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Integ Trans Syste 27634 pair #381737191
details
property
value
status
complete
benchmark
p-15.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n052.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.290726900101 seconds
cpu usage
0.302942644
max memory
9875456.0
stage attributes
key
value
output-size
1967
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = f5#(x1, x2) -> f1#(x1, x2) f3#(I2, I3) -> f2#(I2, I3) f2#(I4, I5) -> f3#(I4, -1 + I5) [-1 * (-1 + I5) <= 0] f1#(I6, I7) -> f2#(I6, I7) R = f5(x1, x2) -> f1(x1, x2) f2(I0, I1) -> f4(rnd1, -1 + I1) [rnd1 = rnd1 /\ 0 <= -1 - (-1 + I1)] f3(I2, I3) -> f2(I2, I3) f2(I4, I5) -> f3(I4, -1 + I5) [-1 * (-1 + I5) <= 0] f1(I6, I7) -> f2(I6, I7) The dependency graph for this problem is: 0 -> 3 1 -> 2 2 -> 1 3 -> 2 Where: 0) f5#(x1, x2) -> f1#(x1, x2) 1) f3#(I2, I3) -> f2#(I2, I3) 2) f2#(I4, I5) -> f3#(I4, -1 + I5) [-1 * (-1 + I5) <= 0] 3) f1#(I6, I7) -> f2#(I6, I7) We have the following SCCs. { 1, 2 } DP problem for innermost termination. P = f3#(I2, I3) -> f2#(I2, I3) f2#(I4, I5) -> f3#(I4, -1 + I5) [-1 * (-1 + I5) <= 0] R = f5(x1, x2) -> f1(x1, x2) f2(I0, I1) -> f4(rnd1, -1 + I1) [rnd1 = rnd1 /\ 0 <= -1 - (-1 + I1)] f3(I2, I3) -> f2(I2, I3) f2(I4, I5) -> f3(I4, -1 + I5) [-1 * (-1 + I5) <= 0] f1(I6, I7) -> f2(I6, I7) We use the basic value criterion with the projection function NU: NU[f2#(z1,z2)] = z2 NU[f3#(z1,z2)] = z2 This gives the following inequalities: ==> I3 (>! \union =) I3 -1 * (-1 + I5) <= 0 ==> I5 >! -1 + I5 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I2, I3) -> f2#(I2, I3) R = f5(x1, x2) -> f1(x1, x2) f2(I0, I1) -> f4(rnd1, -1 + I1) [rnd1 = rnd1 /\ 0 <= -1 - (-1 + I1)] f3(I2, I3) -> f2(I2, I3) f2(I4, I5) -> f3(I4, -1 + I5) [-1 * (-1 + I5) <= 0] f1(I6, I7) -> f2(I6, I7) The dependency graph for this problem is: 1 -> Where: 1) f3#(I2, I3) -> f2#(I2, I3) We have the following SCCs.
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