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ITS pair #487097575
details
property
value
status
complete
benchmark
fuhs-inflasso.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n151.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
6.72216 seconds
cpu usage
6.7958
user time
3.5698
system time
3.226
max virtual memory
743504.0
max residence set size
9036.0
stage attributes
key
value
starexec-result
YES
output
YES DP problem for innermost termination. P = f5#(x1, x2, x3) -> f4#(x1, x2, x3) f4#(I0, I1, I2) -> f2#(0, I1, I2) f2#(I3, I4, I5) -> f1#(I3, 0, I5) [1 + I3 <= I5] f3#(I6, I7, I8) -> f1#(I6, I7, I8) f1#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [1 + I10 <= I9] f1#(I12, I13, I14) -> f2#(1 + I12, I13, I14) [I12 <= I13] R = f5(x1, x2, x3) -> f4(x1, x2, x3) f4(I0, I1, I2) -> f2(0, I1, I2) f2(I3, I4, I5) -> f1(I3, 0, I5) [1 + I3 <= I5] f3(I6, I7, I8) -> f1(I6, I7, I8) f1(I9, I10, I11) -> f3(I9, 1 + I10, I11) [1 + I10 <= I9] f1(I12, I13, I14) -> f2(1 + I12, I13, I14) [I12 <= I13] The dependency graph for this problem is: 0 -> 1 1 -> 2 2 -> 4, 5 3 -> 4, 5 4 -> 3 5 -> 2 Where: 0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 1) f4#(I0, I1, I2) -> f2#(0, I1, I2) 2) f2#(I3, I4, I5) -> f1#(I3, 0, I5) [1 + I3 <= I5] 3) f3#(I6, I7, I8) -> f1#(I6, I7, I8) 4) f1#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [1 + I10 <= I9] 5) f1#(I12, I13, I14) -> f2#(1 + I12, I13, I14) [I12 <= I13] We have the following SCCs. { 2, 3, 4, 5 } DP problem for innermost termination. P = f2#(I3, I4, I5) -> f1#(I3, 0, I5) [1 + I3 <= I5] f3#(I6, I7, I8) -> f1#(I6, I7, I8) f1#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [1 + I10 <= I9] f1#(I12, I13, I14) -> f2#(1 + I12, I13, I14) [I12 <= I13] R = f5(x1, x2, x3) -> f4(x1, x2, x3) f4(I0, I1, I2) -> f2(0, I1, I2) f2(I3, I4, I5) -> f1(I3, 0, I5) [1 + I3 <= I5] f3(I6, I7, I8) -> f1(I6, I7, I8) f1(I9, I10, I11) -> f3(I9, 1 + I10, I11) [1 + I10 <= I9] f1(I12, I13, I14) -> f2(1 + I12, I13, I14) [I12 <= I13] We use the extended value criterion with the projection function NU: NU[f3#(x0,x1,x2)] = -x0 + x2 - 2 NU[f1#(x0,x1,x2)] = -x0 + x2 - 2 NU[f2#(x0,x1,x2)] = -x0 + x2 - 1 This gives the following inequalities: 1 + I3 <= I5 ==> -I3 + I5 - 1 > -I3 + I5 - 2 with -I3 + I5 - 1 >= 0 ==> -I6 + I8 - 2 >= -I6 + I8 - 2 1 + I10 <= I9 ==> -I9 + I11 - 2 >= -I9 + I11 - 2 I12 <= I13 ==> -I12 + I14 - 2 >= -(1 + I12) + I14 - 1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I6, I7, I8) -> f1#(I6, I7, I8) f1#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [1 + I10 <= I9] f1#(I12, I13, I14) -> f2#(1 + I12, I13, I14) [I12 <= I13] R = f5(x1, x2, x3) -> f4(x1, x2, x3) f4(I0, I1, I2) -> f2(0, I1, I2) f2(I3, I4, I5) -> f1(I3, 0, I5) [1 + I3 <= I5] f3(I6, I7, I8) -> f1(I6, I7, I8) f1(I9, I10, I11) -> f3(I9, 1 + I10, I11) [1 + I10 <= I9] f1(I12, I13, I14) -> f2(1 + I12, I13, I14) [I12 <= I13] The dependency graph for this problem is: 3 -> 4, 5 4 -> 3 5 -> Where: 3) f3#(I6, I7, I8) -> f1#(I6, I7, I8) 4) f1#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [1 + I10 <= I9] 5) f1#(I12, I13, I14) -> f2#(1 + I12, I13, I14) [I12 <= I13] We have the following SCCs. { 3, 4 } DP problem for innermost termination. P = f3#(I6, I7, I8) -> f1#(I6, I7, I8) f1#(I9, I10, I11) -> f3#(I9, 1 + I10, I11) [1 + I10 <= I9] R = f5(x1, x2, x3) -> f4(x1, x2, x3) f4(I0, I1, I2) -> f2(0, I1, I2) f2(I3, I4, I5) -> f1(I3, 0, I5) [1 + I3 <= I5] f3(I6, I7, I8) -> f1(I6, I7, I8) f1(I9, I10, I11) -> f3(I9, 1 + I10, I11) [1 + I10 <= I9] f1(I12, I13, I14) -> f2(1 + I12, I13, I14) [I12 <= I13]
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