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Integ Trans Syste 27634 pair #381740446
details
property
value
status
complete
benchmark
n-15.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n050.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
1.13208198547 seconds
cpu usage
1.201028109
max memory
1.4237696E7
stage attributes
key
value
output-size
2692
starexec-result
MAYBE
output
/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = f5#(x1) -> f1#(x1) f4#(I0) -> f2#(I0) f2#(I1) -> f4#(-1 + I1) [0 <= -1 + -1 + I1] f3#(I2) -> f2#(I2) f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0] f1#(I4) -> f2#(I4) R = f5(x1) -> f1(x1) f4(I0) -> f2(I0) f2(I1) -> f4(-1 + I1) [0 <= -1 + -1 + I1] f3(I2) -> f2(I2) f2(I3) -> f3(-1 + I3) [-1 + I3 <= 0] f1(I4) -> f2(I4) The dependency graph for this problem is: 0 -> 5 1 -> 2, 4 2 -> 1 3 -> 2, 4 4 -> 3 5 -> 2, 4 Where: 0) f5#(x1) -> f1#(x1) 1) f4#(I0) -> f2#(I0) 2) f2#(I1) -> f4#(-1 + I1) [0 <= -1 + -1 + I1] 3) f3#(I2) -> f2#(I2) 4) f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0] 5) f1#(I4) -> f2#(I4) We have the following SCCs. { 1, 2, 3, 4 } DP problem for innermost termination. P = f4#(I0) -> f2#(I0) f2#(I1) -> f4#(-1 + I1) [0 <= -1 + -1 + I1] f3#(I2) -> f2#(I2) f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0] R = f5(x1) -> f1(x1) f4(I0) -> f2(I0) f2(I1) -> f4(-1 + I1) [0 <= -1 + -1 + I1] f3(I2) -> f2(I2) f2(I3) -> f3(-1 + I3) [-1 + I3 <= 0] f1(I4) -> f2(I4) We use the reverse value criterion with the projection function NU: NU[f3#(z1)] = -1 + -1 + z1 + -1 * 0 NU[f2#(z1)] = -1 + -1 + z1 + -1 * 0 NU[f4#(z1)] = -1 + -1 + z1 + -1 * 0 This gives the following inequalities: ==> -1 + -1 + I0 + -1 * 0 >= -1 + -1 + I0 + -1 * 0 0 <= -1 + -1 + I1 ==> -1 + -1 + I1 + -1 * 0 > -1 + -1 + (-1 + I1) + -1 * 0 with -1 + -1 + I1 + -1 * 0 >= 0 ==> -1 + -1 + I2 + -1 * 0 >= -1 + -1 + I2 + -1 * 0 -1 + I3 <= 0 ==> -1 + -1 + I3 + -1 * 0 >= -1 + -1 + (-1 + I3) + -1 * 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f4#(I0) -> f2#(I0) f3#(I2) -> f2#(I2) f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0] R = f5(x1) -> f1(x1) f4(I0) -> f2(I0) f2(I1) -> f4(-1 + I1) [0 <= -1 + -1 + I1] f3(I2) -> f2(I2) f2(I3) -> f3(-1 + I3) [-1 + I3 <= 0] f1(I4) -> f2(I4) The dependency graph for this problem is: 1 -> 4 3 -> 4 4 -> 3 Where: 1) f4#(I0) -> f2#(I0) 3) f3#(I2) -> f2#(I2) 4) f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0] We have the following SCCs. { 3, 4 } DP problem for innermost termination. P = f3#(I2) -> f2#(I2) f2#(I3) -> f3#(-1 + I3) [-1 + I3 <= 0] R = f5(x1) -> f1(x1)
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