details

property	value
status	complete
benchmark	e-acqrel-fail.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n084.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	2.27223515511 seconds
cpu usage	2.421790684
max memory	1.8161664E7

stage attributes

key	value
output-size	3254
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) -> f9#(x1, x2, x3, x4) f9#(I0, I1, I2, I3) -> f4#(0, 0, rnd3, I3) [rnd3 = rnd3] f5#(I4, I5, I6, I7) -> f4#(I4, 0, I8, I7) [I7 <= 0 /\ y1 = 1 /\ I8 = I8] f5#(I9, I10, I11, I12) -> f2#(I9, I10, I11, I12) [1 <= I12] f8#(I13, I14, I15, I16) -> f3#(I13, I14, I15, I16) f3#(I17, I18, I19, I20) -> f8#(I17, I18, I19, I20) f2#(I25, I26, I27, I28) -> f5#(I25, I26, I27, I28) f4#(I29, I30, I31, I32) -> f1#(I29, I30, I31, I32) f1#(I33, I34, I35, I36) -> f3#(I33, I34, I35, I36) [1 <= I35] f1#(I37, I38, I39, I40) -> f2#(0, I38, I39, rnd4) [I39 <= 0 /\ I41 = 1 /\ rnd4 = rnd4 /\ 1 <= rnd4] R = f10(x1, x2, x3, x4) -> f9(x1, x2, x3, x4) f9(I0, I1, I2, I3) -> f4(0, 0, rnd3, I3) [rnd3 = rnd3] f5(I4, I5, I6, I7) -> f4(I4, 0, I8, I7) [I7 <= 0 /\ y1 = 1 /\ I8 = I8] f5(I9, I10, I11, I12) -> f2(I9, I10, I11, I12) [1 <= I12] f8(I13, I14, I15, I16) -> f3(I13, I14, I15, I16) f3(I17, I18, I19, I20) -> f8(I17, I18, I19, I20) f6(I21, I22, I23, I24) -> f7(I21, I22, I23, I24) f2(I25, I26, I27, I28) -> f5(I25, I26, I27, I28) f4(I29, I30, I31, I32) -> f1(I29, I30, I31, I32) f1(I33, I34, I35, I36) -> f3(I33, I34, I35, I36) [1 <= I35] f1(I37, I38, I39, I40) -> f2(0, I38, I39, rnd4) [I39 <= 0 /\ I41 = 1 /\ rnd4 = rnd4 /\ 1 <= rnd4] The dependency graph for this problem is: 0 -> 1 1 -> 7 2 -> 7 3 -> 6 4 -> 5 5 -> 4 6 -> 2, 3 7 -> 8, 9 8 -> 5 9 -> 6 Where: 0) f10#(x1, x2, x3, x4) -> f9#(x1, x2, x3, x4) 1) f9#(I0, I1, I2, I3) -> f4#(0, 0, rnd3, I3) [rnd3 = rnd3] 2) f5#(I4, I5, I6, I7) -> f4#(I4, 0, I8, I7) [I7 <= 0 /\ y1 = 1 /\ I8 = I8] 3) f5#(I9, I10, I11, I12) -> f2#(I9, I10, I11, I12) [1 <= I12] 4) f8#(I13, I14, I15, I16) -> f3#(I13, I14, I15, I16) 5) f3#(I17, I18, I19, I20) -> f8#(I17, I18, I19, I20) 6) f2#(I25, I26, I27, I28) -> f5#(I25, I26, I27, I28) 7) f4#(I29, I30, I31, I32) -> f1#(I29, I30, I31, I32) 8) f1#(I33, I34, I35, I36) -> f3#(I33, I34, I35, I36) [1 <= I35] 9) f1#(I37, I38, I39, I40) -> f2#(0, I38, I39, rnd4) [I39 <= 0 /\ I41 = 1 /\ rnd4 = rnd4 /\ 1 <= rnd4] We have the following SCCs. { 2, 3, 6, 7, 9 } { 4, 5 } DP problem for innermost termination. P = f8#(I13, I14, I15, I16) -> f3#(I13, I14, I15, I16) f3#(I17, I18, I19, I20) -> f8#(I17, I18, I19, I20) R = f10(x1, x2, x3, x4) -> f9(x1, x2, x3, x4) f9(I0, I1, I2, I3) -> f4(0, 0, rnd3, I3) [rnd3 = rnd3] f5(I4, I5, I6, I7) -> f4(I4, 0, I8, I7) [I7 <= 0 /\ y1 = 1 /\ I8 = I8] f5(I9, I10, I11, I12) -> f2(I9, I10, I11, I12) [1 <= I12] f8(I13, I14, I15, I16) -> f3(I13, I14, I15, I16) f3(I17, I18, I19, I20) -> f8(I17, I18, I19, I20) f6(I21, I22, I23, I24) -> f7(I21, I22, I23, I24) f2(I25, I26, I27, I28) -> f5(I25, I26, I27, I28) f4(I29, I30, I31, I32) -> f1(I29, I30, I31, I32) f1(I33, I34, I35, I36) -> f3(I33, I34, I35, I36) [1 <= I35] f1(I37, I38, I39, I40) -> f2(0, I38, I39, rnd4) [I39 <= 0 /\ I41 = 1 /\ rnd4 = rnd4 /\ 1 <= rnd4] popout

output may be truncated. 'popout' for the full output.

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