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Integ Trans Syste 27634 pair #381738421
details
property
value
status
complete
benchmark
p-40.t2_fixed.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n038.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
5.0396130085 seconds
cpu usage
4.924393929
max memory
2.5468928E7
stage attributes
key
value
output-size
3879
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 = f6#(x1, x2, x3, x4, x5, x6) -> f1#(x1, x2, x3, x4, x5, x6) f2#(I0, I1, I2, I3, I4, I5) -> f3#(1 + I0, I1, I2, I3, I4, I5) [1 + I0 <= 500] f5#(I12, I13, I14, I15, I16, I17) -> f3#(I12, I13, I14, I15, I16, I17) f3#(I18, I19, I20, I21, I22, I23) -> f5#(1 + I18, I19, I20, I21, I22, I23) [1 + I18 <= 500] f1#(I30, I31, I32, I33, I34, I35) -> f2#(rnd1, I33, I32, I33, I34, I35) [y1 = I34 /\ rnd1 = I33] R = f6(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) f2(I0, I1, I2, I3, I4, I5) -> f3(1 + I0, I1, I2, I3, I4, I5) [1 + I0 <= 500] f2(I6, I7, I8, I9, I10, I11) -> f4(I6, I7, I11, I9, I10, I11) [500 <= I6] f5(I12, I13, I14, I15, I16, I17) -> f3(I12, I13, I14, I15, I16, I17) f3(I18, I19, I20, I21, I22, I23) -> f5(1 + I18, I19, I20, I21, I22, I23) [1 + I18 <= 500] f3(I24, I25, I26, I27, I28, I29) -> f4(I24, I25, I29, I27, I28, I29) [500 <= I24] f1(I30, I31, I32, I33, I34, I35) -> f2(rnd1, I33, I32, I33, I34, I35) [y1 = I34 /\ rnd1 = I33] The dependency graph for this problem is: 0 -> 4 1 -> 3 2 -> 3 3 -> 2 4 -> 1 Where: 0) f6#(x1, x2, x3, x4, x5, x6) -> f1#(x1, x2, x3, x4, x5, x6) 1) f2#(I0, I1, I2, I3, I4, I5) -> f3#(1 + I0, I1, I2, I3, I4, I5) [1 + I0 <= 500] 2) f5#(I12, I13, I14, I15, I16, I17) -> f3#(I12, I13, I14, I15, I16, I17) 3) f3#(I18, I19, I20, I21, I22, I23) -> f5#(1 + I18, I19, I20, I21, I22, I23) [1 + I18 <= 500] 4) f1#(I30, I31, I32, I33, I34, I35) -> f2#(rnd1, I33, I32, I33, I34, I35) [y1 = I34 /\ rnd1 = I33] We have the following SCCs. { 2, 3 } DP problem for innermost termination. P = f5#(I12, I13, I14, I15, I16, I17) -> f3#(I12, I13, I14, I15, I16, I17) f3#(I18, I19, I20, I21, I22, I23) -> f5#(1 + I18, I19, I20, I21, I22, I23) [1 + I18 <= 500] R = f6(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) f2(I0, I1, I2, I3, I4, I5) -> f3(1 + I0, I1, I2, I3, I4, I5) [1 + I0 <= 500] f2(I6, I7, I8, I9, I10, I11) -> f4(I6, I7, I11, I9, I10, I11) [500 <= I6] f5(I12, I13, I14, I15, I16, I17) -> f3(I12, I13, I14, I15, I16, I17) f3(I18, I19, I20, I21, I22, I23) -> f5(1 + I18, I19, I20, I21, I22, I23) [1 + I18 <= 500] f3(I24, I25, I26, I27, I28, I29) -> f4(I24, I25, I29, I27, I28, I29) [500 <= I24] f1(I30, I31, I32, I33, I34, I35) -> f2(rnd1, I33, I32, I33, I34, I35) [y1 = I34 /\ rnd1 = I33] We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4,z5,z6)] = 500 + -1 * (1 + z1) NU[f5#(z1,z2,z3,z4,z5,z6)] = 500 + -1 * (1 + z1) This gives the following inequalities: ==> 500 + -1 * (1 + I12) >= 500 + -1 * (1 + I12) 1 + I18 <= 500 ==> 500 + -1 * (1 + I18) > 500 + -1 * (1 + (1 + I18)) with 500 + -1 * (1 + I18) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f5#(I12, I13, I14, I15, I16, I17) -> f3#(I12, I13, I14, I15, I16, I17) R = f6(x1, x2, x3, x4, x5, x6) -> f1(x1, x2, x3, x4, x5, x6) f2(I0, I1, I2, I3, I4, I5) -> f3(1 + I0, I1, I2, I3, I4, I5) [1 + I0 <= 500] f2(I6, I7, I8, I9, I10, I11) -> f4(I6, I7, I11, I9, I10, I11) [500 <= I6] f5(I12, I13, I14, I15, I16, I17) -> f3(I12, I13, I14, I15, I16, I17) f3(I18, I19, I20, I21, I22, I23) -> f5(1 + I18, I19, I20, I21, I22, I23) [1 + I18 <= 500] f3(I24, I25, I26, I27, I28, I29) -> f4(I24, I25, I29, I27, I28, I29) [500 <= I24] f1(I30, I31, I32, I33, I34, I35) -> f2(rnd1, I33, I32, I33, I34, I35) [y1 = I34 /\ rnd1 = I33] The dependency graph for this problem is: 2 -> Where: 2) f5#(I12, I13, I14, I15, I16, I17) -> f3#(I12, I13, I14, I15, I16, I17) We have the following SCCs.
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