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Integ Trans Syste 27634 pair #381738349
details
property
value
status
complete
benchmark
p-42.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n084.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
6.12099409103 seconds
cpu usage
6.481345497
max memory
2.5563136E7
stage attributes
key
value
output-size
4923
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) -> f1#(I0, I1, I2) f4#(I3, I4, I5) -> f1#(I3, I4, I5) f1#(I6, I7, I8) -> f4#(I6, 1 + I7, 1 + I8) [0 <= 1 - I8] f3#(I9, I10, I11) -> f1#(I9, I10, I11) f1#(I12, I13, I14) -> f3#(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0] R = f6(x1, x2, x3) -> f5(x1, x2, x3) f5(I0, I1, I2) -> f1(I0, I1, I2) f4(I3, I4, I5) -> f1(I3, I4, I5) f1(I6, I7, I8) -> f4(I6, 1 + I7, 1 + I8) [0 <= 1 - I8] f3(I9, I10, I11) -> f1(I9, I10, I11) f1(I12, I13, I14) -> f3(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0] f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ 3 - I16 <= 0 /\ 2 - I17 <= 0] The dependency graph for this problem is: 0 -> 1 1 -> 3, 5 2 -> 3, 5 3 -> 2 4 -> 3, 5 5 -> 4 Where: 0) f6#(x1, x2, x3) -> f5#(x1, x2, x3) 1) f5#(I0, I1, I2) -> f1#(I0, I1, I2) 2) f4#(I3, I4, I5) -> f1#(I3, I4, I5) 3) f1#(I6, I7, I8) -> f4#(I6, 1 + I7, 1 + I8) [0 <= 1 - I8] 4) f3#(I9, I10, I11) -> f1#(I9, I10, I11) 5) f1#(I12, I13, I14) -> f3#(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0] We have the following SCCs. { 2, 3, 4, 5 } DP problem for innermost termination. P = f4#(I3, I4, I5) -> f1#(I3, I4, I5) f1#(I6, I7, I8) -> f4#(I6, 1 + I7, 1 + I8) [0 <= 1 - I8] f3#(I9, I10, I11) -> f1#(I9, I10, I11) f1#(I12, I13, I14) -> f3#(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0] R = f6(x1, x2, x3) -> f5(x1, x2, x3) f5(I0, I1, I2) -> f1(I0, I1, I2) f4(I3, I4, I5) -> f1(I3, I4, I5) f1(I6, I7, I8) -> f4(I6, 1 + I7, 1 + I8) [0 <= 1 - I8] f3(I9, I10, I11) -> f1(I9, I10, I11) f1(I12, I13, I14) -> f3(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0] f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ 3 - I16 <= 0 /\ 2 - I17 <= 0] We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3)] = 2 - z2 + -1 * 0 NU[f1#(z1,z2,z3)] = 2 - z2 + -1 * 0 NU[f4#(z1,z2,z3)] = 2 - z2 + -1 * 0 This gives the following inequalities: ==> 2 - I4 + -1 * 0 >= 2 - I4 + -1 * 0 0 <= 1 - I8 ==> 2 - I7 + -1 * 0 >= 2 - (1 + I7) + -1 * 0 ==> 2 - I10 + -1 * 0 >= 2 - I10 + -1 * 0 0 <= 2 - I13 /\ 2 - I14 <= 0 ==> 2 - I13 + -1 * 0 > 2 - (1 + I13) + -1 * 0 with 2 - I13 + -1 * 0 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f4#(I3, I4, I5) -> f1#(I3, I4, I5) f1#(I6, I7, I8) -> f4#(I6, 1 + I7, 1 + I8) [0 <= 1 - I8] f3#(I9, I10, I11) -> f1#(I9, I10, I11) R = f6(x1, x2, x3) -> f5(x1, x2, x3) f5(I0, I1, I2) -> f1(I0, I1, I2) f4(I3, I4, I5) -> f1(I3, I4, I5) f1(I6, I7, I8) -> f4(I6, 1 + I7, 1 + I8) [0 <= 1 - I8] f3(I9, I10, I11) -> f1(I9, I10, I11) f1(I12, I13, I14) -> f3(I12, 1 + I13, 1 + I14) [0 <= 2 - I13 /\ 2 - I14 <= 0] f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ 3 - I16 <= 0 /\ 2 - I17 <= 0] The dependency graph for this problem is: 2 -> 3 3 -> 2 4 -> 3 Where: 2) f4#(I3, I4, I5) -> f1#(I3, I4, I5) 3) f1#(I6, I7, I8) -> f4#(I6, 1 + I7, 1 + I8) [0 <= 1 - I8] 4) f3#(I9, I10, I11) -> f1#(I9, I10, I11) We have the following SCCs. { 2, 3 } DP problem for innermost termination. P = f4#(I3, I4, I5) -> f1#(I3, I4, I5)
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