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Integ Trans Syste 27634 pair #381740470
details
property
value
status
complete
benchmark
complex_guard.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n071.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
0.54045009613 seconds
cpu usage
0.566665222
max memory
1.1091968E7
stage attributes
key
value
output-size
2834
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 = f9#(x1) -> f8#(x1) f8#(I0) -> f3#(0) f7#(I1) -> f5#(I1) [2 <= I1] f7#(I2) -> f5#(I2) [1 + I2 <= 2] f2#(I3) -> f7#(I3) f3#(I5) -> f4#(I5) f4#(I6) -> f2#(I6) f4#(I7) -> f1#(I7) f4#(I8) -> f1#(I8) f1#(I9) -> f3#(1 + I9) [1 + I9 <= 3] f1#(I10) -> f2#(I10) [3 <= I10] R = f9(x1) -> f8(x1) f8(I0) -> f3(0) f7(I1) -> f5(I1) [2 <= I1] f7(I2) -> f5(I2) [1 + I2 <= 2] f2(I3) -> f7(I3) f5(I4) -> f6(I4) f3(I5) -> f4(I5) f4(I6) -> f2(I6) f4(I7) -> f1(I7) f4(I8) -> f1(I8) f1(I9) -> f3(1 + I9) [1 + I9 <= 3] f1(I10) -> f2(I10) [3 <= I10] The dependency graph for this problem is: 0 -> 1 1 -> 5 2 -> 3 -> 4 -> 2, 3 5 -> 6, 7, 8 6 -> 4 7 -> 9, 10 8 -> 9, 10 9 -> 5 10 -> 4 Where: 0) f9#(x1) -> f8#(x1) 1) f8#(I0) -> f3#(0) 2) f7#(I1) -> f5#(I1) [2 <= I1] 3) f7#(I2) -> f5#(I2) [1 + I2 <= 2] 4) f2#(I3) -> f7#(I3) 5) f3#(I5) -> f4#(I5) 6) f4#(I6) -> f2#(I6) 7) f4#(I7) -> f1#(I7) 8) f4#(I8) -> f1#(I8) 9) f1#(I9) -> f3#(1 + I9) [1 + I9 <= 3] 10) f1#(I10) -> f2#(I10) [3 <= I10] We have the following SCCs. { 5, 7, 8, 9 } DP problem for innermost termination. P = f3#(I5) -> f4#(I5) f4#(I7) -> f1#(I7) f4#(I8) -> f1#(I8) f1#(I9) -> f3#(1 + I9) [1 + I9 <= 3] R = f9(x1) -> f8(x1) f8(I0) -> f3(0) f7(I1) -> f5(I1) [2 <= I1] f7(I2) -> f5(I2) [1 + I2 <= 2] f2(I3) -> f7(I3) f5(I4) -> f6(I4) f3(I5) -> f4(I5) f4(I6) -> f2(I6) f4(I7) -> f1(I7) f4(I8) -> f1(I8) f1(I9) -> f3(1 + I9) [1 + I9 <= 3] f1(I10) -> f2(I10) [3 <= I10] We use the extended value criterion with the projection function NU: NU[f1#(x0)] = -x0 + 2 NU[f4#(x0)] = -x0 + 2 NU[f3#(x0)] = -x0 + 2 This gives the following inequalities: ==> -I5 + 2 >= -I5 + 2 ==> -I7 + 2 >= -I7 + 2 ==> -I8 + 2 >= -I8 + 2 1 + I9 <= 3 ==> -I9 + 2 > -(1 + I9) + 2 with -I9 + 2 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I5) -> f4#(I5) f4#(I7) -> f1#(I7)
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