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Integ Trans Syste 27634 pair #381738889
details
property
value
status
complete
benchmark
heidy5.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n050.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.651078939438 seconds
cpu usage
0.680558613
max memory
1.1808768E7
stage attributes
key
value
output-size
2516
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 = f7#(x1, x2, x3) -> f6#(x1, x2, x3) f6#(I0, I1, I2) -> f4#(0, I1, I2) f4#(I6, I7, I8) -> f3#(I6, I7, I8) [I7 <= 0] f3#(I9, I10, I11) -> f1#(I9, I10, I11) f1#(I12, I13, I14) -> f3#(I12, I13, -1 + I14) [1 <= I14] R = f7(x1, x2, x3) -> f6(x1, x2, x3) f6(I0, I1, I2) -> f4(0, I1, I2) f4(I3, I4, I5) -> f5(1 + I3, I4, I5) [1 <= I4] f4(I6, I7, I8) -> f3(I6, I7, I8) [I7 <= 0] f3(I9, I10, I11) -> f1(I9, I10, I11) f1(I12, I13, I14) -> f3(I12, I13, -1 + I14) [1 <= I14] f1(I15, I16, I17) -> f2(1, I16, I17) [I16 <= 0 /\ I17 <= 0] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 3 3 -> 4 4 -> 3 Where: 0) f7#(x1, x2, x3) -> f6#(x1, x2, x3) 1) f6#(I0, I1, I2) -> f4#(0, I1, I2) 2) f4#(I6, I7, I8) -> f3#(I6, I7, I8) [I7 <= 0] 3) f3#(I9, I10, I11) -> f1#(I9, I10, I11) 4) f1#(I12, I13, I14) -> f3#(I12, I13, -1 + I14) [1 <= I14] We have the following SCCs. { 3, 4 } DP problem for innermost termination. P = f3#(I9, I10, I11) -> f1#(I9, I10, I11) f1#(I12, I13, I14) -> f3#(I12, I13, -1 + I14) [1 <= I14] R = f7(x1, x2, x3) -> f6(x1, x2, x3) f6(I0, I1, I2) -> f4(0, I1, I2) f4(I3, I4, I5) -> f5(1 + I3, I4, I5) [1 <= I4] f4(I6, I7, I8) -> f3(I6, I7, I8) [I7 <= 0] f3(I9, I10, I11) -> f1(I9, I10, I11) f1(I12, I13, I14) -> f3(I12, I13, -1 + I14) [1 <= I14] f1(I15, I16, I17) -> f2(1, I16, I17) [I16 <= 0 /\ I17 <= 0] We use the basic value criterion with the projection function NU: NU[f1#(z1,z2,z3)] = z3 NU[f3#(z1,z2,z3)] = z3 This gives the following inequalities: ==> I11 (>! \union =) I11 1 <= I14 ==> I14 >! -1 + I14 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I9, I10, I11) -> f1#(I9, I10, I11) R = f7(x1, x2, x3) -> f6(x1, x2, x3) f6(I0, I1, I2) -> f4(0, I1, I2) f4(I3, I4, I5) -> f5(1 + I3, I4, I5) [1 <= I4] f4(I6, I7, I8) -> f3(I6, I7, I8) [I7 <= 0] f3(I9, I10, I11) -> f1(I9, I10, I11) f1(I12, I13, I14) -> f3(I12, I13, -1 + I14) [1 <= I14] f1(I15, I16, I17) -> f2(1, I16, I17) [I16 <= 0 /\ I17 <= 0] The dependency graph for this problem is: 3 -> Where: 3) f3#(I9, I10, I11) -> f1#(I9, I10, I11) We have the following SCCs.
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