details

property	value
status	complete
benchmark	heidy10.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n149.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	1.07171 seconds
cpu usage	1.08779
user time	0.56372
system time	0.524067
max virtual memory	350876.0
max residence set size	8388.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = f5#(x1, x2) -> f4#(x1, x2) f4#(I0, I1) -> f3#(I0, I1) f3#(I2, I3) -> f1#(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2] f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0] f2#(I6, I7) -> f1#(I6, I7) f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9] R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f3(I0, I1) f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2] f1(I4, I5) -> f3(I4, I5) [I5 <= 0] f2(I6, I7) -> f1(I6, I7) f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 3, 5 3 -> 2 4 -> 3, 5 5 -> 4 Where: 0) f5#(x1, x2) -> f4#(x1, x2) 1) f4#(I0, I1) -> f3#(I0, I1) 2) f3#(I2, I3) -> f1#(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2] 3) f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0] 4) f2#(I6, I7) -> f1#(I6, I7) 5) f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9] We have the following SCCs. { 2, 3, 4, 5 } DP problem for innermost termination. P = f3#(I2, I3) -> f1#(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2] f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0] f2#(I6, I7) -> f1#(I6, I7) f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9] R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f3(I0, I1) f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2] f1(I4, I5) -> f3(I4, I5) [I5 <= 0] f2(I6, I7) -> f1(I6, I7) f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9] We use the basic value criterion with the projection function NU: NU[f2#(z1,z2)] = z1 NU[f1#(z1,z2)] = z1 NU[f3#(z1,z2)] = z1 This gives the following inequalities: rnd2 = rnd2 /\ 1 <= I2 ==> I2 >! -1 + I2 I5 <= 0 ==> I4 (>! \union =) I4 ==> I6 (>! \union =) I6 1 <= I9 ==> I8 (>! \union =) I8 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0] f2#(I6, I7) -> f1#(I6, I7) f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9] R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f3(I0, I1) f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2] f1(I4, I5) -> f3(I4, I5) [I5 <= 0] f2(I6, I7) -> f1(I6, I7) f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9] The dependency graph for this problem is: 3 -> 4 -> 3, 5 5 -> 4 Where: 3) f1#(I4, I5) -> f3#(I4, I5) [I5 <= 0] 4) f2#(I6, I7) -> f1#(I6, I7) 5) f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9] We have the following SCCs. { 4, 5 } DP problem for innermost termination. P = f2#(I6, I7) -> f1#(I6, I7) f1#(I8, I9) -> f2#(I8, -1 + I9) [1 <= I9] R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f3(I0, I1) f3(I2, I3) -> f1(-1 + I2, rnd2) [rnd2 = rnd2 /\ 1 <= I2] f1(I4, I5) -> f3(I4, I5) [I5 <= 0] f2(I6, I7) -> f1(I6, I7) f1(I8, I9) -> f2(I8, -1 + I9) [1 <= I9] popout

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