details

property	value
status	complete
benchmark	small12.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n091.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.925797939301 seconds
cpu usage	0.965648482
max memory	1.34144E7

stage attributes

key	value
output-size	2021
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 = f4#(x1, x2) -> f3#(x1, x2) f3#(I0, I1) -> f1#(I0, I1) f3#(I2, I3) -> f2#(I2, -1 + I3) f2#(I4, I5) -> f1#(-1 + I4, I5) f1#(I6, I7) -> f2#(I6, I7) [1 <= I7 /\ 1 <= I6] R = f4(x1, x2) -> f3(x1, x2) f3(I0, I1) -> f1(I0, I1) f3(I2, I3) -> f2(I2, -1 + I3) f2(I4, I5) -> f1(-1 + I4, I5) f1(I6, I7) -> f2(I6, I7) [1 <= I7 /\ 1 <= I6] The dependency graph for this problem is: 0 -> 1, 2 1 -> 4 2 -> 3 3 -> 4 4 -> 3 Where: 0) f4#(x1, x2) -> f3#(x1, x2) 1) f3#(I0, I1) -> f1#(I0, I1) 2) f3#(I2, I3) -> f2#(I2, -1 + I3) 3) f2#(I4, I5) -> f1#(-1 + I4, I5) 4) f1#(I6, I7) -> f2#(I6, I7) [1 <= I7 /\ 1 <= I6] We have the following SCCs. { 3, 4 } DP problem for innermost termination. P = f2#(I4, I5) -> f1#(-1 + I4, I5) f1#(I6, I7) -> f2#(I6, I7) [1 <= I7 /\ 1 <= I6] R = f4(x1, x2) -> f3(x1, x2) f3(I0, I1) -> f1(I0, I1) f3(I2, I3) -> f2(I2, -1 + I3) f2(I4, I5) -> f1(-1 + I4, I5) f1(I6, I7) -> f2(I6, I7) [1 <= I7 /\ 1 <= I6] We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2)] = z1 + -1 * 1 NU[f2#(z1,z2)] = -1 + z1 + -1 * 1 This gives the following inequalities: ==> -1 + I4 + -1 * 1 >= -1 + I4 + -1 * 1 1 <= I7 /\ 1 <= I6 ==> I6 + -1 * 1 > -1 + I6 + -1 * 1 with I6 + -1 * 1 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I4, I5) -> f1#(-1 + I4, I5) R = f4(x1, x2) -> f3(x1, x2) f3(I0, I1) -> f1(I0, I1) f3(I2, I3) -> f2(I2, -1 + I3) f2(I4, I5) -> f1(-1 + I4, I5) f1(I6, I7) -> f2(I6, I7) [1 <= I7 /\ 1 <= I6] The dependency graph for this problem is: 3 -> Where: 3) f2#(I4, I5) -> f1#(-1 + I4, I5) We have the following SCCs. popout

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all output return to Integ Trans Syste 27634