details

property	value
status	complete
benchmark	CyclicList.jar-obl-9.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n076.star.cs.uiowa.edu
space	From_AProVE_2014

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.988296985626 seconds
cpu usage	1.04774197
max memory	1.175552E7

stage attributes

key	value
output-size	7211
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) -> f1#(rnd1, rnd2, rnd3, rnd4) f4#(I0, I1, I2, I3) -> f4#(I0 - 1, 1, I4, I5) [0 <= I0 - 1 /\ y1 <= I1 - 1 /\ -1 <= I1 - 1] f4#(I6, I7, I8, I9) -> f4#(I6 - 1, I10, I11, I12) [-1 <= I7 - 1 /\ 0 <= I13 - 1 /\ I13 <= I7 - 1 /\ 0 <= I6 - 1 /\ I13 <= I10 - 1] f3#(I14, I15, I16, I17) -> f4#(I18, I15, I19, I20) [0 <= I16 - 1 /\ -1 <= I21 - 1 /\ 0 <= I14 - 1 /\ I21 - 1 = I18] f2#(I22, I23, I24, I25) -> f2#(I26, I23 - 1, I27, I25) [0 <= I26 - 1 /\ 0 <= I22 - 1 /\ 0 <= I23 - 1 /\ I26 <= I22] f2#(I28, I29, I30, I31) -> f3#(I32, 1, I31, I33) [0 <= I32 - 1 /\ 0 <= I28 - 1 /\ I29 <= 0 /\ I32 <= I28] f2#(I34, I35, I36, I37) -> f3#(I38, I39, I37, I40) [0 <= I38 - 1 /\ 0 <= I34 - 1 /\ I38 <= I34 /\ I36 <= I39 - 1 /\ 0 <= I36 - 1 /\ I35 <= 0] f1#(I41, I42, I43, I44) -> f2#(I45, I42 - 1, 0, I42) [0 <= I45 - 1 /\ 0 <= I41 - 1 /\ -1 <= I42 - 1 /\ I45 <= I41] R = init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) f4(I0, I1, I2, I3) -> f4(I0 - 1, 1, I4, I5) [0 <= I0 - 1 /\ y1 <= I1 - 1 /\ -1 <= I1 - 1] f4(I6, I7, I8, I9) -> f4(I6 - 1, I10, I11, I12) [-1 <= I7 - 1 /\ 0 <= I13 - 1 /\ I13 <= I7 - 1 /\ 0 <= I6 - 1 /\ I13 <= I10 - 1] f3(I14, I15, I16, I17) -> f4(I18, I15, I19, I20) [0 <= I16 - 1 /\ -1 <= I21 - 1 /\ 0 <= I14 - 1 /\ I21 - 1 = I18] f2(I22, I23, I24, I25) -> f2(I26, I23 - 1, I27, I25) [0 <= I26 - 1 /\ 0 <= I22 - 1 /\ 0 <= I23 - 1 /\ I26 <= I22] f2(I28, I29, I30, I31) -> f3(I32, 1, I31, I33) [0 <= I32 - 1 /\ 0 <= I28 - 1 /\ I29 <= 0 /\ I32 <= I28] f2(I34, I35, I36, I37) -> f3(I38, I39, I37, I40) [0 <= I38 - 1 /\ 0 <= I34 - 1 /\ I38 <= I34 /\ I36 <= I39 - 1 /\ 0 <= I36 - 1 /\ I35 <= 0] f1(I41, I42, I43, I44) -> f2(I45, I42 - 1, 0, I42) [0 <= I45 - 1 /\ 0 <= I41 - 1 /\ -1 <= I42 - 1 /\ I45 <= I41] The dependency graph for this problem is: 0 -> 7 1 -> 1 2 -> 1, 2 3 -> 1, 2 4 -> 4, 5, 6 5 -> 3 6 -> 3 7 -> 4, 5 Where: 0) init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 1) f4#(I0, I1, I2, I3) -> f4#(I0 - 1, 1, I4, I5) [0 <= I0 - 1 /\ y1 <= I1 - 1 /\ -1 <= I1 - 1] 2) f4#(I6, I7, I8, I9) -> f4#(I6 - 1, I10, I11, I12) [-1 <= I7 - 1 /\ 0 <= I13 - 1 /\ I13 <= I7 - 1 /\ 0 <= I6 - 1 /\ I13 <= I10 - 1] 3) f3#(I14, I15, I16, I17) -> f4#(I18, I15, I19, I20) [0 <= I16 - 1 /\ -1 <= I21 - 1 /\ 0 <= I14 - 1 /\ I21 - 1 = I18] 4) f2#(I22, I23, I24, I25) -> f2#(I26, I23 - 1, I27, I25) [0 <= I26 - 1 /\ 0 <= I22 - 1 /\ 0 <= I23 - 1 /\ I26 <= I22] 5) f2#(I28, I29, I30, I31) -> f3#(I32, 1, I31, I33) [0 <= I32 - 1 /\ 0 <= I28 - 1 /\ I29 <= 0 /\ I32 <= I28] 6) f2#(I34, I35, I36, I37) -> f3#(I38, I39, I37, I40) [0 <= I38 - 1 /\ 0 <= I34 - 1 /\ I38 <= I34 /\ I36 <= I39 - 1 /\ 0 <= I36 - 1 /\ I35 <= 0] 