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ITS pair #487097620
details
property
value
status
complete
benchmark
ex7.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n144.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
6.24691 seconds
cpu usage
6.31377
user time
3.20112
system time
3.11265
max virtual memory
720476.0
max residence set size
8772.0
stage attributes
key
value
starexec-result
YES
output
YES DP problem for innermost termination. P = f5#(x1, x2, x3, x4, x5, x6) -> f4#(x1, x2, x3, x4, x5, x6) f4#(I0, I1, I2, I3, I4, I5) -> f3#(I0, 0, I0, rnd4, rnd5, I5) [rnd4 = rnd5 /\ rnd5 = rnd5] f3#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) f1#(I12, I13, I14, I15, I16, I17) -> f3#(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14] R = f5(x1, x2, x3, x4, x5, x6) -> f4(x1, x2, x3, x4, x5, x6) f4(I0, I1, I2, I3, I4, I5) -> f3(I0, 0, I0, rnd4, rnd5, I5) [rnd4 = rnd5 /\ rnd5 = rnd5] f3(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) f1(I12, I13, I14, I15, I16, I17) -> f3(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14] f1(I18, I19, I20, I21, I22, I23) -> f2(I18, I19, I20, I21, I22, I23) [I20 <= I19] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 3 3 -> 2 Where: 0) f5#(x1, x2, x3, x4, x5, x6) -> f4#(x1, x2, x3, x4, x5, x6) 1) f4#(I0, I1, I2, I3, I4, I5) -> f3#(I0, 0, I0, rnd4, rnd5, I5) [rnd4 = rnd5 /\ rnd5 = rnd5] 2) f3#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) 3) f1#(I12, I13, I14, I15, I16, I17) -> f3#(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14] We have the following SCCs. { 2, 3 } DP problem for innermost termination. P = f3#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) f1#(I12, I13, I14, I15, I16, I17) -> f3#(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14] R = f5(x1, x2, x3, x4, x5, x6) -> f4(x1, x2, x3, x4, x5, x6) f4(I0, I1, I2, I3, I4, I5) -> f3(I0, 0, I0, rnd4, rnd5, I5) [rnd4 = rnd5 /\ rnd5 = rnd5] f3(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) f1(I12, I13, I14, I15, I16, I17) -> f3(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14] f1(I18, I19, I20, I21, I22, I23) -> f2(I18, I19, I20, I21, I22, I23) [I20 <= I19] We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3,z4,z5,z6)] = z3 + -1 * (1 + z2) NU[f3#(z1,z2,z3,z4,z5,z6)] = z3 + -1 * (1 + z2) This gives the following inequalities: ==> I8 + -1 * (1 + I7) >= I8 + -1 * (1 + I7) rnd6 = rnd6 /\ 1 + I13 <= I14 ==> I14 + -1 * (1 + I13) > I14 + -1 * (1 + (1 + I13)) with I14 + -1 * (1 + I13) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) R = f5(x1, x2, x3, x4, x5, x6) -> f4(x1, x2, x3, x4, x5, x6) f4(I0, I1, I2, I3, I4, I5) -> f3(I0, 0, I0, rnd4, rnd5, I5) [rnd4 = rnd5 /\ rnd5 = rnd5] f3(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) f1(I12, I13, I14, I15, I16, I17) -> f3(I12, 1 + I13, I14, I15, I16, rnd6) [rnd6 = rnd6 /\ 1 + I13 <= I14] f1(I18, I19, I20, I21, I22, I23) -> f2(I18, I19, I20, I21, I22, I23) [I20 <= I19] The dependency graph for this problem is: 2 -> Where: 2) f3#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) We have the following SCCs.
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