details

property	value
status	complete
benchmark	p-49.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n151.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.383439064026 seconds
cpu usage	0.375095853
max memory	9883648.0

stage attributes

key	value
output-size	2051
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 = f5#(x1, x2) -> f4#(x1, x2) f4#(I0, I1) -> f1#(I0, I1) f2#(I4, I5) -> f1#(I4, I5) f1#(I6, I7) -> f2#(I6, rnd2) [0 <= I7 /\ y1 = 1 + I7 /\ rnd2 = -2 + y1] R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f1(I0, I1) f1(I2, I3) -> f3(rnd1, I3) [rnd1 = rnd1 /\ 1 + I3 <= 0] f2(I4, I5) -> f1(I4, I5) f1(I6, I7) -> f2(I6, rnd2) [0 <= I7 /\ y1 = 1 + I7 /\ rnd2 = -2 + y1] The dependency graph for this problem is: 0 -> 1 1 -> 3 2 -> 3 3 -> 2 Where: 0) f5#(x1, x2) -> f4#(x1, x2) 1) f4#(I0, I1) -> f1#(I0, I1) 2) f2#(I4, I5) -> f1#(I4, I5) 3) f1#(I6, I7) -> f2#(I6, rnd2) [0 <= I7 /\ y1 = 1 + I7 /\ rnd2 = -2 + y1] We have the following SCCs. { 2, 3 } DP problem for innermost termination. P = f2#(I4, I5) -> f1#(I4, I5) f1#(I6, I7) -> f2#(I6, rnd2) [0 <= I7 /\ y1 = 1 + I7 /\ rnd2 = -2 + y1] R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f1(I0, I1) f1(I2, I3) -> f3(rnd1, I3) [rnd1 = rnd1 /\ 1 + I3 <= 0] f2(I4, I5) -> f1(I4, I5) f1(I6, I7) -> f2(I6, rnd2) [0 <= I7 /\ y1 = 1 + I7 /\ rnd2 = -2 + y1] We use the basic value criterion with the projection function NU: NU[f1#(z1,z2)] = z2 NU[f2#(z1,z2)] = z2 This gives the following inequalities: ==> I5 (>! \union =) I5 0 <= I7 /\ y1 = 1 + I7 /\ rnd2 = -2 + y1 ==> I7 >! rnd2 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I4, I5) -> f1#(I4, I5) R = f5(x1, x2) -> f4(x1, x2) f4(I0, I1) -> f1(I0, I1) f1(I2, I3) -> f3(rnd1, I3) [rnd1 = rnd1 /\ 1 + I3 <= 0] f2(I4, I5) -> f1(I4, I5) f1(I6, I7) -> f2(I6, rnd2) [0 <= I7 /\ y1 = 1 + I7 /\ rnd2 = -2 + y1] The dependency graph for this problem is: 2 -> Where: 2) f2#(I4, I5) -> f1#(I4, I5) We have the following SCCs. popout

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