7) f1#(I41, I42, I43, I44) -> f2#(I45, I42 - 1, 0, I42) [0 <= I45 - 1 /\ 0 <= I41 - 1 /\ -1 <= I42 - 1 /\ I45 <= I41] We have the following SCCs. { 4 } { 2 } { 1 } DP problem for innermost termination. P = f4#(I0, I1, I2, I3) -> f4#(I0 - 1, 1, I4, I5) [0 <= I0 - 1 /\ y1 <= I1 - 1 /\ -1 <= I1 - 1] R = init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) f4(I0, I1, I2, I3) -> f4(I0 - 1, 1, I4, I5) [0 <= I0 - 1 /\ y1 <= I1 - 1 /\ -1 <= I1 - 1] f4(I6, I7, I8, I9) -> f4(I6 - 1, I10, I11, I12) [-1 <= I7 - 1 /\ 0 <= I13 - 1 /\ I13 <= I7 - 1 /\ 0 <= I6 - 1 /\ I13 <= I10 - 1] f3(I14, I15, I16, I17) -> f4(I18, I15, I19, I20) [0 <= I16 - 1 /\ -1 <= I21 - 1 /\ 0 <= I14 - 1 /\ I21 - 1 = I18] f2(I22, I23, I24, I25) -> f2(I26, I23 - 1, I27, I25) [0 <= I26 - 1 /\ 0 <= I22 - 1 /\ 0 <= I23 - 1 /\ I26 <= I22] f2(I28, I29, I30, I31) -> f3(I32, 1, I31, I33) [0 <= I32 - 1 /\ 0 <= I28 - 1 /\ I29 <= 0 /\ I32 <= I28] f2(I34, I35, I36, I37) -> f3(I38, I39, I37, I40) [0 <= I38 - 1 /\ 0 <= I34 - 1 /\ I38 <= I34 /\ I36 <= I39 - 1 /\ 0 <= I36 - 1 /\ I35 <= 0] f1(I41, I42, I43, I44) -> f2(I45, I42 - 1, 0, I42) [0 <= I45 - 1 /\ 0 <= I41 - 1 /\ -1 <= I42 - 1 /\ I45 <= I41] We use the basic value criterion with the projection function NU: NU[f4#(z1,z2,z3,z4)] = z1 This gives the following inequalities: 0 <= I0 - 1 /\ y1 <= I1 - 1 /\ -1 <= I1 - 1 ==> I0 >! I0 - 1 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f4#(I6, I7, I8, I9) -> f4#(I6 - 1, I10, I11, I12) [-1 <= I7 - 1 /\ 0 <= I13 - 1 /\ I13 <= I7 - 1 /\ 0 <= I6 - 1 /\ I13 <= I10 - 1] R = init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) f4(I0, I1, I2, I3) -> f4(I0 - 1, 1, I4, I5) [0 <= I0 - 1 /\ y1 <= I1 - 1 /\ -1 <= I1 - 1] f4(I6, I7, I8, I9) -> f4(I6 - 1, I10, I11, I12) [-1 <= I7 - 1 /\ 0 <= I13 - 1 /\ I13 <= I7 - 1 /\ 0 <= I6 - 1 /\ I13 <= I10 - 1] f3(I14, I15, I16, I17) -> f4(I18, I15, I19, I20) [0 <= I16 - 1 /\ -1 <= I21 - 1 /\ 0 <= I14 - 1 /\ I21 - 1 = I18] f2(I22, I23, I24, I25) -> f2(I26, I23 - 1, I27, I25) [0 <= I26 - 1 /\ 0 <= I22 - 1 /\ 0 <= I23 - 1 /\ I26 <= I22] f2(I28, I29, I30, I31) -> f3(I32, 1, I31, I33) [0 <= I32 - 1 /\ 0 <= I28 - 1 /\ I29 <= 0 /\ I32 <= I28] f2(I34, I35, I36, I37) -> f3(I38, I39, I37, I40) [0 <= I38 - 1 /\ 0 <= I34 - 1 /\ I38 <= I34 /\ I36 <= I39 - 1 /\ 0 <= I36 - 1 /\ I35 <= 0] f1(I41, I42, I43, I44) -> f2(I45, I42 - 1, 0, I42) [0 <= I45 - 1 /\ 0 <= I41 - 1 /\ -1 <= I42 - 1 /\ I45 <= I41] We use the basic value criterion with the projection function NU: NU[f4#(z1,z2,z3,z4)] = z1 This gives the following inequalities: -1 <= I7 - 1 /\ 0 <= I13 - 1 /\ I13 <= I7 - 1 /\ 0 <= I6 - 1 /\ I13 <= I10 - 1 ==> I6 >! I6 - 1 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = f2#(I22, I23, I24, I25) -> f2#(I26, I23 - 1, I27, I25) [0 <= I26 - 1 /\ 0 <= I22 - 1 /\ 0 <= I23 - 1 /\ I26 <= I22] R = init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) f4(I0, I1, I2, I3) -> f4(I0 - 1, 1, I4, I5) [0 <= I0 - 1 /\ y1 <= I1 - 1 /\ -1 <= I1 - 1] popout

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