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Integer_Transition_Systems 2019-03-29 01.54 pair #432274992
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
n157.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
1.58773 seconds
cpu usage
1.18725
user time
0.570779
system time
0.61647
max virtual memory
720980.0
max residence set size
8400.0
stage attributes
key
value
starexec-result
MAYBE
output
1.12/1.58 MAYBE 1.12/1.58 1.12/1.58 DP problem for innermost termination. 1.12/1.58 P = 1.12/1.58 f5#(x1) -> f4#(x1) 1.12/1.58 f4#(I0) -> f1#(300) 1.12/1.58 f3#(I1) -> f1#(I1) 1.12/1.58 f2#(I2) -> f3#(I2) [101 <= I2] 1.12/1.58 f2#(I3) -> f3#(I3) [1 + I3 <= 100] 1.12/1.58 f1#(I4) -> f2#(-1 + I4) 1.12/1.58 R = 1.12/1.58 f5(x1) -> f4(x1) 1.12/1.58 f4(I0) -> f1(300) 1.12/1.58 f3(I1) -> f1(I1) 1.12/1.58 f2(I2) -> f3(I2) [101 <= I2] 1.12/1.58 f2(I3) -> f3(I3) [1 + I3 <= 100] 1.12/1.58 f1(I4) -> f2(-1 + I4) 1.12/1.58 1.12/1.58 The dependency graph for this problem is: 1.12/1.58 0 -> 1 1.12/1.58 1 -> 5 1.12/1.58 2 -> 5 1.12/1.58 3 -> 2 1.12/1.58 4 -> 2 1.12/1.58 5 -> 3, 4 1.12/1.58 Where: 1.12/1.58 0) f5#(x1) -> f4#(x1) 1.12/1.58 1) f4#(I0) -> f1#(300) 1.12/1.58 2) f3#(I1) -> f1#(I1) 1.12/1.58 3) f2#(I2) -> f3#(I2) [101 <= I2] 1.12/1.58 4) f2#(I3) -> f3#(I3) [1 + I3 <= 100] 1.12/1.58 5) f1#(I4) -> f2#(-1 + I4) 1.12/1.58 1.12/1.58 We have the following SCCs. 1.12/1.58 { 2, 3, 4, 5 } 1.12/1.58 1.12/1.58 DP problem for innermost termination. 1.12/1.58 P = 1.12/1.58 f3#(I1) -> f1#(I1) 1.12/1.58 f2#(I2) -> f3#(I2) [101 <= I2] 1.12/1.58 f2#(I3) -> f3#(I3) [1 + I3 <= 100] 1.12/1.58 f1#(I4) -> f2#(-1 + I4) 1.12/1.58 R = 1.12/1.58 f5(x1) -> f4(x1) 1.12/1.58 f4(I0) -> f1(300) 1.12/1.58 f3(I1) -> f1(I1) 1.12/1.58 f2(I2) -> f3(I2) [101 <= I2] 1.12/1.58 f2(I3) -> f3(I3) [1 + I3 <= 100] 1.12/1.58 f1(I4) -> f2(-1 + I4) 1.12/1.58 1.12/1.58 We use the extended value criterion with the projection function NU: 1.12/1.58 NU[f2#(x0)] = x0 1.12/1.58 NU[f1#(x0)] = x0 - 1 1.12/1.58 NU[f3#(x0)] = x0 - 1 1.12/1.58 1.12/1.58 This gives the following inequalities: 1.12/1.58 ==> I1 - 1 >= I1 - 1 1.12/1.58 101 <= I2 ==> I2 > I2 - 1 with I2 >= 0 1.12/1.58 1 + I3 <= 100 ==> I3 >= I3 - 1 1.12/1.58 ==> I4 - 1 >= (-1 + I4) 1.12/1.58 1.12/1.58 We remove all the strictly oriented dependency pairs. 1.12/1.58 1.12/1.58 DP problem for innermost termination. 1.12/1.58 P = 1.12/1.58 f3#(I1) -> f1#(I1) 1.12/1.58 f2#(I3) -> f3#(I3) [1 + I3 <= 100] 1.12/1.58 f1#(I4) -> f2#(-1 + I4) 1.12/1.58 R = 1.12/1.58 f5(x1) -> f4(x1) 1.12/1.58 f4(I0) -> f1(300) 1.12/1.58 f3(I1) -> f1(I1) 1.12/1.58 f2(I2) -> f3(I2) [101 <= I2] 1.12/1.58 f2(I3) -> f3(I3) [1 + I3 <= 100] 1.12/1.58 f1(I4) -> f2(-1 + I4) 1.12/1.58 1.12/4.56 EOF
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