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Integer_Transition_Systems 2019-03-29 01.54 pair #432275883
details
property
value
status
complete
benchmark
p-56.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n139.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
8.93828 seconds
cpu usage
9.0875
user time
4.4841
system time
4.6034
max virtual memory
698828.0
max residence set size
9240.0
stage attributes
key
value
starexec-result
YES
output
8.99/8.93 YES 8.99/8.93 8.99/8.93 DP problem for innermost termination. 8.99/8.93 P = 8.99/8.93 f6#(x1, x2, x3, x4, x5, x6) -> f5#(x1, x2, x3, x4, x5, x6) 8.99/8.93 f5#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) [0 <= -2 + I9 /\ 0 <= -2 + I8] 8.99/8.93 f3#(I19, I20, I21, I22, I23, I24) -> f1#(I19, I20, I21, I22, I23, I24) 8.99/8.93 f2#(I25, I26, I27, I28, I29, I30) -> f3#(I25, I26, I27, I28, I29, I30) [I26 = I26] 8.99/8.93 f1#(I31, I32, I33, I34, I35, I36) -> f2#(I31, I32, I33, I34, 1 + I35, 1 + I36) [0 <= -2 - I35] 8.99/8.93 R = 8.99/8.93 f6(x1, x2, x3, x4, x5, x6) -> f5(x1, x2, x3, x4, x5, x6) 8.99/8.93 f5(I0, I1, I2, I3, I4, I5) -> f4(rnd1, I1, I2, I3, I4, I5) [rnd1 = rnd1 /\ -1 + I2 <= 0] 8.99/8.93 f5(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) [0 <= -2 + I9 /\ 0 <= -2 + I8] 8.99/8.93 f1(I12, I13, I14, I15, I16, I17) -> f4(I18, I13, I14, I15, I16, I17) [I18 = I18 /\ -1 - I16 <= 0] 8.99/8.93 f3(I19, I20, I21, I22, I23, I24) -> f1(I19, I20, I21, I22, I23, I24) 8.99/8.93 f2(I25, I26, I27, I28, I29, I30) -> f3(I25, I26, I27, I28, I29, I30) [I26 = I26] 8.99/8.93 f1(I31, I32, I33, I34, I35, I36) -> f2(I31, I32, I33, I34, 1 + I35, 1 + I36) [0 <= -2 - I35] 8.99/8.93 8.99/8.93 The dependency graph for this problem is: 8.99/8.93 0 -> 1 8.99/8.93 1 -> 4 8.99/8.93 2 -> 4 8.99/8.93 3 -> 2 8.99/8.93 4 -> 3 8.99/8.93 Where: 8.99/8.93 0) f6#(x1, x2, x3, x4, x5, x6) -> f5#(x1, x2, x3, x4, x5, x6) 8.99/8.93 1) f5#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) [0 <= -2 + I9 /\ 0 <= -2 + I8] 8.99/8.93 2) f3#(I19, I20, I21, I22, I23, I24) -> f1#(I19, I20, I21, I22, I23, I24) 8.99/8.93 3) f2#(I25, I26, I27, I28, I29, I30) -> f3#(I25, I26, I27, I28, I29, I30) [I26 = I26] 8.99/8.93 4) f1#(I31, I32, I33, I34, I35, I36) -> f2#(I31, I32, I33, I34, 1 + I35, 1 + I36) [0 <= -2 - I35] 8.99/8.93 8.99/8.93 We have the following SCCs. 8.99/8.93 { 2, 3, 4 } 8.99/8.93 8.99/8.93 DP problem for innermost termination. 