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Integ Trans Syste 27634 pair #381738223
details
property
value
status
complete
benchmark
n-6a.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n013.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
12.6222069263 seconds
cpu usage
13.457908049
max memory
2.7529216E7
stage attributes
key
value
output-size
6195
starexec-result
MAYBE
output
/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) f6#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, I3, I4) f5#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, I13, I14) [I11 = I11] f1#(I15, I16, I17, I18, I19) -> f4#(I15, I16, rnd3, I18, I19) [rnd3 = rnd3 /\ 0 <= -1 - I18 + I19] f3#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) f1#(I25, I26, I27, I28, I29) -> f3#(I25, I26, I30, I28, -1 + I29) [0 <= I30 /\ I30 <= 0 /\ I30 = I30 /\ 0 <= -1 - I28 + I29] R = f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) f6(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) f5(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f4(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I13, I14) [I11 = I11] f1(I15, I16, I17, I18, I19) -> f4(I15, I16, rnd3, I18, I19) [rnd3 = rnd3 /\ 0 <= -1 - I18 + I19] f3(I20, I21, I22, I23, I24) -> f1(I20, I21, I22, I23, I24) f1(I25, I26, I27, I28, I29) -> f3(I25, I26, I30, I28, -1 + I29) [0 <= I30 /\ I30 <= 0 /\ I30 = I30 /\ 0 <= -1 - I28 + I29] f1(I31, I32, I33, I34, I35) -> f2(rnd1, I32, I33, I34, I35) [rnd1 = rnd1 /\ -1 * I34 + I35 <= 0] The dependency graph for this problem is: 0 -> 1 1 -> 4, 6 2 -> 4, 6 3 -> 2 4 -> 3 5 -> 4, 6 6 -> 5 Where: 0) f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) 1) f6#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, I3, I4) 2) f5#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 3) f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, I13, I14) [I11 = I11] 4) f1#(I15, I16, I17, I18, I19) -> f4#(I15, I16, rnd3, I18, I19) [rnd3 = rnd3 /\ 0 <= -1 - I18 + I19] 5) f3#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) 6) f1#(I25, I26, I27, I28, I29) -> f3#(I25, I26, I30, I28, -1 + I29) [0 <= I30 /\ I30 <= 0 /\ I30 = I30 /\ 0 <= -1 - I28 + I29] We have the following SCCs. { 2, 3, 4, 5, 6 } DP problem for innermost termination. P = f5#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, I13, I14) [I11 = I11] f1#(I15, I16, I17, I18, I19) -> f4#(I15, I16, rnd3, I18, I19) [rnd3 = rnd3 /\ 0 <= -1 - I18 + I19] f3#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) f1#(I25, I26, I27, I28, I29) -> f3#(I25, I26, I30, I28, -1 + I29) [0 <= I30 /\ I30 <= 0 /\ I30 = I30 /\ 0 <= -1 - I28 + I29] R = f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) f6(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) f5(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f4(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I13, I14) [I11 = I11] f1(I15, I16, I17, I18, I19) -> f4(I15, I16, rnd3, I18, I19) [rnd3 = rnd3 /\ 0 <= -1 - I18 + I19] f3(I20, I21, I22, I23, I24) -> f1(I20, I21, I22, I23, I24) f1(I25, I26, I27, I28, I29) -> f3(I25, I26, I30, I28, -1 + I29) [0 <= I30 /\ I30 <= 0 /\ I30 = I30 /\ 0 <= -1 - I28 + I29] f1(I31, I32, I33, I34, I35) -> f2(rnd1, I32, I33, I34, I35) [rnd1 = rnd1 /\ -1 * I34 + I35 <= 0] We use the extended value criterion with the projection function NU: NU[f3#(x0,x1,x2,x3,x4)] = -x3 + x4 - 1 NU[f4#(x0,x1,x2,x3,x4)] = -x3 + x4 - 1 NU[f1#(x0,x1,x2,x3,x4)] = -x3 + x4 - 1 NU[f5#(x0,x1,x2,x3,x4)] = -x3 + x4 - 1 This gives the following inequalities: ==> -I8 + I9 - 1 >= -I8 + I9 - 1 I11 = I11 ==> -I13 + I14 - 1 >= -I13 + I14 - 1 rnd3 = rnd3 /\ 0 <= -1 - I18 + I19 ==> -I18 + I19 - 1 >= -I18 + I19 - 1 ==> -I23 + I24 - 1 >= -I23 + I24 - 1 0 <= I30 /\ I30 <= 0 /\ I30 = I30 /\ 0 <= -1 - I28 + I29 ==> -I28 + I29 - 1 > -I28 + (-1 + I29) - 1 with -I28 + I29 - 1 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f5#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, I13, I14) [I11 = I11] f1#(I15, I16, I17, I18, I19) -> f4#(I15, I16, rnd3, I18, I19) [rnd3 = rnd3 /\ 0 <= -1 - I18 + I19] f3#(I20, I21, I22, I23, I24) -> f1#(I20, I21, I22, I23, I24) R = f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) f6(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, I3, I4) f5(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) f4(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I13, I14) [I11 = I11] f1(I15, I16, I17, I18, I19) -> f4(I15, I16, rnd3, I18, I19) [rnd3 = rnd3 /\ 0 <= -1 - I18 + I19] f3(I20, I21, I22, I23, I24) -> f1(I20, I21, I22, I23, I24) f1(I25, I26, I27, I28, I29) -> f3(I25, I26, I30, I28, -1 + I29) [0 <= I30 /\ I30 <= 0 /\ I30 = I30 /\ 0 <= -1 - I28 + I29] f1(I31, I32, I33, I34, I35) -> f2(rnd1, I32, I33, I34, I35) [rnd1 = rnd1 /\ -1 * I34 + I35 <= 0] The dependency graph for this problem is: 2 -> 4 3 -> 2 4 -> 3 5 -> 4
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