination. P = f4#(I67, I68, I69, I70, I71, I72, I73, I74, I75) -> f3#(I73, I74, I75, I70, I71, I72, I73, I74, I75) [1 + I74 <= I73] f3#(I85, I86, I87, I88, I89, I90, I91, I92, I93) -> f4#(I91, I92, I93, I88, I89, I90, I91, 1 + I92, 1 + I93) R = f7(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f6(x1, x2, x3, x4, x5, x6, x7, x8, x9) f6(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f5(I0, I1, I2, I3, I4, I5, I6, I7, I8) f6(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f4(I9, I10, I11, I12, I13, I14, I15, I16, I17) f6(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3(I18, I19, I20, I21, I22, I23, I24, I25, I26) f6(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f1(I27, I28, I29, I30, I31, I32, I33, I34, I35) f6(I36, I37, I38, I39, I40, I41, I42, I43, I44) -> f2(I36, I37, I38, I39, I40, I41, I42, I43, I44) f6(I45, I46, I47, I48, I49, I50, I51, I52, I53) -> f5(I51, I52, I53, rnd4, rnd5, rnd6, rnd7, rnd8, rnd9) [rnd9 = rnd6 /\ rnd8 = rnd5 /\ rnd7 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5 /\ rnd4 = rnd4] f5(I54, I55, I56, I57, I58, I59, I60, I61, I62) -> f4(I60, I61, I62, I63, I64, I59, I65, I66, 0) [I66 = I64 /\ I65 = I63 /\ I64 = I64 /\ I63 = I63] f4(I67, I68, I69, I70, I71, I72, I73, I74, I75) -> f3(I73, I74, I75, I70, I71, I72, I73, I74, I75) [1 + I74 <= I73] f4(I76, I77, I78, I79, I80, I81, I82, I83, I84) -> f1(I82, I83, I84, I79, I80, I81, I82, I83, I84) [I82 <= I83] f3(I85, I86, I87, I88, I89, I90, I91, I92, I93) -> f4(I91, I92, I93, I88, I89, I90, I91, 1 + I92, 1 + I93) f1(I94, I95, I96, I97, I98, I99, I100, I101, I102) -> f2(I100, I101, I102, I103, I104, I105, I106, I107, I108) [I108 = I105 /\ I107 = I104 /\ I106 = I103 /\ I105 = I105 /\ I104 = I104 /\ I103 = I103] We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4,z5,z6,z7,z8,z9)] = z7 + -1 * (1 + (1 + z8)) NU[f4#(z1,z2,z3,z4,z5,z6,z7,z8,z9)] = z7 + -1 * (1 + z8) This gives the following inequalities: 1 + I74 <= I73 ==> I73 + -1 * (1 + I74) > I73 + -1 * (1 + (1 + I74)) with I73 + -1 * (1 + I74) >= 0 ==> I91 + -1 * (1 + (1 + I92)) >= I91 + -1 * (1 + (1 + I92)) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I85, I86, I87, I88, I89, I90, I91, I92, I93) -> f4#(I91, I92, I93, I88, I89, I90, I91, 1 + I92, 1 + I93) R = f7(x1, x2, x3, x4, x5, x6, x7, x8, x9) -> f6(x1, x2, x3, x4, x5, x6, x7, x8, x9) f6(I0, I1, I2, I3, I4, I5, I6, I7, I8) -> f5(I0, I1, I2, I3, I4, I5, I6, I7, I8) f6(I9, I10, I11, I12, I13, I14, I15, I16, I17) -> f4(I9, I10, I11, I12, I13, I14, I15, I16, I17) f6(I18, I19, I20, I21, I22, I23, I24, I25, I26) -> f3(I18, I19, I20, I21, I22, I23, I24, I25, I26) f6(I27, I28, I29, I30, I31, I32, I33, I34, I35) -> f1(I27, I28, I29, I30, I31, I32, I33, I34, I35) f6(I36, I37, I38, I39, I40, I41, I42, I43, I44) -> f2(I36, I37, I38, I39, I40, I41, I42, I43, I44) f6(I45, I46, I47, I48, I49, I50, I51, I52, I53) -> f5(I51, I52, I53, rnd4, rnd5, rnd6, rnd7, rnd8, rnd9) [rnd9 = rnd6 /\ rnd8 = rnd5 /\ rnd7 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5 /\ rnd4 = rnd4] f5(I54, I55, I56, I57, I58, I59, I60, I61, I62) -> f4(I60, I61, I62, I63, I64, I59, I65, I66, 0) [I66 = I64 /\ I65 = I63 /\ I64 = I64 /\ I63 = I63] f4(I67, I68, I69, I70, I71, I72, I73, I74, I75) -> f3(I73, I74, I75, I70, I71, I72, I73, I74, I75) [1 + I74 <= I73] f4(I76, I77, I78, I79, I80, I81, I82, I83, I84) -> f1(I82, I83, I84, I79, I80, I81, I82, I83, I84) [I82 <= I83] f3(I85, I86, I87, I88, I89, I90, I91, I92, I93) -> f4(I91, I92, I93, I88, I89, I90, I91, 1 + I92, 1 + I93) f1(I94, I95, I96, I97, I98, I99, I100, I101, I102) -> f2(I100, I101, I102, I103, I104, I105, I106, I107, I108) [I108 = I105 /\ I107 = I104 /\ I106 = I103 /\ I105 = I105 /\ I104 = I104 /\ I103 = I103] The dependency graph for this problem is:
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