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Integ Trans Syste 27634 pair #381737593
details
property
value
status
complete
benchmark
seq.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n096.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.108434915543 seconds
cpu usage
0.110363008
max memory
8556544.0
stage attributes
key
value
output-size
1654
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 = f4#(x1) -> f1#(x1) f3#(I0) -> f2#(I0) f2#(I1) -> f3#(rnd1) [0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] f1#(I2) -> f2#(I2) R = f4(x1) -> f1(x1) f3(I0) -> f2(I0) f2(I1) -> f3(rnd1) [0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] f1(I2) -> f2(I2) The dependency graph for this problem is: 0 -> 3 1 -> 2 2 -> 1 3 -> 2 Where: 0) f4#(x1) -> f1#(x1) 1) f3#(I0) -> f2#(I0) 2) f2#(I1) -> f3#(rnd1) [0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] 3) f1#(I2) -> f2#(I2) We have the following SCCs. { 1, 2 } DP problem for innermost termination. P = f3#(I0) -> f2#(I0) f2#(I1) -> f3#(rnd1) [0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] R = f4(x1) -> f1(x1) f3(I0) -> f2(I0) f2(I1) -> f3(rnd1) [0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] f1(I2) -> f2(I2) We use the basic value criterion with the projection function NU: NU[f2#(z1)] = z1 NU[f3#(z1)] = z1 This gives the following inequalities: ==> I0 (>! \union =) I0 0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1 ==> I1 >! rnd1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I0) -> f2#(I0) R = f4(x1) -> f1(x1) f3(I0) -> f2(I0) f2(I1) -> f3(rnd1) [0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1] f1(I2) -> f2(I2) The dependency graph for this problem is: 1 -> Where: 1) f3#(I0) -> f2#(I0) We have the following SCCs.
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