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Integ Trans Syste 27634 pair #381740182
details
property
value
status
complete
benchmark
s2.t2_fixed.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n028.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
3.85004591942 seconds
cpu usage
4.098498774
max memory
2.53952E7
stage attributes
key
value
output-size
3782
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 = f5#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) f4#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) f2#(I4, I5, I6, I7) -> f4#(1 + I4, I5, I6, I7) [1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10] f1#(I12, I13, I14, I15) -> f2#(rnd1, I13, I14, I15) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= 10 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= 10 /\ rnd1 = 1 + y2 /\ 2 <= rnd1 /\ rnd1 <= 2] R = f5(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) f4(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) f2(I4, I5, I6, I7) -> f4(1 + I4, I5, I6, I7) [1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10] f2(I8, I9, I10, I11) -> f3(I8, I9, I11, I11) [10 <= I8] f1(I12, I13, I14, I15) -> f2(rnd1, I13, I14, I15) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= 10 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= 10 /\ rnd1 = 1 + y2 /\ 2 <= rnd1 /\ rnd1 <= 2] The dependency graph for this problem is: 0 -> 3 1 -> 2 2 -> 1 3 -> 2 Where: 0) f5#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) 1) f4#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 2) f2#(I4, I5, I6, I7) -> f4#(1 + I4, I5, I6, I7) [1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10] 3) f1#(I12, I13, I14, I15) -> f2#(rnd1, I13, I14, I15) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= 10 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= 10 /\ rnd1 = 1 + y2 /\ 2 <= rnd1 /\ rnd1 <= 2] We have the following SCCs. { 1, 2 } DP problem for innermost termination. P = f4#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) f2#(I4, I5, I6, I7) -> f4#(1 + I4, I5, I6, I7) [1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10] R = f5(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) f4(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) f2(I4, I5, I6, I7) -> f4(1 + I4, I5, I6, I7) [1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10] f2(I8, I9, I10, I11) -> f3(I8, I9, I11, I11) [10 <= I8] f1(I12, I13, I14, I15) -> f2(rnd1, I13, I14, I15) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= 10 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= 10 /\ rnd1 = 1 + y2 /\ 2 <= rnd1 /\ rnd1 <= 2] We use the reverse value criterion with the projection function NU: NU[f2#(z1,z2,z3,z4)] = 1 + z2 + -1 * (1 + z1) NU[f4#(z1,z2,z3,z4)] = 1 + z2 + -1 * (1 + z1) This gives the following inequalities: ==> 1 + I1 + -1 * (1 + I0) >= 1 + I1 + -1 * (1 + I0) 1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10 ==> 1 + I5 + -1 * (1 + I4) > 1 + I5 + -1 * (1 + (1 + I4)) with 1 + I5 + -1 * (1 + I4) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f4#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) R = f5(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) f4(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) f2(I4, I5, I6, I7) -> f4(1 + I4, I5, I6, I7) [1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10] f2(I8, I9, I10, I11) -> f3(I8, I9, I11, I11) [10 <= I8] f1(I12, I13, I14, I15) -> f2(rnd1, I13, I14, I15) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= 10 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= 10 /\ rnd1 = 1 + y2 /\ 2 <= rnd1 /\ rnd1 <= 2] The dependency graph for this problem is: 1 -> Where: 1) f4#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) We have the following SCCs.
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