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	n144.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	2.53272 seconds
cpu usage	2.57045
user time	1.29076
system time	1.27969
max virtual memory	689884.0
max residence set size	8560.0

stage attributes

key	value
starexec-result	MAYBE

output

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.

job log

popout

actions

all output return to ITS