details

property	value
status	complete
benchmark	DivMinus2.jar-obl-8.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n071.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	6.47203493118 seconds
cpu usage	6.300737307
max memory	2.4158208E7

stage attributes

key	value
output-size	7609
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = init#(x1, x2, x3, x4, x5) -> f1#(rnd1, rnd2, rnd3, rnd4, rnd5) f3#(I0, I1, I2, I3, I4) -> f2#(I1 + 1, I2, I0, I5, I6) [0 = I4 /\ 0 = I3 /\ -1 <= I1 - 1] f3#(I7, I8, I9, I10, I11) -> f3#(I7, I8, I9 - 1, I10 - 1, I10 - 1) [I10 = I11 /\ 0 <= I10 - 1] f3#(I12, I13, I14, I15, I16) -> f3#(I12, I13, I14 + 1, I15 + 1, I15 + 1) [I15 = I16 /\ I15 <= 0 /\ 0 <= I15 - 1] f3#(I17, I18, I19, I20, I21) -> f3#(I17, I18, I19 + 1, I20 + 1, I20 + 1) [I20 = I21 /\ I20 <= 0 /\ I20 <= -1] f2#(I22, I23, I24, I25, I26) -> f3#(I24, I22, I23, I24, I24) [0 <= I24 - 1 /\ I24 <= I23] f1#(I27, I28, I29, I30, I31) -> f2#(0, I32, I33, I34, I35) [0 <= I27 - 1 /\ -1 <= I32 - 1 /\ -1 <= I28 - 1 /\ -1 <= I33 - 1] R = init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) f3(I0, I1, I2, I3, I4) -> f2(I1 + 1, I2, I0, I5, I6) [0 = I4 /\ 0 = I3 /\ -1 <= I1 - 1] f3(I7, I8, I9, I10, I11) -> f3(I7, I8, I9 - 1, I10 - 1, I10 - 1) [I10 = I11 /\ 0 <= I10 - 1] f3(I12, I13, I14, I15, I16) -> f3(I12, I13, I14 + 1, I15 + 1, I15 + 1) [I15 = I16 /\ I15 <= 0 /\ 0 <= I15 - 1] f3(I17, I18, I19, I20, I21) -> f3(I17, I18, I19 + 1, I20 + 1, I20 + 1) [I20 = I21 /\ I20 <= 0 /\ I20 <= -1] f2(I22, I23, I24, I25, I26) -> f3(I24, I22, I23, I24, I24) [0 <= I24 - 1 /\ I24 <= I23] f1(I27, I28, I29, I30, I31) -> f2(0, I32, I33, I34, I35) [0 <= I27 - 1 /\ -1 <= I32 - 1 /\ -1 <= I28 - 1 /\ -1 <= I33 - 1] The dependency graph for this problem is: 0 -> 6 1 -> 5 2 -> 1, 2 3 -> 4 -> 1, 4 5 -> 2 6 -> 5 Where: 0) init#(x1, x2, x3, x4, x5) -> f1#(rnd1, rnd2, rnd3, rnd4, rnd5) 1) f3#(I0, I1, I2, I3, I4) -> f2#(I1 + 1, I2, I0, I5, I6) [0 = I4 /\ 0 = I3 /\ -1 <= I1 - 1] 2) f3#(I7, I8, I9, I10, I11) -> f3#(I7, I8, I9 - 1, I10 - 1, I10 - 1) [I10 = I11 /\ 0 <= I10 - 1] 3) f3#(I12, I13, I14, I15, I16) -> f3#(I12, I13, I14 + 1, I15 + 1, I15 + 1) [I15 = I16 /\ I15 <= 0 /\ 0 <= I15 - 1] 4) f3#(I17, I18, I19, I20, I21) -> f3#(I17, I18, I19 + 1, I20 + 1, I20 + 1) [I20 = I21 /\ I20 <= 0 /\ I20 <= -1] 5) f2#(I22, I23, I24, I25, I26) -> f3#(I24, I22, I23, I24, I24) [0 <= I24 - 1 /\ I24 <= I23] 6) f1#(I27, I28, I29, I30, I31) -> f2#(0, I32, I33, I34, I35) [0 <= I27 - 1 /\ -1 <= I32 - 1 /\ -1 <= I28 - 1 /\ -1 <= I33 - 1] We have the following SCCs. { 4 } { 1, 2, 5 } DP problem for innermost termination. P = f3#(I0, I1, I2, I3, I4) -> f2#(I1 + 1, I2, I0, I5, I6) [0 = I4 /\ 0 = I3 /\ -1 <= I1 - 1] f3#(I7, I8, I9, I10, I11) -> f3#(I7, I8, I9 - 1, I10 - 1, I10 - 1) [I10 = I11 /\ 0 <= I10 - 1] f2#(I22, I23, I24, I25, I26) -> f3#(I24, I22, I23, I24, I24) [0 <= I24 - 1 /\ I24 <= I23] R = init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) f3(I0, I1, I2, I3, I4) -> f2(I1 + 1, I2, I0, I5, I6) [0 = I4 /\ 0 = I3 /\ -1 <= I1 - 1] f3(I7, I8, I9, I10, I11) -> f3(I7, I8, I9 - 1, I10 - 1, I10 - 1) [I10 = I11 /\ 0 <= I10 - 1] f3(I12, I13, I14, I15, I16) -> f3(I12, I13, I14 + 1, I15 + 1, I15 + 1) [I15 = I16 /\ I15 <= 0 /\ 0 <= I15 - 1] f3(I17, I18, I19, I20, I21) -> f3(I17, I18, I19 + 1, I20 + 1, I20 + 1) [I20 = I21 /\ I20 <= 0 /\ I20 <= -1] f2(I22, I23, I24, I25, I26) -> f3(I24, I22, I23, I24, I24) [0 <= I24 - 1 /\ I24 <= I23] f1(I27, I28, I29, I30, I31) -> f2(0, I32, I33, I34, I35) [0 <= I27 - 1 /\ -1 <= I32 - 1 /\ -1 <= I28 - 1 /\ -1 <= I33 - 1] We use the extended value criterion with the projection function NU: NU[f2#(x0,x1,x2,x3,x4)] = x1 - x2 NU[f3#(x0,x1,x2,x3,x4)] = -x0 + x2 - x3 This gives the following inequalities: 0 = I4 /\ 0 = I3 /\ -1 <= I1 - 1 ==> -I0 + I2 - I3 >= I2 - I0 I10 = I11 /\ 0 <= I10 - 1 ==> -I7 + I9 - I10 >= -I7 + (I9 - 1) - (I10 - 1) 0 <= I24 - 1 /\ I24 <= I23 ==> I23 - I24 > -I24 + I23 - I24 with I23 - I24 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I0, I1, I2, I3, I4) -> f2#(I1 + 1, I2, I0, I5, I6) [0 = I4 /\ 0 = I3 /\ -1 <= I1 - 1] f3#(I7, I8, I9, I10, I11) -> f3#(I7, I8, I9 - 1, I10 - 1, I10 - 1) [I10 = I11 /\ 0 <= I10 - 1] R = init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) f3(I0, I1, I2, I3, I4) -> f2(I1 + 1, I2, I0, I5, I6) [0 = I4 /\ 0 = I3 /\ -1 <= I1 - 1] f3(I7, I8, I9, I10, I11) -> f3(I7, I8, I9 - 1, I10 - 1, I10 - 1) [I10 = I11 /\ 0 <= I10 - 1] f3(I12, I13, I14, I15, I16) -> f3(I12, I13, I14 + 1, I15 + 1, I15 + 1) [I15 = I16 /\ I15 <= 0 /\ 0 <= I15 - 1] f3(I17, I18, I19, I20, I21) -> f3(I17, I18, I19 + 1, I20 + 1, I20 + 1) [I20 = I21 /\ I20 <= 0 /\ I20 <= -1] f2(I22, I23, I24, I25, I26) -> f3(I24, I22, I23, I24, I24) [0 <= I24 - 1 /\ I24 <= I23] f1(I27, I28, I29, I30, I31) -> f2(0, I32, I33, I34, I35) [0 <= I27 - 1 /\ -1 <= I32 - 1 /\ -1 <= I28 - 1 /\ -1 <= I33 - 1] The dependency graph for this problem is: 1 -> 2 -> 1, 2 Where: 1) f3#(I0, I1, I2, I3, I4) -> f2#(I1 + 1, I2, I0, I5, I6) [0 = I4 /\ 0 = I3 /\ -1 <= I1 - 1] 2) f3#(I7, I8, I9, I10, I11) -> f3#(I7, I8, I9 - 1, I10 - 1, I10 - 1) [I10 = I11 /\ 0 <= I10 - 1] We have the following SCCs. { 2 } DP problem for innermost termination. P = f3#(I7, I8, I9, I10, I11) -> f3#(I7, I8, I9 - 1, I10 - 1, I10 - 1) [I10 = I11 /\ 0 <= I10 - 1] R = init(x1, x2, x3, x4, x5) -> f1(rnd1, rnd2, rnd3, rnd4, rnd5) popout

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all output return to Integ Trans Syste 27634