details

property	value
status	complete
benchmark	heidy8.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n137.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	6.11604 seconds
cpu usage	6.18523
user time	3.34465
system time	2.84058
max virtual memory	720348.0
max residence set size	8640.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = f6#(x1, x2, x3, x4, x5) -> f5#(x1, x2, x3, x4, x5) f5#(I0, I1, I2, I3, I4) -> f4#(I0, I1, 0, 0, I4) [y1 = 0 /\ y2 = 1] f4#(I5, I6, I7, I8, I9) -> f3#(rnd1, I6, I7, I8, I9) [rnd1 = rnd1 /\ 1 <= I9] f4#(I10, I11, I12, I13, I14) -> f2#(I10, I11, I12, I13, I14) [I14 <= 0] f3#(I15, I16, I17, I18, I19) -> f4#(I15, I16, I17, I18, -1 + I19) [1 <= I15] f3#(I20, I21, I22, I23, I24) -> f4#(I20, I21, I22, I23, -2 + I24) [I20 <= 0] f2#(I25, I26, I27, I28, I29) -> f1#(I25, rnd2, I27, I28, I29) [rnd2 = rnd2] f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, 1, I33, I34) [1 <= I31] f1#(I35, I36, I37, I38, I39) -> f2#(I35, I36, 0, I38, I39) [I36 <= 0] R = f6(x1, x2, x3, x4, x5) -> f5(x1, x2, x3, x4, x5) f5(I0, I1, I2, I3, I4) -> f4(I0, I1, 0, 0, I4) [y1 = 0 /\ y2 = 1] f4(I5, I6, I7, I8, I9) -> f3(rnd1, I6, I7, I8, I9) [rnd1 = rnd1 /\ 1 <= I9] f4(I10, I11, I12, I13, I14) -> f2(I10, I11, I12, I13, I14) [I14 <= 0] f3(I15, I16, I17, I18, I19) -> f4(I15, I16, I17, I18, -1 + I19) [1 <= I15] f3(I20, I21, I22, I23, I24) -> f4(I20, I21, I22, I23, -2 + I24) [I20 <= 0] f2(I25, I26, I27, I28, I29) -> f1(I25, rnd2, I27, I28, I29) [rnd2 = rnd2] f1(I30, I31, I32, I33, I34) -> f2(I30, I31, 1, I33, I34) [1 <= I31] f1(I35, I36, I37, I38, I39) -> f2(I35, I36, 0, I38, I39) [I36 <= 0] The dependency graph for this problem is: 0 -> 1 1 -> 2, 3 2 -> 4, 5 3 -> 6 4 -> 2, 3 5 -> 2, 3 6 -> 7, 8 7 -> 6 8 -> 6 Where: 0) f6#(x1, x2, x3, x4, x5) -> f5#(x1, x2, x3, x4, x5) 1) f5#(I0, I1, I2, I3, I4) -> f4#(I0, I1, 0, 0, I4) [y1 = 0 /\ y2 = 1] 2) f4#(I5, I6, I7, I8, I9) -> f3#(rnd1, I6, I7, I8, I9) [rnd1 = rnd1 /\ 1 <= I9] 3) f4#(I10, I11, I12, I13, I14) -> f2#(I10, I11, I12, I13, I14) [I14 <= 0] 4) f3#(I15, I16, I17, I18, I19) -> f4#(I15, I16, I17, I18, -1 + I19) [1 <= I15] 5) f3#(I20, I21, I22, I23, I24) -> f4#(I20, I21, I22, I23, -2 + I24) [I20 <= 0] 6) f2#(I25, I26, I27, I28, I29) -> f1#(I25, rnd2, I27, I28, I29) [rnd2 = rnd2] 7) f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, 1, I33, I34) [1 <= I31] 8) f1#(I35, I36, I37, I38, I39) -> f2#(I35, I36, 0, I38, I39) [I36 <= 0] We have the following SCCs. { 2, 4, 5 } { 6, 7, 8 } DP problem for innermost termination. P = f2#(I25, I26, I27, I28, I29) -> f1#(I25, rnd2, I27, I28, I29) [rnd2 = rnd2] f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, 1, I33, I34) [1 <= I31] f1#(I35, I36, I37, I38, I39) -> f2#(I35, I36, 0, I38, I39) [I36 <= 0] R = f6(x1, x2, x3, x4, x5) -> f5(x1, x2, x3, x4, x5) f5(I0, I1, I2, I3, I4) -> f4(I0, I1, 0, 0, I4) [y1 = 0 /\ y2 = 1] f4(I5, I6, I7, I8, I9) -> f3(rnd1, I6, I7, I8, I9) [rnd1 = rnd1 /\ 1 <= I9] f4(I10, I11, I12, I13, I14) -> f2(I10, I11, I12, I13, I14) [I14 <= 0] f3(I15, I16, I17, I18, I19) -> f4(I15, I16, I17, I18, -1 + I19) [1 <= I15] f3(I20, I21, I22, I23, I24) -> f4(I20, I21, I22, I23, -2 + I24) [I20 <= 0] f2(I25, I26, I27, I28, I29) -> f1(I25, rnd2, I27, I28, I29) [rnd2 = rnd2] f1(I30, I31, I32, I33, I34) -> f2(I30, I31, 1, I33, I34) [1 <= I31] f1(I35, I36, I37, I38, I39) -> f2(I35, I36, 0, I38, I39) [I36 <= 0] popout

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

job log

popout

actions

all output return to ITS