details

property	value
status	complete
benchmark	n-14.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n090.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	10.5809760094 seconds
cpu usage	11.315760143
max memory	2.4670208E7

stage attributes

key	value
output-size	8039
starexec-result	MAYBE

output

/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = f10#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) f9#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) f8#(I5, I6, I7, I8, I9) -> f9#(I5, I6, I7, I8, -1 + I9) [0 <= -1 + -1 + I9] f7#(I10, I11, I12, I13, I14) -> f8#(I10, I11, I12, I13, I14) [I12 = I12] f2#(I15, I16, I17, I18, I19) -> f7#(I15, I16, I17, rnd4, I19) [rnd4 = rnd4] f5#(I25, I26, I27, I28, I29) -> f6#(I25, I26, I27, I28, I29) [I26 = I26] f2#(I30, I31, I32, I33, I34) -> f5#(I30, I31, I32, I35, I34) [I35 = I35] f4#(I36, I37, I38, I39, I40) -> f2#(I36, I37, I38, I39, I40) f2#(I41, I42, I43, I44, I45) -> f4#(I41, I42, I43, I46, I45) [0 <= -1 + I45 /\ 0 <= I46 /\ I46 <= 0 /\ I46 = I46] f1#(I54, I55, I56, I57, I58) -> f2#(I54, I55, I56, I57, I58) R = f10(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) f9(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) f8(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, -1 + I9) [0 <= -1 + -1 + I9] f7(I10, I11, I12, I13, I14) -> f8(I10, I11, I12, I13, I14) [I12 = I12] f2(I15, I16, I17, I18, I19) -> f7(I15, I16, I17, rnd4, I19) [rnd4 = rnd4] f6(I20, I21, I22, I23, I24) -> f3(rnd1, I21, I22, I23, -1 + I24) [rnd1 = rnd1 /\ -1 + I24 <= 0] f5(I25, I26, I27, I28, I29) -> f6(I25, I26, I27, I28, I29) [I26 = I26] f2(I30, I31, I32, I33, I34) -> f5(I30, I31, I32, I35, I34) [I35 = I35] f4(I36, I37, I38, I39, I40) -> f2(I36, I37, I38, I39, I40) f2(I41, I42, I43, I44, I45) -> f4(I41, I42, I43, I46, I45) [0 <= -1 + I45 /\ 0 <= I46 /\ I46 <= 0 /\ I46 = I46] f2(I47, I48, I49, I50, I51) -> f3(I52, I48, I49, I53, I51) [I52 = I52 /\ I51 <= 0 /\ 0 <= I53 /\ I53 <= 0 /\ I53 = I53] f1(I54, I55, I56, I57, I58) -> f2(I54, I55, I56, I57, I58) The dependency graph for this problem is: 0 -> 9 1 -> 4, 6, 8 2 -> 1 3 -> 2 4 -> 3 5 -> 6 -> 5 7 -> 4, 6, 8 8 -> 7 9 -> 4, 6, 8 Where: 0) f10#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) 1) f9#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 2) f8#(I5, I6, I7, I8, I9) -> f9#(I5, I6, I7, I8, -1 + I9) [0 <= -1 + -1 + I9] 3) f7#(I10, I11, I12, I13, I14) -> f8#(I10, I11, I12, I13, I14) [I12 = I12] 4) f2#(I15, I16, I17, I18, I19) -> f7#(I15, I16, I17, rnd4, I19) [rnd4 = rnd4] 5) f5#(I25, I26, I27, I28, I29) -> f6#(I25, I26, I27, I28, I29) [I26 = I26] 6) f2#(I30, I31, I32, I33, I34) -> f5#(I30, I31, I32, I35, I34) [I35 = I35] 7) f4#(I36, I37, I38, I39, I40) -> f2#(I36, I37, I38, I39, I40) 8) f2#(I41, I42, I43, I44, I45) -> f4#(I41, I42, I43, I46, I45) [0 <= -1 + I45 /\ 0 <= I46 /\ I46 <= 0 /\ I46 = I46] 9) f1#(I54, I55, I56, I57, I58) -> f2#(I54, I55, I56, I57, I58) We have the following SCCs. { 1, 2, 3, 4, 7, 8 } DP problem for innermost termination. P = f9#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) f8#(I5, I6, I7, I8, I9) -> f9#(I5, I6, I7, I8, -1 + I9) [0 <= -1 + -1 + I9] f7#(I10, I11, I12, I13, I14) -> f8#(I10, I11, I12, I13, I14) [I12 = I12] f2#(I15, I16, I17, I18, I19) -> f7#(I15, I16, I17, rnd4, I19) [rnd4 = rnd4] f4#(I36, I37, I38, I39, I40) -> f2#(I36, I37, I38, I39, I40) f2#(I41, I42, I43, I44, I45) -> f4#(I41, I42, I43, I46, I45) [0 <= -1 + I45 /\ 0 <= I46 /\ I46 <= 0 /\ I46 = I46] R = f10(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) f9(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) f8(I5, I6, I7, I8, I9) -> f9(I5, I6, I7, I8, -1 + I9) [0 <= -1 + -1 + I9] f7(I10, I11, I12, I13, I14) -> f8(I10, I11, I12, I13, I14) [I12 = I12] f2(I15, I16, I17, I18, I19) -> f7(I15, I16, I17, rnd4, I19) [rnd4 = rnd4] f6(I20, I21, I22, I23, I24) -> f3(rnd1, I21, I22, I23, -1 + I24) [rnd1 = rnd1 /\ -1 + I24 <= 0] f5(I25, I26, I27, I28, I29) -> f6(I25, I26, I27, I28, I29) [I26 = I26] f2(I30, I31, I32, I33, I34) -> f5(I30, I31, I32, I35, I34) [I35 = I35] f4(I36, I37, I38, I39, I40) -> f2(I36, I37, I38, I39, I40) f2(I41, I42, I43, I44, I45) -> f4(I41, I42, I43, I46, I45) [0 <= -1 + I45 /\ 0 <= I46 /\ I46 <= 0 /\ I46 = I46] f2(I47, I48, I49, I50, I51) -> f3(I52, I48, I49, I53, I51) [I52 = I52 /\ I51 <= 0 /\ 0 <= I53 /\ I53 <= 0 /\ I53 = I53] f1(I54, I55, I56, I57, I58) -> f2(I54, I55, I56, I57, I58) We use the basic value criterion with the projection function NU: NU[f4#(z1,z2,z3,z4,z5)] = z5 NU[f7#(z1,z2,z3,z4,z5)] = z5 NU[f8#(z1,z2,z3,z4,z5)] = z5 NU[f2#(z1,z2,z3,z4,z5)] = z5 NU[f9#(z1,z2,z3,z4,z5)] = z5 This gives the following inequalities: ==> I4 (>! \union =) I4 0 <= -1 + -1 + I9 ==> I9 >! -1 + I9 I12 = I12 ==> I14 (>! \union =) I14 rnd4 = rnd4 ==> I19 (>! \union =) I19 ==> I40 (>! \union =) I40 0 <= -1 + I45 /\ 0 <= I46 /\ I46 <= 0 /\ I46 = I46 ==> I45 (>! \union =) I45 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. popout

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job log

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