details

property	value
status	complete
benchmark	n-15.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)	1.5155 seconds
cpu usage	1.54307
user time	0.772843
system time	0.770232
max virtual memory	723560.0
max residence set size	8564.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = f5#(x1) -> f1#(x1) f4#(I0) -> f2#(I0) f2#(I1) -> f4#(-1 + I1) [0 <= -1 + -1 + I1] f3#(I2) -> f2#(I2) f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0] f1#(I4) -> f2#(I4) R = f5(x1) -> f1(x1) f4(I0) -> f2(I0) f2(I1) -> f4(-1 + I1) [0 <= -1 + -1 + I1] f3(I2) -> f2(I2) f2(I3) -> f3(-1 + I3) [-1 + I3 <= 0] f1(I4) -> f2(I4) The dependency graph for this problem is: 0 -> 5 1 -> 2, 4 2 -> 1 3 -> 2, 4 4 -> 3 5 -> 2, 4 Where: 0) f5#(x1) -> f1#(x1) 1) f4#(I0) -> f2#(I0) 2) f2#(I1) -> f4#(-1 + I1) [0 <= -1 + -1 + I1] 3) f3#(I2) -> f2#(I2) 4) f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0] 5) f1#(I4) -> f2#(I4) We have the following SCCs. { 1, 2, 3, 4 } DP problem for innermost termination. P = f4#(I0) -> f2#(I0) f2#(I1) -> f4#(-1 + I1) [0 <= -1 + -1 + I1] f3#(I2) -> f2#(I2) f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0] R = f5(x1) -> f1(x1) f4(I0) -> f2(I0) f2(I1) -> f4(-1 + I1) [0 <= -1 + -1 + I1] f3(I2) -> f2(I2) f2(I3) -> f3(-1 + I3) [-1 + I3 <= 0] f1(I4) -> f2(I4) We use the reverse value criterion with the projection function NU: NU[f3#(z1)] = -1 + -1 + z1 + -1 * 0 NU[f2#(z1)] = -1 + -1 + z1 + -1 * 0 NU[f4#(z1)] = -1 + -1 + z1 + -1 * 0 This gives the following inequalities: ==> -1 + -1 + I0 + -1 * 0 >= -1 + -1 + I0 + -1 * 0 0 <= -1 + -1 + I1 ==> -1 + -1 + I1 + -1 * 0 > -1 + -1 + (-1 + I1) + -1 * 0 with -1 + -1 + I1 + -1 * 0 >= 0 ==> -1 + -1 + I2 + -1 * 0 >= -1 + -1 + I2 + -1 * 0 -1 + I3 <= 0 ==> -1 + -1 + I3 + -1 * 0 >= -1 + -1 + (-1 + I3) + -1 * 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f4#(I0) -> f2#(I0) f3#(I2) -> f2#(I2) f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0] R = f5(x1) -> f1(x1) f4(I0) -> f2(I0) f2(I1) -> f4(-1 + I1) [0 <= -1 + -1 + I1] f3(I2) -> f2(I2) f2(I3) -> f3(-1 + I3) [-1 + I3 <= 0] f1(I4) -> f2(I4) The dependency graph for this problem is: 1 -> 4 3 -> 4 4 -> 3 Where: 1) f4#(I0) -> f2#(I0) 3) f3#(I2) -> f2#(I2) 4) f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0] We have the following SCCs. { 3, 4 } DP problem for innermost termination. P = f3#(I2) -> f2#(I2) f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0] R = f5(x1) -> f1(x1) f4(I0) -> f2(I0) f2(I1) -> f4(-1 + I1) [0 <= -1 + -1 + I1] f3(I2) -> f2(I2) f2(I3) -> f3(-1 + I3) [-1 + I3 <= 0] f1(I4) -> f2(I4) popout

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