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ITS pair #487099120
details
property
value
status
complete
benchmark
ex32.t2_fixed.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n139.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
4.87333 seconds
cpu usage
4.95696
user time
2.46698
system time
2.48998
max virtual memory
735264.0
max residence set size
8404.0
stage attributes
key
value
starexec-result
YES
output
YES DP problem for innermost termination. P = f11#(x1, x2, x3, x4) -> f10#(x1, x2, x3, x4) f10#(I0, I1, I2, I3) -> f4#(1, I1, I2, I3) f2#(I4, I5, I6, I7) -> f5#(I4, I5, 999, I7) f6#(I8, I9, I10, I11) -> f5#(I8, I9, -1 + I10, I11) [0 <= I10] f6#(I12, I13, I14, I15) -> f9#(I12, I13, I14, 999) [1 + I14 <= 0] f9#(I16, I17, I18, I19) -> f7#(I16, I17, I18, I19) f7#(I20, I21, I22, I23) -> f9#(I20, I21, I22, -1 + I23) [0 <= I23] f5#(I28, I29, I30, I31) -> f6#(I28, I29, I30, I31) f3#(I32, I33, I34, I35) -> f1#(I32, I33, I34, I35) f4#(I36, I37, I38, I39) -> f3#(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0] f4#(I40, I41, I42, I43) -> f2#(I40, I41, I42, I43) [1 + I40 <= 0] f4#(I44, I45, I46, I47) -> f2#(I44, I45, I46, I47) [1 <= I44] f1#(I48, I49, I50, I51) -> f3#(I48, -1 + I49, I50, I51) [0 <= I49] f1#(I52, I53, I54, I55) -> f2#(I52, I53, I54, I55) [1 + I53 <= 0] R = f11(x1, x2, x3, x4) -> f10(x1, x2, x3, x4) f10(I0, I1, I2, I3) -> f4(1, I1, I2, I3) f2(I4, I5, I6, I7) -> f5(I4, I5, 999, I7) f6(I8, I9, I10, I11) -> f5(I8, I9, -1 + I10, I11) [0 <= I10] f6(I12, I13, I14, I15) -> f9(I12, I13, I14, 999) [1 + I14 <= 0] f9(I16, I17, I18, I19) -> f7(I16, I17, I18, I19) f7(I20, I21, I22, I23) -> f9(I20, I21, I22, -1 + I23) [0 <= I23] f7(I24, I25, I26, I27) -> f8(I24, I25, I26, I27) [1 + I27 <= 0] f5(I28, I29, I30, I31) -> f6(I28, I29, I30, I31) f3(I32, I33, I34, I35) -> f1(I32, I33, I34, I35) f4(I36, I37, I38, I39) -> f3(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0] f4(I40, I41, I42, I43) -> f2(I40, I41, I42, I43) [1 + I40 <= 0] f4(I44, I45, I46, I47) -> f2(I44, I45, I46, I47) [1 <= I44] f1(I48, I49, I50, I51) -> f3(I48, -1 + I49, I50, I51) [0 <= I49] f1(I52, I53, I54, I55) -> f2(I52, I53, I54, I55) [1 + I53 <= 0] The dependency graph for this problem is: 0 -> 1 1 -> 11 2 -> 7 3 -> 7 4 -> 5 5 -> 6 6 -> 5 7 -> 3, 4 8 -> 12, 13 9 -> 8 10 -> 2 11 -> 2 12 -> 8 13 -> 2 Where: 0) f11#(x1, x2, x3, x4) -> f10#(x1, x2, x3, x4) 1) f10#(I0, I1, I2, I3) -> f4#(1, I1, I2, I3) 2) f2#(I4, I5, I6, I7) -> f5#(I4, I5, 999, I7) 3) f6#(I8, I9, I10, I11) -> f5#(I8, I9, -1 + I10, I11) [0 <= I10] 4) f6#(I12, I13, I14, I15) -> f9#(I12, I13, I14, 999) [1 + I14 <= 0] 5) f9#(I16, I17, I18, I19) -> f7#(I16, I17, I18, I19) 6) f7#(I20, I21, I22, I23) -> f9#(I20, I21, I22, -1 + I23) [0 <= I23] 7) f5#(I28, I29, I30, I31) -> f6#(I28, I29, I30, I31) 8) f3#(I32, I33, I34, I35) -> f1#(I32, I33, I34, I35) 9) f4#(I36, I37, I38, I39) -> f3#(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0] 10) f4#(I40, I41, I42, I43) -> f2#(I40, I41, I42, I43) [1 + I40 <= 0] 11) f4#(I44, I45, I46, I47) -> f2#(I44, I45, I46, I47) [1 <= I44] 12) f1#(I48, I49, I50, I51) -> f3#(I48, -1 + I49, I50, I51) [0 <= I49] 13) f1#(I52, I53, I54, I55) -> f2#(I52, I53, I54, I55) [1 + I53 <= 0] We have the following SCCs. { 8, 12 } { 3, 7 } { 5, 6 } DP problem for innermost termination. P = f9#(I16, I17, I18, I19) -> f7#(I16, I17, I18, I19) f7#(I20, I21, I22, I23) -> f9#(I20, I21, I22, -1 + I23) [0 <= I23] R = f11(x1, x2, x3, x4) -> f10(x1, x2, x3, x4) f10(I0, I1, I2, I3) -> f4(1, I1, I2, I3) f2(I4, I5, I6, I7) -> f5(I4, I5, 999, I7) f6(I8, I9, I10, I11) -> f5(I8, I9, -1 + I10, I11) [0 <= I10] f6(I12, I13, I14, I15) -> f9(I12, I13, I14, 999) [1 + I14 <= 0] f9(I16, I17, I18, I19) -> f7(I16, I17, I18, I19) f7(I20, I21, I22, I23) -> f9(I20, I21, I22, -1 + I23) [0 <= I23] f7(I24, I25, I26, I27) -> f8(I24, I25, I26, I27) [1 + I27 <= 0] f5(I28, I29, I30, I31) -> f6(I28, I29, I30, I31) f3(I32, I33, I34, I35) -> f1(I32, I33, I34, I35) f4(I36, I37, I38, I39) -> f3(I36, 999, I38, I39) [0 <= I36 /\ I36 <= 0] f4(I40, I41, I42, I43) -> f2(I40, I41, I42, I43) [1 + I40 <= 0] f4(I44, I45, I46, I47) -> f2(I44, I45, I46, I47) [1 <= I44] f1(I48, I49, I50, I51) -> f3(I48, -1 + I49, I50, I51) [0 <= I49] f1(I52, I53, I54, I55) -> f2(I52, I53, I54, I55) [1 + I53 <= 0] We use the basic value criterion with the projection function NU: NU[f7#(z1,z2,z3,z4)] = z4 NU[f9#(z1,z2,z3,z4)] = z4 This gives the following inequalities: ==> I19 (>! \union =) I19 0 <= I23 ==> I23 >! -1 + I23
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