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ITS pair #487099327
details
property
value
status
complete
benchmark
Fibonacci.jar-obl-8.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n150.star.cs.uiowa.edu
space
From_AProVE_2014
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
1.40021 seconds
cpu usage
1.41695
user time
0.769151
system time
0.6478
max virtual memory
351452.0
max residence set size
8500.0
stage attributes
key
value
starexec-result
YES
output
YES DP problem for innermost termination. P = init#(x1, x2) -> f1#(rnd1, rnd2) f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1] f4#(I3, I4) -> f3#(I3, I3 - 2) [0 <= I3 - 1] f2#(I5, I6) -> f3#(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1] f2#(I7, I8) -> f3#(2, 0) [2 = I7] f2#(I9, I10) -> f2#(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1] f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1] f4(I3, I4) -> f3(I3, I3 - 2) [0 <= I3 - 1] f2(I5, I6) -> f3(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1] f2(I7, I8) -> f3(2, 0) [2 = I7] f2(I9, I10) -> f2(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1] f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] The dependency graph for this problem is: 0 -> 6 1 -> 3, 4, 5 2 -> 1 3 -> 1 4 -> 1 5 -> 3, 4, 5 6 -> 3, 4, 5 Where: 0) init#(x1, x2) -> f1#(rnd1, rnd2) 1) f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1] 2) f4#(I3, I4) -> f3#(I3, I3 - 2) [0 <= I3 - 1] 3) f2#(I5, I6) -> f3#(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1] 4) f2#(I7, I8) -> f3#(2, 0) [2 = I7] 5) f2#(I9, I10) -> f2#(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1] 6) f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] We have the following SCCs. { 1, 3, 4, 5 } DP problem for innermost termination. P = f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1] f2#(I5, I6) -> f3#(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1] f2#(I7, I8) -> f3#(2, 0) [2 = I7] f2#(I9, I10) -> f2#(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1] f4(I3, I4) -> f3(I3, I3 - 2) [0 <= I3 - 1] f2(I5, I6) -> f3(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1] f2(I7, I8) -> f3(2, 0) [2 = I7] f2(I9, I10) -> f2(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1] f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] We use the basic value criterion with the projection function NU: NU[f2#(z1,z2)] = z1 NU[f3#(z1,z2)] = z2 This gives the following inequalities: I1 <= I0 - 1 /\ 1 <= I0 - 1 ==> I1 (>! \union =) I1 I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1 ==> I5 >! I5 - 2 2 = I7 ==> I7 >! 0 I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1 ==> I9 >! I9 - 1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1] R = init(x1, x2) -> f1(rnd1, rnd2) f3(I0, I1) -> f2(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1] f4(I3, I4) -> f3(I3, I3 - 2) [0 <= I3 - 1] f2(I5, I6) -> f3(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1] f2(I7, I8) -> f3(2, 0) [2 = I7] f2(I9, I10) -> f2(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1] f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] The dependency graph for this problem is: 1 -> Where: 1) f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1] We have the following SCCs.
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