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Integ Trans Syste 27634 pair #381739822
details
property
value
status
complete
benchmark
simple.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n075.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.59915804863 seconds
cpu usage
0.628565867
max memory
1.140736E7
stage attributes
key
value
output-size
2265
starexec-result
MAYBE
output
/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = f5#(x1) -> f4#(x1) f4#(I0) -> f1#(I0) f3#(I1) -> f1#(I1) f1#(I2) -> f3#(-1 + I2) [1 <= -1 + I2] f2#(I3) -> f1#(I3) f1#(I4) -> f2#(-1 + I4) R = f5(x1) -> f4(x1) f4(I0) -> f1(I0) f3(I1) -> f1(I1) f1(I2) -> f3(-1 + I2) [1 <= -1 + I2] f2(I3) -> f1(I3) f1(I4) -> f2(-1 + I4) The dependency graph for this problem is: 0 -> 1 1 -> 3, 5 2 -> 3, 5 3 -> 2 4 -> 3, 5 5 -> 4 Where: 0) f5#(x1) -> f4#(x1) 1) f4#(I0) -> f1#(I0) 2) f3#(I1) -> f1#(I1) 3) f1#(I2) -> f3#(-1 + I2) [1 <= -1 + I2] 4) f2#(I3) -> f1#(I3) 5) f1#(I4) -> f2#(-1 + I4) We have the following SCCs. { 2, 3, 4, 5 } DP problem for innermost termination. P = f3#(I1) -> f1#(I1) f1#(I2) -> f3#(-1 + I2) [1 <= -1 + I2] f2#(I3) -> f1#(I3) f1#(I4) -> f2#(-1 + I4) R = f5(x1) -> f4(x1) f4(I0) -> f1(I0) f3(I1) -> f1(I1) f1(I2) -> f3(-1 + I2) [1 <= -1 + I2] f2(I3) -> f1(I3) f1(I4) -> f2(-1 + I4) We use the reverse value criterion with the projection function NU: NU[f2#(z1)] = z1 NU[f1#(z1)] = z1 NU[f3#(z1)] = z1 This gives the following inequalities: ==> I1 >= I1 1 <= -1 + I2 ==> I2 > -1 + I2 with I2 >= 0 ==> I3 >= I3 ==> I4 >= -1 + I4 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I1) -> f1#(I1) f2#(I3) -> f1#(I3) f1#(I4) -> f2#(-1 + I4) R = f5(x1) -> f4(x1) f4(I0) -> f1(I0) f3(I1) -> f1(I1) f1(I2) -> f3(-1 + I2) [1 <= -1 + I2] f2(I3) -> f1(I3) f1(I4) -> f2(-1 + I4) The dependency graph for this problem is: 2 -> 5 4 -> 5 5 -> 4 Where: 2) f3#(I1) -> f1#(I1) 4) f2#(I3) -> f1#(I3) 5) f1#(I4) -> f2#(-1 + I4) We have the following SCCs. { 4, 5 } DP problem for innermost termination. P = f2#(I3) -> f1#(I3) f1#(I4) -> f2#(-1 + I4) R = f5(x1) -> f4(x1)
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