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Integ Trans Syste 27634 pair #381737623
details
property
value
status
complete
benchmark
p-21.t2_fixed.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n047.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
3.30685710907 seconds
cpu usage
3.111897445
max memory
2.2540288E7
stage attributes
key
value
output-size
4827
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 = f6#(x1, x2, x3) -> f4#(x1, x2, x3) f5#(I0, I1, I2) -> f3#(I0, I1, I2) f3#(I3, I4, I5) -> f5#(I3, I4, -1 + I5) [0 <= -31 + I5] f3#(I6, I7, I8) -> f1#(I6, -1 + I7, I8) [-30 + I8 <= 0] f4#(I9, I10, I11) -> f1#(I9, I10, I11) f1#(I12, I13, I14) -> f3#(I12, I13, I14) [0 <= -21 + I13] R = f6(x1, x2, x3) -> f4(x1, x2, x3) f5(I0, I1, I2) -> f3(I0, I1, I2) f3(I3, I4, I5) -> f5(I3, I4, -1 + I5) [0 <= -31 + I5] f3(I6, I7, I8) -> f1(I6, -1 + I7, I8) [-30 + I8 <= 0] f4(I9, I10, I11) -> f1(I9, I10, I11) f1(I12, I13, I14) -> f3(I12, I13, I14) [0 <= -21 + I13] f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ -20 + I16 <= 0] The dependency graph for this problem is: 0 -> 4 1 -> 2, 3 2 -> 1 3 -> 5 4 -> 5 5 -> 2, 3 Where: 0) f6#(x1, x2, x3) -> f4#(x1, x2, x3) 1) f5#(I0, I1, I2) -> f3#(I0, I1, I2) 2) f3#(I3, I4, I5) -> f5#(I3, I4, -1 + I5) [0 <= -31 + I5] 3) f3#(I6, I7, I8) -> f1#(I6, -1 + I7, I8) [-30 + I8 <= 0] 4) f4#(I9, I10, I11) -> f1#(I9, I10, I11) 5) f1#(I12, I13, I14) -> f3#(I12, I13, I14) [0 <= -21 + I13] We have the following SCCs. { 1, 2, 3, 5 } DP problem for innermost termination. P = f5#(I0, I1, I2) -> f3#(I0, I1, I2) f3#(I3, I4, I5) -> f5#(I3, I4, -1 + I5) [0 <= -31 + I5] f3#(I6, I7, I8) -> f1#(I6, -1 + I7, I8) [-30 + I8 <= 0] f1#(I12, I13, I14) -> f3#(I12, I13, I14) [0 <= -21 + I13] R = f6(x1, x2, x3) -> f4(x1, x2, x3) f5(I0, I1, I2) -> f3(I0, I1, I2) f3(I3, I4, I5) -> f5(I3, I4, -1 + I5) [0 <= -31 + I5] f3(I6, I7, I8) -> f1(I6, -1 + I7, I8) [-30 + I8 <= 0] f4(I9, I10, I11) -> f1(I9, I10, I11) f1(I12, I13, I14) -> f3(I12, I13, I14) [0 <= -21 + I13] f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ -20 + I16 <= 0] We use the basic value criterion with the projection function NU: NU[f1#(z1,z2,z3)] = z3 NU[f3#(z1,z2,z3)] = z3 NU[f5#(z1,z2,z3)] = z3 This gives the following inequalities: ==> I2 (>! \union =) I2 0 <= -31 + I5 ==> I5 >! -1 + I5 -30 + I8 <= 0 ==> I8 (>! \union =) I8 0 <= -21 + I13 ==> I14 (>! \union =) I14 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f5#(I0, I1, I2) -> f3#(I0, I1, I2) f3#(I6, I7, I8) -> f1#(I6, -1 + I7, I8) [-30 + I8 <= 0] f1#(I12, I13, I14) -> f3#(I12, I13, I14) [0 <= -21 + I13] R = f6(x1, x2, x3) -> f4(x1, x2, x3) f5(I0, I1, I2) -> f3(I0, I1, I2) f3(I3, I4, I5) -> f5(I3, I4, -1 + I5) [0 <= -31 + I5] f3(I6, I7, I8) -> f1(I6, -1 + I7, I8) [-30 + I8 <= 0] f4(I9, I10, I11) -> f1(I9, I10, I11) f1(I12, I13, I14) -> f3(I12, I13, I14) [0 <= -21 + I13] f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ -20 + I16 <= 0] The dependency graph for this problem is: 1 -> 3 3 -> 5 5 -> 3 Where: 1) f5#(I0, I1, I2) -> f3#(I0, I1, I2) 3) f3#(I6, I7, I8) -> f1#(I6, -1 + I7, I8) [-30 + I8 <= 0] 5) f1#(I12, I13, I14) -> f3#(I12, I13, I14) [0 <= -21 + I13] We have the following SCCs. { 3, 5 } DP problem for innermost termination. P = f3#(I6, I7, I8) -> f1#(I6, -1 + I7, I8) [-30 + I8 <= 0]
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