details

property	value
status	complete
benchmark	consts1.t2_fixed.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n148.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	1.14337 seconds
cpu usage	1.1629
user time	0.569045
system time	0.593852
max virtual memory	127128.0
max residence set size	8564.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = f5#(x1) -> f4#(x1) f4#(I0) -> f1#(300) f3#(I1) -> f1#(I1) f2#(I2) -> f3#(I2) [101 <= I2] f2#(I3) -> f3#(I3) [1 + I3 <= 100] f1#(I4) -> f2#(-1 + I4) R = f5(x1) -> f4(x1) f4(I0) -> f1(300) f3(I1) -> f1(I1) f2(I2) -> f3(I2) [101 <= I2] f2(I3) -> f3(I3) [1 + I3 <= 100] f1(I4) -> f2(-1 + I4) The dependency graph for this problem is: 0 -> 1 1 -> 5 2 -> 5 3 -> 2 4 -> 2 5 -> 3, 4 Where: 0) f5#(x1) -> f4#(x1) 1) f4#(I0) -> f1#(300) 2) f3#(I1) -> f1#(I1) 3) f2#(I2) -> f3#(I2) [101 <= I2] 4) f2#(I3) -> f3#(I3) [1 + I3 <= 100] 5) f1#(I4) -> f2#(-1 + I4) We have the following SCCs. { 2, 3, 4, 5 } DP problem for innermost termination. P = f3#(I1) -> f1#(I1) f2#(I2) -> f3#(I2) [101 <= I2] f2#(I3) -> f3#(I3) [1 + I3 <= 100] f1#(I4) -> f2#(-1 + I4) R = f5(x1) -> f4(x1) f4(I0) -> f1(300) f3(I1) -> f1(I1) f2(I2) -> f3(I2) [101 <= I2] f2(I3) -> f3(I3) [1 + I3 <= 100] f1(I4) -> f2(-1 + I4) We use the extended value criterion with the projection function NU: NU[f2#(x0)] = x0 NU[f1#(x0)] = x0 - 1 NU[f3#(x0)] = x0 - 1 This gives the following inequalities: ==> I1 - 1 >= I1 - 1 101 <= I2 ==> I2 > I2 - 1 with I2 >= 0 1 + I3 <= 100 ==> I3 >= I3 - 1 ==> I4 - 1 >= (-1 + I4) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I1) -> f1#(I1) f2#(I3) -> f3#(I3) [1 + I3 <= 100] f1#(I4) -> f2#(-1 + I4) R = f5(x1) -> f4(x1) f4(I0) -> f1(300) f3(I1) -> f1(I1) f2(I2) -> f3(I2) [101 <= I2] f2(I3) -> f3(I3) [1 + I3 <= 100] f1(I4) -> f2(-1 + I4) popout

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