details

property	value
status	complete
benchmark	heidy5.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n141.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	0.798043 seconds
cpu usage	0.814035
user time	0.434929
system time	0.379106
max virtual memory	720860.0
max residence set size	8392.0

stage attributes

key	value
starexec-result	YES

output

YES DP problem for innermost termination. P = f7#(x1, x2, x3) -> f6#(x1, x2, x3) f6#(I0, I1, I2) -> f4#(0, I1, I2) f4#(I6, I7, I8) -> f3#(I6, I7, I8) [I7 <= 0] f3#(I9, I10, I11) -> f1#(I9, I10, I11) f1#(I12, I13, I14) -> f3#(I12, I13, -1 + I14) [1 <= I14] R = f7(x1, x2, x3) -> f6(x1, x2, x3) f6(I0, I1, I2) -> f4(0, I1, I2) f4(I3, I4, I5) -> f5(1 + I3, I4, I5) [1 <= I4] f4(I6, I7, I8) -> f3(I6, I7, I8) [I7 <= 0] f3(I9, I10, I11) -> f1(I9, I10, I11) f1(I12, I13, I14) -> f3(I12, I13, -1 + I14) [1 <= I14] f1(I15, I16, I17) -> f2(1, I16, I17) [I16 <= 0 /\ I17 <= 0] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 3 3 -> 4 4 -> 3 Where: 0) f7#(x1, x2, x3) -> f6#(x1, x2, x3) 1) f6#(I0, I1, I2) -> f4#(0, I1, I2) 2) f4#(I6, I7, I8) -> f3#(I6, I7, I8) [I7 <= 0] 3) f3#(I9, I10, I11) -> f1#(I9, I10, I11) 4) f1#(I12, I13, I14) -> f3#(I12, I13, -1 + I14) [1 <= I14] We have the following SCCs. { 3, 4 } DP problem for innermost termination. P = f3#(I9, I10, I11) -> f1#(I9, I10, I11) f1#(I12, I13, I14) -> f3#(I12, I13, -1 + I14) [1 <= I14] R = f7(x1, x2, x3) -> f6(x1, x2, x3) f6(I0, I1, I2) -> f4(0, I1, I2) f4(I3, I4, I5) -> f5(1 + I3, I4, I5) [1 <= I4] f4(I6, I7, I8) -> f3(I6, I7, I8) [I7 <= 0] f3(I9, I10, I11) -> f1(I9, I10, I11) f1(I12, I13, I14) -> f3(I12, I13, -1 + I14) [1 <= I14] f1(I15, I16, I17) -> f2(1, I16, I17) [I16 <= 0 /\ I17 <= 0] We use the basic value criterion with the projection function NU: NU[f1#(z1,z2,z3)] = z3 NU[f3#(z1,z2,z3)] = z3 This gives the following inequalities: ==> I11 (>! \union =) I11 1 <= I14 ==> I14 >! -1 + I14 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I9, I10, I11) -> f1#(I9, I10, I11) R = f7(x1, x2, x3) -> f6(x1, x2, x3) f6(I0, I1, I2) -> f4(0, I1, I2) f4(I3, I4, I5) -> f5(1 + I3, I4, I5) [1 <= I4] f4(I6, I7, I8) -> f3(I6, I7, I8) [I7 <= 0] f3(I9, I10, I11) -> f1(I9, I10, I11) f1(I12, I13, I14) -> f3(I12, I13, -1 + I14) [1 <= I14] f1(I15, I16, I17) -> f2(1, I16, I17) [I16 <= 0 /\ I17 <= 0] The dependency graph for this problem is: 3 -> Where: 3) f3#(I9, I10, I11) -> f1#(I9, I10, I11) We have the following SCCs. popout

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job log

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actions

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