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Integ Trans Syste 27634 pair #381740296
details
property
value
status
complete
benchmark
ex23.t2_fixed.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n092.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
1.35644888878 seconds
cpu usage
1.436429165
max memory
1.5597568E7
stage attributes
key
value
output-size
2733
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = f6#(x1, x2, x3) -> f5#(x1, x2, x3) f5#(I0, I1, I2) -> f4#(0, I1, I2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) f4#(I9, I10, I11) -> f3#(I9, I10, rnd3) [rnd3 = rnd3 /\ I10 <= 127] f1#(I12, I13, I14) -> f3#(1 + I12, I13, 1 + I14) [1 + I12 <= 36] R = f6(x1, x2, x3) -> f5(x1, x2, x3) f5(I0, I1, I2) -> f4(0, I1, I2) f3(I3, I4, I5) -> f1(I3, I4, I5) f4(I6, I7, I8) -> f2(I6, I7, I8) [128 <= I7] f4(I9, I10, I11) -> f3(I9, I10, rnd3) [rnd3 = rnd3 /\ I10 <= 127] f1(I12, I13, I14) -> f3(1 + I12, I13, 1 + I14) [1 + I12 <= 36] f1(I15, I16, I17) -> f2(I15, I16, I17) [36 <= I15] The dependency graph for this problem is: 0 -> 1 1 -> 3 2 -> 4 3 -> 2 4 -> 2 Where: 0) f6#(x1, x2, x3) -> f5#(x1, x2, x3) 1) f5#(I0, I1, I2) -> f4#(0, I1, I2) 2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 3) f4#(I9, I10, I11) -> f3#(I9, I10, rnd3) [rnd3 = rnd3 /\ I10 <= 127] 4) f1#(I12, I13, I14) -> f3#(1 + I12, I13, 1 + I14) [1 + I12 <= 36] We have the following SCCs. { 2, 4 } DP problem for innermost termination. P = f3#(I3, I4, I5) -> f1#(I3, I4, I5) f1#(I12, I13, I14) -> f3#(1 + I12, I13, 1 + I14) [1 + I12 <= 36] R = f6(x1, x2, x3) -> f5(x1, x2, x3) f5(I0, I1, I2) -> f4(0, I1, I2) f3(I3, I4, I5) -> f1(I3, I4, I5) f4(I6, I7, I8) -> f2(I6, I7, I8) [128 <= I7] f4(I9, I10, I11) -> f3(I9, I10, rnd3) [rnd3 = rnd3 /\ I10 <= 127] f1(I12, I13, I14) -> f3(1 + I12, I13, 1 + I14) [1 + I12 <= 36] f1(I15, I16, I17) -> f2(I15, I16, I17) [36 <= I15] We use the reverse value criterion with the projection function NU: NU[f1#(z1,z2,z3)] = 36 + -1 * (1 + z1) NU[f3#(z1,z2,z3)] = 36 + -1 * (1 + z1) This gives the following inequalities: ==> 36 + -1 * (1 + I3) >= 36 + -1 * (1 + I3) 1 + I12 <= 36 ==> 36 + -1 * (1 + I12) > 36 + -1 * (1 + (1 + I12)) with 36 + -1 * (1 + I12) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I3, I4, I5) -> f1#(I3, I4, I5) R = f6(x1, x2, x3) -> f5(x1, x2, x3) f5(I0, I1, I2) -> f4(0, I1, I2) f3(I3, I4, I5) -> f1(I3, I4, I5) f4(I6, I7, I8) -> f2(I6, I7, I8) [128 <= I7] f4(I9, I10, I11) -> f3(I9, I10, rnd3) [rnd3 = rnd3 /\ I10 <= 127] f1(I12, I13, I14) -> f3(1 + I12, I13, 1 + I14) [1 + I12 <= 36] f1(I15, I16, I17) -> f2(I15, I16, I17) [36 <= I15] The dependency graph for this problem is: 2 -> Where: 2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) We have the following SCCs.
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