8.99/8.93 P = 8.99/8.93 f3#(I19, I20, I21, I22, I23, I24) -> f1#(I19, I20, I21, I22, I23, I24) 8.99/8.93 f2#(I25, I26, I27, I28, I29, I30) -> f3#(I25, I26, I27, I28, I29, I30) [I26 = I26] 8.99/8.93 f1#(I31, I32, I33, I34, I35, I36) -> f2#(I31, I32, I33, I34, 1 + I35, 1 + I36) [0 <= -2 - I35] 8.99/8.93 R = 8.99/8.93 f6(x1, x2, x3, x4, x5, x6) -> f5(x1, x2, x3, x4, x5, x6) 8.99/8.93 f5(I0, I1, I2, I3, I4, I5) -> f4(rnd1, I1, I2, I3, I4, I5) [rnd1 = rnd1 /\ -1 + I2 <= 0] 8.99/8.93 f5(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) [0 <= -2 + I9 /\ 0 <= -2 + I8] 8.99/8.93 f1(I12, I13, I14, I15, I16, I17) -> f4(I18, I13, I14, I15, I16, I17) [I18 = I18 /\ -1 - I16 <= 0] 8.99/8.93 f3(I19, I20, I21, I22, I23, I24) -> f1(I19, I20, I21, I22, I23, I24) 8.99/8.93 f2(I25, I26, I27, I28, I29, I30) -> f3(I25, I26, I27, I28, I29, I30) [I26 = I26] 8.99/8.93 f1(I31, I32, I33, I34, I35, I36) -> f2(I31, I32, I33, I34, 1 + I35, 1 + I36) [0 <= -2 - I35] 8.99/8.93 8.99/8.93 We use the extended value criterion with the projection function NU: 8.99/8.93 NU[f2#(x0,x1,x2,x3,x4,x5)] = -x4 - 2 8.99/8.93 NU[f1#(x0,x1,x2,x3,x4,x5)] = -x4 - 2 8.99/8.93 NU[f3#(x0,x1,x2,x3,x4,x5)] = -x4 - 2 8.99/8.93 8.99/8.93 This gives the following inequalities: 8.99/8.93 ==> -I23 - 2 >= -I23 - 2 8.99/8.93 I26 = I26 ==> -I29 - 2 >= -I29 - 2 8.99/8.93 0 <= -2 - I35 ==> -I35 - 2 > -(1 + I35) - 2 with -I35 - 2 >= 0 8.99/8.93 8.99/8.93 We remove all the strictly oriented dependency pairs. 8.99/8.93 8.99/8.93 DP problem for innermost termination. 8.99/8.93 P = 8.99/8.93 f3#(I19, I20, I21, I22, I23, I24) -> f1#(I19, I20, I21, I22, I23, I24) 8.99/8.93 f2#(I25, I26, I27, I28, I29, I30) -> f3#(I25, I26, I27, I28, I29, I30) [I26 = I26] 8.99/8.93 R = 8.99/8.93 f6(x1, x2, x3, x4, x5, x6) -> f5(x1, x2, x3, x4, x5, x6) 8.99/8.93 f5(I0, I1, I2, I3, I4, I5) -> f4(rnd1, I1, I2, I3, I4, I5) [rnd1 = rnd1 /\ -1 + I2 <= 0] 8.99/8.93 f5(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) [0 <= -2 + I9 /\ 0 <= -2 + I8] 8.99/8.93 f1(I12, I13, I14, I15, I16, I17) -> f4(I18, I13, I14, I15, I16, I17) [I18 = I18 /\ -1 - I16 <= 0] 8.99/8.93 f3(I19, I20, I21, I22, I23, I24) -> f1(I19, I20, I21, I22, I23, I24) 8.99/8.93 f2(I25, I26, I27, I28, I29, I30) -> f3(I25, I26, I27, I28, I29, I30) [I26 = I26] 8.99/8.93 f1(I31, I32, I33, I34, I35, I36) -> f2(I31, I32, I33, I34, 1 + I35, 1 + I36) [0 <= -2 - I35] 8.99/8.93 8.99/8.93 The dependency graph for this problem is: 8.99/8.93 2 -> 8.99/8.93 3 -> 2 8.99/8.93 Where: 8.99/8.93 2) f3#(I19, I20, I21, I22, I23, I24) -> f1#(I19, I20, I21, I22, I23, I24) 8.99/8.93 3) f2#(I25, I26, I27, I28, I29, I30) -> f3#(I25, I26, I27, I28, I29, I30) [I26 = I26] 8.99/8.93 8.99/8.93 We have the following SCCs. 8.99/8.93 8.99/11.91 EOF
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