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Integer_Transition_Systems 2019-03-29 01.54 pair #432275220
details
property
value
status
complete
benchmark
n-12.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n083.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
10.93 seconds
cpu usage
11.1048
user time
5.42953
system time
5.67523
max virtual memory
698596.0
max residence set size
9288.0
stage attributes
key
value
starexec-result
MAYBE
output
11.04/10.92 MAYBE 11.04/10.92 11.04/10.92 DP problem for innermost termination. 11.04/10.92 P = 11.04/10.92 f8#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) 11.04/10.92 f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 11.04/10.92 f6#(I4, I5, I6, I7) -> f7#(I4, I5, I6, -1 + I7) 11.04/10.92 f5#(I8, I9, I10, I11) -> f6#(I8, I9, I10, I11) [I9 = I9] 11.04/10.92 f2#(I12, I13, I14, I15) -> f5#(I12, I13, rnd3, I15) [rnd3 = rnd3 /\ -1 * I15 <= 0] 11.04/10.92 f4#(I16, I17, I18, I19) -> f2#(I16, I17, I18, I19) 11.04/10.92 f2#(I20, I21, I22, I23) -> f4#(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0] 11.04/10.92 f1#(I29, I30, I31, I32) -> f2#(I29, I30, I31, I32) 11.04/10.92 R = 11.04/10.92 f8(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 11.04/10.92 f7(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 11.04/10.92 f6(I4, I5, I6, I7) -> f7(I4, I5, I6, -1 + I7) 11.04/10.92 f5(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) [I9 = I9] 11.04/10.92 f2(I12, I13, I14, I15) -> f5(I12, I13, rnd3, I15) [rnd3 = rnd3 /\ -1 * I15 <= 0] 11.04/10.92 f4(I16, I17, I18, I19) -> f2(I16, I17, I18, I19) 11.04/10.92 f2(I20, I21, I22, I23) -> f4(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0] 11.04/10.92 f2(I25, I26, I27, I28) -> f3(rnd1, I26, I27, I28) [rnd1 = rnd1 /\ 0 <= -1 - I28] 11.04/10.92 f1(I29, I30, I31, I32) -> f2(I29, I30, I31, I32) 11.04/10.92 11.04/10.92 The dependency graph for this problem is: 11.04/10.92 0 -> 7 11.04/10.92 1 -> 4, 6 11.04/10.92 2 -> 1 11.04/10.92 3 -> 2 11.04/10.92 4 -> 3 11.04/10.92 5 -> 4, 6 11.04/10.92 6 -> 5 11.04/10.92 7 -> 4, 6 11.04/10.92 Where: 11.04/10.92 0) f8#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) 11.04/10.92 1) f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 11.04/10.92 2) f6#(I4, I5, I6, I7) -> f7#(I4, I5, I6, -1 + I7) 11.04/10.92 3) f5#(I8, I9, I10, I11) -> f6#(I8, I9, I10, I11) [I9 = I9] 11.04/10.92 4) f2#(I12, I13, I14, I15) -> f5#(I12, I13, rnd3, I15) [rnd3 = rnd3 /\ -1 * I15 <= 0] 11.04/10.92 5) f4#(I16, I17, I18, I19) -> f2#(I16, I17, I18, I19) 11.04/10.92 6) f2#(I20, I21, I22, I23) -> f4#(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0] 11.04/10.92 7) f1#(I29, I30, I31, I32) -> f2#(I29, I30, I31, I32) 11.04/10.92 11.04/10.92 We have the following SCCs. 11.04/10.92 { 1, 2, 3, 4, 5, 6 } 11.04/10.92 11.04/10.92 DP problem for innermost termination. 