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Integ Trans Syste 27634 pair #381737599
details
property
value
status
complete
benchmark
p-4.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n103.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
5.89586400986 seconds
cpu usage
6.203873644
max memory
2.6783744E7
stage attributes
key
value
output-size
4533
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 = f5#(x1, x2, x3) -> f1#(x1, x2, x3) f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1] f4#(I3, I4, I5) -> f3#(I3, I4, I5) f3#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [1 + I7 <= I8] f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10] f1#(I12, I13, I14) -> f2#(I12, I13, I14) R = f5(x1, x2, x3) -> f1(x1, x2, x3) f2(I0, I1, I2) -> f3(I0, I1, I2) [1 + I0 <= I1] f4(I3, I4, I5) -> f3(I3, I4, I5) f3(I6, I7, I8) -> f4(I6, 1 + I7, I8) [1 + I7 <= I8] f3(I9, I10, I11) -> f2(1 + I9, I10, I11) [I11 <= I10] f1(I12, I13, I14) -> f2(I12, I13, I14) The dependency graph for this problem is: 0 -> 5 1 -> 3, 4 2 -> 3, 4 3 -> 2 4 -> 1 5 -> 1 Where: 0) f5#(x1, x2, x3) -> f1#(x1, x2, x3) 1) f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1] 2) f4#(I3, I4, I5) -> f3#(I3, I4, I5) 3) f3#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [1 + I7 <= I8] 4) f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10] 5) f1#(I12, I13, I14) -> f2#(I12, I13, I14) We have the following SCCs. { 1, 2, 3, 4 } DP problem for innermost termination. P = f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1] f4#(I3, I4, I5) -> f3#(I3, I4, I5) f3#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [1 + I7 <= I8] f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10] R = f5(x1, x2, x3) -> f1(x1, x2, x3) f2(I0, I1, I2) -> f3(I0, I1, I2) [1 + I0 <= I1] f4(I3, I4, I5) -> f3(I3, I4, I5) f3(I6, I7, I8) -> f4(I6, 1 + I7, I8) [1 + I7 <= I8] f3(I9, I10, I11) -> f2(1 + I9, I10, I11) [I11 <= I10] f1(I12, I13, I14) -> f2(I12, I13, I14) We use the reverse value criterion with the projection function NU: NU[f4#(z1,z2,z3)] = z3 + -1 * (1 + z2) NU[f3#(z1,z2,z3)] = z3 + -1 * (1 + z2) NU[f2#(z1,z2,z3)] = z3 + -1 * (1 + z2) This gives the following inequalities: 1 + I0 <= I1 ==> I2 + -1 * (1 + I1) >= I2 + -1 * (1 + I1) ==> I5 + -1 * (1 + I4) >= I5 + -1 * (1 + I4) 1 + I7 <= I8 ==> I8 + -1 * (1 + I7) > I8 + -1 * (1 + (1 + I7)) with I8 + -1 * (1 + I7) >= 0 I11 <= I10 ==> I11 + -1 * (1 + I10) >= I11 + -1 * (1 + I10) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1] f4#(I3, I4, I5) -> f3#(I3, I4, I5) f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10] R = f5(x1, x2, x3) -> f1(x1, x2, x3) f2(I0, I1, I2) -> f3(I0, I1, I2) [1 + I0 <= I1] f4(I3, I4, I5) -> f3(I3, I4, I5) f3(I6, I7, I8) -> f4(I6, 1 + I7, I8) [1 + I7 <= I8] f3(I9, I10, I11) -> f2(1 + I9, I10, I11) [I11 <= I10] f1(I12, I13, I14) -> f2(I12, I13, I14) The dependency graph for this problem is: 1 -> 4 2 -> 4 4 -> 1 Where: 1) f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1] 2) f4#(I3, I4, I5) -> f3#(I3, I4, I5) 4) f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10] We have the following SCCs. { 1, 4 } DP problem for innermost termination. P = f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1] f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10] R = f5(x1, x2, x3) -> f1(x1, x2, x3)
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