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ITS pair #487098376
details
property
value
status
complete
benchmark
n-1d.t2_fixed.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n146.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
11.4808 seconds
cpu usage
11.6705
user time
6.46481
system time
5.20572
max virtual memory
605844.0
max residence set size
10352.0
stage attributes
key
value
starexec-result
MAYBE
output
MAYBE DP problem for innermost termination. P = f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) f7#(I0, I1, I2, I3) -> f1#(I0, I1, I2, I3) f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13] f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0] f1#(I20, I21, I22, I23) -> f4#(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23] f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31] R = f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) f7(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) f6(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) f5(I8, I9, I10, I11) -> f6(I8, I9, 1 + I10, I11) f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 <= I13] f4(I16, I17, I18, I19) -> f5(I16, I17, I18, I19) [1 + I17 <= 0] f1(I20, I21, I22, I23) -> f4(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23] f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) f1(I28, I29, I30, I31) -> f3(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31] f1(I33, I34, I35, I36) -> f2(rnd1, I34, I35, I36) [rnd1 = rnd1 /\ -1 * I35 + I36 <= 0] The dependency graph for this problem is: 0 -> 1 1 -> 6, 8 2 -> 6, 8 3 -> 2 4 -> 3 5 -> 3 6 -> 4, 5 7 -> 6, 8 8 -> 7 Where: 0) f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 1) f7#(I0, I1, I2, I3) -> f1#(I0, I1, I2, I3) 2) f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) 3) f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) 4) f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13] 5) f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0] 6) f1#(I20, I21, I22, I23) -> f4#(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23] 7) f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 8) f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31] We have the following SCCs. { 2, 3, 4, 5, 6, 7, 8 } DP problem for innermost termination. P = f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13] f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0] f1#(I20, I21, I22, I23) -> f4#(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23] f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31] R = f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) f7(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) f6(I4, I5, I6, I7) -> f1(I4, I5, I6, I7) f5(I8, I9, I10, I11) -> f6(I8, I9, 1 + I10, I11) f4(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) [1 <= I13] f4(I16, I17, I18, I19) -> f5(I16, I17, I18, I19) [1 + I17 <= 0] f1(I20, I21, I22, I23) -> f4(I20, rnd2, I22, I23) [rnd2 = rnd2 /\ 0 <= -1 - I22 + I23] f3(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) f1(I28, I29, I30, I31) -> f3(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31] f1(I33, I34, I35, I36) -> f2(rnd1, I34, I35, I36) [rnd1 = rnd1 /\ -1 * I35 + I36 <= 0] We use the extended value criterion with the projection function NU: NU[f3#(x0,x1,x2,x3)] = -x2 + x3 + 1 NU[f4#(x0,x1,x2,x3)] = -x2 + x3 NU[f5#(x0,x1,x2,x3)] = -x2 + x3 NU[f1#(x0,x1,x2,x3)] = -x2 + x3 + 1 NU[f6#(x0,x1,x2,x3)] = -x2 + x3 + 1 This gives the following inequalities: ==> -I6 + I7 + 1 >= -I6 + I7 + 1 ==> -I10 + I11 >= -(1 + I10) + I11 + 1 1 <= I13 ==> -I14 + I15 >= -I14 + I15 1 + I17 <= 0 ==> -I18 + I19 >= -I18 + I19 rnd2 = rnd2 /\ 0 <= -1 - I22 + I23 ==> -I22 + I23 + 1 > -I22 + I23 with -I22 + I23 + 1 >= 0 ==> -I26 + I27 + 1 >= -I26 + I27 + 1 0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31 ==> -I30 + I31 + 1 >= -I30 + I31 + 1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f6#(I4, I5, I6, I7) -> f1#(I4, I5, I6, I7) f5#(I8, I9, I10, I11) -> f6#(I8, I9, 1 + I10, I11) f4#(I12, I13, I14, I15) -> f5#(I12, I13, I14, I15) [1 <= I13] f4#(I16, I17, I18, I19) -> f5#(I16, I17, I18, I19) [1 + I17 <= 0] f3#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) f1#(I28, I29, I30, I31) -> f3#(I28, I32, I30, I31) [0 <= I32 /\ I32 <= 0 /\ I32 = I32 /\ 0 <= -1 - I30 + I31] R = f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) f7(I0, I1, I2, I3) -> f1(I0, I1, I2, I3) f6(I4, I5, I6, I7) -> f1(I4, I5, I6, I7)
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