11.04/10.92 P = 11.04/10.92 f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 11.04/10.92 f6#(I4, I5, I6, I7) -> f7#(I4, I5, I6, -1 + I7) 11.04/10.92 f5#(I8, I9, I10, I11) -> f6#(I8, I9, I10, I11) [I9 = I9] 11.04/10.92 f2#(I12, I13, I14, I15) -> f5#(I12, I13, rnd3, I15) [rnd3 = rnd3 /\ -1 * I15 <= 0] 11.04/10.92 f4#(I16, I17, I18, I19) -> f2#(I16, I17, I18, I19) 11.04/10.92 f2#(I20, I21, I22, I23) -> f4#(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0] 11.04/10.92 R = 11.04/10.92 f8(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 11.04/10.92 f7(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 11.04/10.92 f6(I4, I5, I6, I7) -> f7(I4, I5, I6, -1 + I7) 11.04/10.92 f5(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) [I9 = I9] 11.04/10.92 f2(I12, I13, I14, I15) -> f5(I12, I13, rnd3, I15) [rnd3 = rnd3 /\ -1 * I15 <= 0] 11.04/10.92 f4(I16, I17, I18, I19) -> f2(I16, I17, I18, I19) 11.04/10.92 f2(I20, I21, I22, I23) -> f4(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0] 11.04/10.92 f2(I25, I26, I27, I28) -> f3(rnd1, I26, I27, I28) [rnd1 = rnd1 /\ 0 <= -1 - I28] 11.04/10.92 f1(I29, I30, I31, I32) -> f2(I29, I30, I31, I32) 11.04/10.92 11.04/10.92 We use the extended value criterion with the projection function NU: 11.04/10.92 NU[f4#(x0,x1,x2,x3)] = x3 11.04/10.92 NU[f5#(x0,x1,x2,x3)] = x3 - 1 11.04/10.92 NU[f6#(x0,x1,x2,x3)] = x3 - 1 11.04/10.92 NU[f2#(x0,x1,x2,x3)] = x3 11.04/10.92 NU[f7#(x0,x1,x2,x3)] = x3 11.04/10.92 11.04/10.92 This gives the following inequalities: 11.04/10.92 ==> I3 >= I3 11.04/10.92 ==> I7 - 1 >= (-1 + I7) 11.04/10.92 I9 = I9 ==> I11 - 1 >= I11 - 1 11.04/10.92 rnd3 = rnd3 /\ -1 * I15 <= 0 ==> I15 > I15 - 1 with I15 >= 0 11.04/10.92 ==> I19 >= I19 11.04/10.92 0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0 ==> I23 >= I23 11.04/10.92 11.04/10.92 We remove all the strictly oriented dependency pairs. 11.04/10.92 11.04/10.92 DP problem for innermost termination. 11.04/10.92 P = 11.04/10.92 f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 11.04/10.92 f6#(I4, I5, I6, I7) -> f7#(I4, I5, I6, -1 + I7) 11.04/10.92 f5#(I8, I9, I10, I11) -> f6#(I8, I9, I10, I11) [I9 = I9] 11.04/10.92 f4#(I16, I17, I18, I19) -> f2#(I16, I17, I18, I19) 11.04/10.92 f2#(I20, I21, I22, I23) -> f4#(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0] 11.04/10.92 R = 11.04/10.92 f8(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 11.04/10.92 f7(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 11.04/10.92 f6(I4, I5, I6, I7) -> f7(I4, I5, I6, -1 + I7) 11.04/10.92 f5(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) [I9 = I9] 11.04/10.92 f2(I12, I13, I14, I15) -> f5(I12, I13, rnd3, I15) [rnd3 = rnd3 /\ -1 * I15 <= 0] 11.04/10.92 f4(I16, I17, I18, I19) -> f2(I16, I17, I18, I19) 11.04/10.92 f2(I20, I21, I22, I23) -> f4(I20, I21, I24, I23) [0 <= I24 /\ I24 <= 0 /\ I24 = I24 /\ -1 * I23 <= 0] 11.04/10.92 f2(I25, I26, I27, I28) -> f3(rnd1, I26, I27, I28) [rnd1 = rnd1 /\ 0 <= -1 - I28] 11.04/10.92 f1(I29, I30, I31, I32) -> f2(I29, I30, I31, I32) 11.04/10.92 11.04/10.92 The dependency graph for this problem is:
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