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Integ Trans Syste 27634 pair #381739921
details
property
value
status
complete
benchmark
andrey.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n030.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.0908551216125 seconds
cpu usage
0.09025571
max memory
8552448.0
stage attributes
key
value
output-size
1754
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = f4#(x1) -> f3#(x1) f3#(I0) -> f1#(I0) f1#(I1) -> f3#(rnd1) [1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] R = f4(x1) -> f3(x1) f3(I0) -> f1(I0) f1(I1) -> f3(rnd1) [1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] f1(I2) -> f2(I3) [I2 <= 0 /\ I4 = I2 /\ y2 = I4 /\ y3 = y2 /\ I3 = y3] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 1 Where: 0) f4#(x1) -> f3#(x1) 1) f3#(I0) -> f1#(I0) 2) f1#(I1) -> f3#(rnd1) [1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] We have the following SCCs. { 1, 2 } DP problem for innermost termination. P = f3#(I0) -> f1#(I0) f1#(I1) -> f3#(rnd1) [1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] R = f4(x1) -> f3(x1) f3(I0) -> f1(I0) f1(I1) -> f3(rnd1) [1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] f1(I2) -> f2(I3) [I2 <= 0 /\ I4 = I2 /\ y2 = I4 /\ y3 = y2 /\ I3 = y3] We use the basic value criterion with the projection function NU: NU[f1#(z1)] = z1 NU[f3#(z1)] = z1 This gives the following inequalities: ==> I0 (>! \union =) I0 1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1 ==> I1 >! rnd1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I0) -> f1#(I0) R = f4(x1) -> f3(x1) f3(I0) -> f1(I0) f1(I1) -> f3(rnd1) [1 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] f1(I2) -> f2(I3) [I2 <= 0 /\ I4 = I2 /\ y2 = I4 /\ y3 = y2 /\ I3 = y3] The dependency graph for this problem is: 1 -> Where: 1) f3#(I0) -> f1#(I0) We have the following SCCs.
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