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Integ Trans Syste 27634 pair #381739417
details
property
value
status
complete
benchmark
array_free.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n078.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
1.5051009655 seconds
cpu usage
1.595542337
max memory
1.6400384E7
stage attributes
key
value
output-size
3614
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 = f8#(x1, x2) -> f7#(x1, x2) f7#(I0, I1) -> f2#(I0, 0) f2#(I2, I3) -> f5#(I2, I3) f5#(I4, I5) -> f4#(I4, I5) [1 + I5 <= I4] f4#(I8, I9) -> f3#(I8, I9) f4#(I10, I11) -> f1#(I10, I11) f4#(I12, I13) -> f3#(I12, I13) f3#(I14, I15) -> f1#(I14, I15) f1#(I16, I17) -> f2#(I16, 1 + I17) R = f8(x1, x2) -> f7(x1, x2) f7(I0, I1) -> f2(I0, 0) f2(I2, I3) -> f5(I2, I3) f5(I4, I5) -> f4(I4, I5) [1 + I5 <= I4] f5(I6, I7) -> f6(I6, I7) [I6 <= I7] f4(I8, I9) -> f3(I8, I9) f4(I10, I11) -> f1(I10, I11) f4(I12, I13) -> f3(I12, I13) f3(I14, I15) -> f1(I14, I15) f1(I16, I17) -> f2(I16, 1 + I17) The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 3 3 -> 4, 5, 6 4 -> 7 5 -> 8 6 -> 7 7 -> 8 8 -> 2 Where: 0) f8#(x1, x2) -> f7#(x1, x2) 1) f7#(I0, I1) -> f2#(I0, 0) 2) f2#(I2, I3) -> f5#(I2, I3) 3) f5#(I4, I5) -> f4#(I4, I5) [1 + I5 <= I4] 4) f4#(I8, I9) -> f3#(I8, I9) 5) f4#(I10, I11) -> f1#(I10, I11) 6) f4#(I12, I13) -> f3#(I12, I13) 7) f3#(I14, I15) -> f1#(I14, I15) 8) f1#(I16, I17) -> f2#(I16, 1 + I17) We have the following SCCs. { 2, 3, 4, 5, 6, 7, 8 } DP problem for innermost termination. P = f2#(I2, I3) -> f5#(I2, I3) f5#(I4, I5) -> f4#(I4, I5) [1 + I5 <= I4] f4#(I8, I9) -> f3#(I8, I9) f4#(I10, I11) -> f1#(I10, I11) f4#(I12, I13) -> f3#(I12, I13) f3#(I14, I15) -> f1#(I14, I15) f1#(I16, I17) -> f2#(I16, 1 + I17) R = f8(x1, x2) -> f7(x1, x2) f7(I0, I1) -> f2(I0, 0) f2(I2, I3) -> f5(I2, I3) f5(I4, I5) -> f4(I4, I5) [1 + I5 <= I4] f5(I6, I7) -> f6(I6, I7) [I6 <= I7] f4(I8, I9) -> f3(I8, I9) f4(I10, I11) -> f1(I10, I11) f4(I12, I13) -> f3(I12, I13) f3(I14, I15) -> f1(I14, I15) f1(I16, I17) -> f2(I16, 1 + I17) We use the extended value criterion with the projection function NU: NU[f1#(x0,x1)] = x0 - x1 - 2 NU[f3#(x0,x1)] = x0 - x1 - 2 NU[f4#(x0,x1)] = x0 - x1 - 2 NU[f5#(x0,x1)] = x0 - x1 - 1 NU[f2#(x0,x1)] = x0 - x1 - 1 This gives the following inequalities: ==> I2 - I3 - 1 >= I2 - I3 - 1 1 + I5 <= I4 ==> I4 - I5 - 1 > I4 - I5 - 2 with I4 - I5 - 1 >= 0 ==> I8 - I9 - 2 >= I8 - I9 - 2 ==> I10 - I11 - 2 >= I10 - I11 - 2 ==> I12 - I13 - 2 >= I12 - I13 - 2 ==> I14 - I15 - 2 >= I14 - I15 - 2 ==> I16 - I17 - 2 >= I16 - (1 + I17) - 1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f2#(I2, I3) -> f5#(I2, I3) f4#(I8, I9) -> f3#(I8, I9) f4#(I10, I11) -> f1#(I10, I11) f4#(I12, I13) -> f3#(I12, I13)
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