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Integ Trans Syste 27634 pair #381737863
details
property
value
status
complete
benchmark
java_Continue1.c.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n033.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
2.23436617851 seconds
cpu usage
2.372535476
max memory
2.043904E7
stage attributes
key
value
output-size
4524
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, x4) -> f6#(x1, x2, x3, x4) f6#(I0, I1, I2, I3) -> f5#(I0, I1, I2, I3) f6#(I4, I5, I6, I7) -> f4#(I4, I5, I6, I7) f6#(I8, I9, I10, I11) -> f3#(I8, I9, I10, I11) f6#(I12, I13, I14, I15) -> f1#(I12, I13, I14, I15) f6#(I20, I21, I22, I23) -> f5#(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3] f5#(I24, I25, I26, I27) -> f4#(I24, I27, I26, 0) f4#(I28, I29, I30, I31) -> f3#(I28, I31, I30, I31) [I31 <= I28] f4#(I32, I33, I34, I35) -> f1#(I32, I35, I34, I35) [1 + I32 <= I35] f3#(I36, I37, I38, I39) -> f4#(I36, I39, I38, 1 + I39) R = f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) f6(I0, I1, I2, I3) -> f5(I0, I1, I2, I3) f6(I4, I5, I6, I7) -> f4(I4, I5, I6, I7) f6(I8, I9, I10, I11) -> f3(I8, I9, I10, I11) f6(I12, I13, I14, I15) -> f1(I12, I13, I14, I15) f6(I16, I17, I18, I19) -> f2(I16, I17, I18, I19) f6(I20, I21, I22, I23) -> f5(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3] f5(I24, I25, I26, I27) -> f4(I24, I27, I26, 0) f4(I28, I29, I30, I31) -> f3(I28, I31, I30, I31) [I31 <= I28] f4(I32, I33, I34, I35) -> f1(I32, I35, I34, I35) [1 + I32 <= I35] f3(I36, I37, I38, I39) -> f4(I36, I39, I38, 1 + I39) f1(I40, I41, I42, I43) -> f2(I40, I43, I44, I45) [I45 = I44 /\ I44 = I44] The dependency graph for this problem is: 0 -> 1, 2, 3, 4, 5 1 -> 6 2 -> 7, 8 3 -> 9 4 -> 5 -> 6 6 -> 7, 8 7 -> 9 8 -> 9 -> 7, 8 Where: 0) f7#(x1, x2, x3, x4) -> f6#(x1, x2, x3, x4) 1) f6#(I0, I1, I2, I3) -> f5#(I0, I1, I2, I3) 2) f6#(I4, I5, I6, I7) -> f4#(I4, I5, I6, I7) 3) f6#(I8, I9, I10, I11) -> f3#(I8, I9, I10, I11) 4) f6#(I12, I13, I14, I15) -> f1#(I12, I13, I14, I15) 5) f6#(I20, I21, I22, I23) -> f5#(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3] 6) f5#(I24, I25, I26, I27) -> f4#(I24, I27, I26, 0) 7) f4#(I28, I29, I30, I31) -> f3#(I28, I31, I30, I31) [I31 <= I28] 8) f4#(I32, I33, I34, I35) -> f1#(I32, I35, I34, I35) [1 + I32 <= I35] 9) f3#(I36, I37, I38, I39) -> f4#(I36, I39, I38, 1 + I39) We have the following SCCs. { 7, 9 } DP problem for innermost termination. P = f4#(I28, I29, I30, I31) -> f3#(I28, I31, I30, I31) [I31 <= I28] f3#(I36, I37, I38, I39) -> f4#(I36, I39, I38, 1 + I39) R = f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) f6(I0, I1, I2, I3) -> f5(I0, I1, I2, I3) f6(I4, I5, I6, I7) -> f4(I4, I5, I6, I7) f6(I8, I9, I10, I11) -> f3(I8, I9, I10, I11) f6(I12, I13, I14, I15) -> f1(I12, I13, I14, I15) f6(I16, I17, I18, I19) -> f2(I16, I17, I18, I19) f6(I20, I21, I22, I23) -> f5(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3] f5(I24, I25, I26, I27) -> f4(I24, I27, I26, 0) f4(I28, I29, I30, I31) -> f3(I28, I31, I30, I31) [I31 <= I28] f4(I32, I33, I34, I35) -> f1(I32, I35, I34, I35) [1 + I32 <= I35] f3(I36, I37, I38, I39) -> f4(I36, I39, I38, 1 + I39) f1(I40, I41, I42, I43) -> f2(I40, I43, I44, I45) [I45 = I44 /\ I44 = I44] We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4)] = z1 + -1 * (1 + z4) NU[f4#(z1,z2,z3,z4)] = z1 + -1 * z4 This gives the following inequalities: I31 <= I28 ==> I28 + -1 * I31 > I28 + -1 * (1 + I31) with I28 + -1 * I31 >= 0 ==> I36 + -1 * (1 + I39) >= I36 + -1 * (1 + I39) We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I36, I37, I38, I39) -> f4#(I36, I39, I38, 1 + I39) R = f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) f6(I0, I1, I2, I3) -> f5(I0, I1, I2, I3) f6(I4, I5, I6, I7) -> f4(I4, I5, I6, I7) f6(I8, I9, I10, I11) -> f3(I8, I9, I10, I11) f6(I12, I13, I14, I15) -> f1(I12, I13, I14, I15) f6(I16, I17, I18, I19) -> f2(I16, I17, I18, I19) f6(I20, I21, I22, I23) -> f5(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3] f5(I24, I25, I26, I27) -> f4(I24, I27, I26, 0)
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