details

property	value
status	complete
benchmark	small28.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n094.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	1.22289 seconds
cpu usage	1.24525
user time	0.644777
system time	0.600469
max virtual memory	353336.0
max residence set size	8672.0

stage attributes

key	value
starexec-result	YES

output

1.22/1.22 YES 1.22/1.22 1.22/1.22 DP problem for innermost termination. 1.22/1.22 P = 1.22/1.22 f4#(x1, x2) -> f3#(x1, x2) 1.22/1.22 f3#(I0, I1) -> f1#(I0, I1) 1.22/1.22 f2#(I2, I3) -> f1#(I2, I3) 1.22/1.22 f1#(I4, I5) -> f2#(1 + I4, I5) [I5 <= I4 /\ I4 <= I5] 1.22/1.22 R = 1.22/1.22 f4(x1, x2) -> f3(x1, x2) 1.22/1.22 f3(I0, I1) -> f1(I0, I1) 1.22/1.22 f2(I2, I3) -> f1(I2, I3) 1.22/1.22 f1(I4, I5) -> f2(1 + I4, I5) [I5 <= I4 /\ I4 <= I5] 1.22/1.22 1.22/1.22 The dependency graph for this problem is: 1.22/1.22 0 -> 1 1.22/1.22 1 -> 3 1.22/1.22 2 -> 3 1.22/1.22 3 -> 2 1.22/1.22 Where: 1.22/1.22 0) f4#(x1, x2) -> f3#(x1, x2) 1.22/1.22 1) f3#(I0, I1) -> f1#(I0, I1) 1.22/1.22 2) f2#(I2, I3) -> f1#(I2, I3) 1.22/1.22 3) f1#(I4, I5) -> f2#(1 + I4, I5) [I5 <= I4 /\ I4 <= I5] 1.22/1.22 1.22/1.22 We have the following SCCs. 1.22/1.22 { 2, 3 } 1.22/1.22 1.22/1.22 DP problem for innermost termination. 1.22/1.22 P = 1.22/1.22 f2#(I2, I3) -> f1#(I2, I3) 1.22/1.22 f1#(I4, I5) -> f2#(1 + I4, I5) [I5 <= I4 /\ I4 <= I5] 1.22/1.22 R = 1.22/1.22 f4(x1, x2) -> f3(x1, x2) 1.22/1.22 f3(I0, I1) -> f1(I0, I1) 1.22/1.22 f2(I2, I3) -> f1(I2, I3) 1.22/1.22 f1(I4, I5) -> f2(1 + I4, I5) [I5 <= I4 /\ I4 <= I5] 1.22/1.22 1.22/1.22 We use the reverse value criterion with the projection function NU: 1.22/1.22 NU[f1#(z1,z2)] = z2 + -1 * z1 1.22/1.22 NU[f2#(z1,z2)] = z2 + -1 * z1 1.22/1.22 1.22/1.22 This gives the following inequalities: 1.22/1.22 ==> I3 + -1 * I2 >= I3 + -1 * I2 1.22/1.22 I5 <= I4 /\ I4 <= I5 ==> I5 + -1 * I4 > I5 + -1 * (1 + I4) with I5 + -1 * I4 >= 0 1.22/1.22 1.22/1.22 We remove all the strictly oriented dependency pairs. 1.22/1.22 1.22/1.22 DP problem for innermost termination. 1.22/1.22 P = 1.22/1.22 f2#(I2, I3) -> f1#(I2, I3) 1.22/1.22 R = 1.22/1.22 f4(x1, x2) -> f3(x1, x2) 1.22/1.22 f3(I0, I1) -> f1(I0, I1) 1.22/1.22 f2(I2, I3) -> f1(I2, I3) 1.22/1.22 f1(I4, I5) -> f2(1 + I4, I5) [I5 <= I4 /\ I4 <= I5] 1.22/1.22 1.22/1.22 The dependency graph for this problem is: 1.22/1.22 2 -> 1.22/1.22 Where: 1.22/1.22 2) f2#(I2, I3) -> f1#(I2, I3) 1.22/1.22 1.22/1.22 We have the following SCCs. 1.22/1.22 1.22/4.20 EOF popout

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all output return to Integer_Transition_Systems 2019-03-29 01.54