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Integer_Transition_Systems 2019-03-29 01.54 pair #432273129
details
property
value
status
complete
benchmark
Continue1.jar-obl-8.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n103.star.cs.uiowa.edu
space
From_AProVE_2014
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.381273 seconds
cpu usage
0.388905
user time
0.215174
system time
0.173731
max virtual memory
113176.0
max residence set size
8000.0
stage attributes
key
value
starexec-result
YES
output
0.00/0.38 YES 0.00/0.38 0.00/0.38 DP problem for innermost termination. 0.00/0.38 P = 0.00/0.38 init#(x1) -> f1#(rnd1) 0.00/0.38 f2#(I0) -> f2#(I0 + 1) [I0 <= 19 /\ I0 <= 9] 0.00/0.38 f2#(I1) -> f2#(I1 + 1) [I1 <= 19 /\ 9 <= I1 - 1] 0.00/0.38 f1#(I2) -> f2#(0) 0.00/0.38 R = 0.00/0.38 init(x1) -> f1(rnd1) 0.00/0.38 f2(I0) -> f2(I0 + 1) [I0 <= 19 /\ I0 <= 9] 0.00/0.38 f2(I1) -> f2(I1 + 1) [I1 <= 19 /\ 9 <= I1 - 1] 0.00/0.38 f1(I2) -> f2(0) 0.00/0.38 0.00/0.38 The dependency graph for this problem is: 0.00/0.38 0 -> 3 0.00/0.38 1 -> 1, 2 0.00/0.38 2 -> 2 0.00/0.38 3 -> 1 0.00/0.38 Where: 0.00/0.38 0) init#(x1) -> f1#(rnd1) 0.00/0.38 1) f2#(I0) -> f2#(I0 + 1) [I0 <= 19 /\ I0 <= 9] 0.00/0.38 2) f2#(I1) -> f2#(I1 + 1) [I1 <= 19 /\ 9 <= I1 - 1] 0.00/0.38 3) f1#(I2) -> f2#(0) 0.00/0.38 0.00/0.38 We have the following SCCs. 0.00/0.38 { 1 } 0.00/0.38 { 2 } 0.00/0.38 0.00/0.38 DP problem for innermost termination. 0.00/0.38 P = 0.00/0.38 f2#(I1) -> f2#(I1 + 1) [I1 <= 19 /\ 9 <= I1 - 1] 0.00/0.38 R = 0.00/0.38 init(x1) -> f1(rnd1) 0.00/0.38 f2(I0) -> f2(I0 + 1) [I0 <= 19 /\ I0 <= 9] 0.00/0.38 f2(I1) -> f2(I1 + 1) [I1 <= 19 /\ 9 <= I1 - 1] 0.00/0.38 f1(I2) -> f2(0) 0.00/0.38 0.00/0.38 We use the reverse value criterion with the projection function NU: 0.00/0.38 NU[f2#(z1)] = 19 + -1 * z1 0.00/0.38 0.00/0.38 This gives the following inequalities: 0.00/0.38 I1 <= 19 /\ 9 <= I1 - 1 ==> 19 + -1 * I1 > 19 + -1 * (I1 + 1) with 19 + -1 * I1 >= 0 0.00/0.38 0.00/0.38 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/0.38 0.00/0.38 DP problem for innermost termination. 0.00/0.38 P = 0.00/0.38 f2#(I0) -> f2#(I0 + 1) [I0 <= 19 /\ I0 <= 9] 0.00/0.38 R = 0.00/0.38 init(x1) -> f1(rnd1) 0.00/0.38 f2(I0) -> f2(I0 + 1) [I0 <= 19 /\ I0 <= 9] 0.00/0.38 f2(I1) -> f2(I1 + 1) [I1 <= 19 /\ 9 <= I1 - 1] 0.00/0.38 f1(I2) -> f2(0) 0.00/0.38 0.00/0.38 We use the reverse value criterion with the projection function NU: 0.00/0.38 NU[f2#(z1)] = 9 + -1 * z1 0.00/0.38 0.00/0.38 This gives the following inequalities: 0.00/0.38 I0 <= 19 /\ I0 <= 9 ==> 9 + -1 * I0 > 9 + -1 * (I0 + 1) with 9 + -1 * I0 >= 0 0.00/0.38 0.00/0.38 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/3.36 EOF
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