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Integer_Transition_Systems 2019-03-29 01.54 pair #432275850
details
property
value
status
complete
benchmark
slayer-2-filtered.t2_fixed.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n123.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
4.20221 seconds
cpu usage
4.28805
user time
2.29043
system time
1.99762
max virtual memory
725772.0
max residence set size
8732.0
stage attributes
key
value
starexec-result
YES
output
4.23/4.20 YES 4.23/4.20 4.23/4.20 DP problem for innermost termination. 4.23/4.20 P = 4.23/4.20 f7#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) 4.23/4.20 f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 4.23/4.20 f2#(I5, I6, I7, I8, I9) -> f6#(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3] 4.23/4.20 f4#(I15, I16, I17, I18, I19) -> f2#(I15, 1 + I16, I17, I18, I19) [1 + I16 <= 1 /\ 1 <= 1 + I16 /\ 1 + I16 <= 10] 4.23/4.20 f3#(I20, I21, I22, I23, I24) -> f2#(rnd1, rnd2, I22, I23, I24) [1 + I21 <= 10 /\ y1 = 1 + I21 /\ rnd2 = rnd2 /\ 2 <= rnd2 /\ rnd2 <= 2 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1] 4.23/4.20 f1#(I25, I26, I27, I28, I29) -> f2#(I25, 0, I27, I28, I29) [0 <= 0 /\ 0 <= 0] 4.23/4.20 R = 4.23/4.20 f7(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 4.23/4.20 f6(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 4.23/4.20 f2(I5, I6, I7, I8, I9) -> f6(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3] 4.23/4.20 f2(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I14, I14) [10 <= I11] 4.23/4.20 f4(I15, I16, I17, I18, I19) -> f2(I15, 1 + I16, I17, I18, I19) [1 + I16 <= 1 /\ 1 <= 1 + I16 /\ 1 + I16 <= 10] 4.23/4.20 f3(I20, I21, I22, I23, I24) -> f2(rnd1, rnd2, I22, I23, I24) [1 + I21 <= 10 /\ y1 = 1 + I21 /\ rnd2 = rnd2 /\ 2 <= rnd2 /\ rnd2 <= 2 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1] 4.23/4.20 f1(I25, I26, I27, I28, I29) -> f2(I25, 0, I27, I28, I29) [0 <= 0 /\ 0 <= 0] 4.23/4.20 4.23/4.20 The dependency graph for this problem is: 4.23/4.20 0 -> 5 4.23/4.20 1 -> 2 4.23/4.20 2 -> 1 4.23/4.20 3 -> 2 4.23/4.20 4 -> 2 4.23/4.20 5 -> 2 4.23/4.20 Where: 4.23/4.20 0) f7#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) 4.23/4.20 1) f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 4.23/4.20 2) f2#(I5, I6, I7, I8, I9) -> f6#(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3] 4.23/4.20 3) f4#(I15, I16, I17, I18, I19) -> f2#(I15, 1 + I16, I17, I18, I19) [1 + I16 <= 1 /\ 1 <= 1 + I16 /\ 1 + I16 <= 10] 4.23/4.20 4) f3#(I20, I21, I22, I23, I24) -> f2#(rnd1, rnd2, I22, I23, I24) [1 + I21 <= 10 /\ y1 = 1 + I21 /\ rnd2 = rnd2 /\ 2 <= rnd2 /\ rnd2 <= 2 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1] 4.23/4.20 5) f1#(I25, I26, I27, I28, I29) -> f2#(I25, 0, I27, I28, I29) [0 <= 0 /\ 0 <= 0] 4.23/4.20 4.23/4.20 We have the following SCCs. 4.23/4.20 { 1, 2 } 4.23/4.20 4.23/4.20 DP problem for innermost termination. 4.23/4.20 P = 4.23/4.20 f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 4.23/4.20 f2#(I5, I6, I7, I8, I9) -> f6#(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3] 4.23/4.20 R = 4.23/4.20 f7(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 4.23/4.20 f6(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 4.23/4.20 f2(I5, I6, I7, I8, I9) -> f6(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3] 4.23/4.20 f2(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I14, I14) [10 <= I11] 4.23/4.20 f4(I15, I16, I17, I18, I19) -> f2(I15, 1 + I16, I17, I18, I19) [1 + I16 <= 1 /\ 1 <= 1 + I16 /\ 1 + I16 <= 10] 4.23/4.20 f3(I20, I21, I22, I23, I24) -> f2(rnd1, rnd2, I22, I23, I24) [1 + I21 <= 10 /\ y1 = 1 + I21 /\ rnd2 = rnd2 /\ 2 <= rnd2 /\ rnd2 <= 2 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1] 4.23/4.20 f1(I25, I26, I27, I28, I29) -> f2(I25, 0, I27, I28, I29) [0 <= 0 /\ 0 <= 0] 4.23/4.20 4.23/4.20 We use the reverse value criterion with the projection function NU: 4.23/4.20 NU[f2#(z1,z2,z3,z4,z5)] = 10 + -1 * (1 + z2) 4.23/4.20 NU[f6#(z1,z2,z3,z4,z5)] = 10 + -1 * (1 + z2) 4.23/4.20 4.23/4.20 This gives the following inequalities: 4.23/4.20 ==> 10 + -1 * (1 + I1) >= 10 + -1 * (1 + I1) 4.23/4.20 1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3 ==> 10 + -1 * (1 + I6) > 10 + -1 * (1 + (1 + I6)) with 10 + -1 * (1 + I6) >= 0 4.23/4.20 4.23/4.20 We remove all the strictly oriented dependency pairs. 4.23/4.20 4.23/4.20 DP problem for innermost termination. 4.23/4.20 P = 4.23/4.20 f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 4.23/4.20 R = 4.23/4.20 f7(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 4.23/4.20 f6(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 4.23/4.20 f2(I5, I6, I7, I8, I9) -> f6(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3] 4.23/4.20 f2(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I14, I14) [10 <= I11] 4.23/4.20 f4(I15, I16, I17, I18, I19) -> f2(I15, 1 + I16, I17, I18, I19) [1 + I16 <= 1 /\ 1 <= 1 + I16 /\ 1 + I16 <= 10] 4.23/4.20 f3(I20, I21, I22, I23, I24) -> f2(rnd1, rnd2, I22, I23, I24) [1 + I21 <= 10 /\ y1 = 1 + I21 /\ rnd2 = rnd2 /\ 2 <= rnd2 /\ rnd2 <= 2 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1] 4.23/4.20 f1(I25, I26, I27, I28, I29) -> f2(I25, 0, I27, I28, I29) [0 <= 0 /\ 0 <= 0] 4.23/4.20 4.23/4.20 The dependency graph for this problem is: 4.23/4.20 1 -> 4.23/4.20 Where: 4.23/4.20 1) f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 4.23/4.20 4.23/4.20 We have the following SCCs. 4.23/4.20 4.23/7.17 EOF
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