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Integer_Transition_Systems 2019-03-29 01.54 pair #432276204
details
property
value
status
complete
benchmark
p-55.t2_fixed.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n039.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
13.086 seconds
cpu usage
13.2704
user time
6.67235
system time
6.59806
max virtual memory
711416.0
max residence set size
11292.0
stage attributes
key
value
starexec-result
YES
output
13.16/13.08 YES 13.16/13.08 13.16/13.08 DP problem for innermost termination. 13.16/13.08 P = 13.16/13.08 f8#(x1, x2, x3) -> f7#(x1, x2, x3) 13.16/13.08 f7#(I0, I1, I2) -> f1#(I0, I1, I2) 13.16/13.08 f6#(I3, I4, I5) -> f1#(I3, I4, I5) 13.16/13.08 f1#(I6, I7, I8) -> f6#(I6, -99 + I7, 1 + I8) [-1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7] 13.16/13.08 f5#(I9, I10, I11) -> f1#(I9, I10, I11) 13.16/13.08 f4#(I12, I13, I14) -> f5#(I12, 1 + I13, 1 + I14) 13.16/13.08 f3#(I15, I16, I17) -> f4#(I15, I16, I17) [0 <= I17] 13.16/13.08 f3#(I18, I19, I20) -> f4#(I18, I19, I20) [1 + I20 <= -1] 13.16/13.08 f1#(I21, I22, I23) -> f3#(I21, I22, I23) [0 <= -1 - I22] 13.16/13.08 R = 13.16/13.08 f8(x1, x2, x3) -> f7(x1, x2, x3) 13.16/13.08 f7(I0, I1, I2) -> f1(I0, I1, I2) 13.16/13.08 f6(I3, I4, I5) -> f1(I3, I4, I5) 13.16/13.08 f1(I6, I7, I8) -> f6(I6, -99 + I7, 1 + I8) [-1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7] 13.16/13.08 f5(I9, I10, I11) -> f1(I9, I10, I11) 13.16/13.08 f4(I12, I13, I14) -> f5(I12, 1 + I13, 1 + I14) 13.16/13.08 f3(I15, I16, I17) -> f4(I15, I16, I17) [0 <= I17] 13.16/13.08 f3(I18, I19, I20) -> f4(I18, I19, I20) [1 + I20 <= -1] 13.16/13.08 f1(I21, I22, I23) -> f3(I21, I22, I23) [0 <= -1 - I22] 13.16/13.08 f1(I24, I25, I26) -> f2(rnd1, I25, I26) [rnd1 = rnd1 /\ -1 * I25 <= 0] 13.16/13.08 13.16/13.08 The dependency graph for this problem is: 13.16/13.08 0 -> 1 13.16/13.08 1 -> 3, 8 13.16/13.08 2 -> 3, 8 13.16/13.08 3 -> 2 13.16/13.08 4 -> 3, 8 13.16/13.08 5 -> 4 13.16/13.08 6 -> 5 13.16/13.08 7 -> 5 13.16/13.08 8 -> 6, 7 13.16/13.08 Where: 13.16/13.08 0) f8#(x1, x2, x3) -> f7#(x1, x2, x3) 13.16/13.08 1) f7#(I0, I1, I2) -> f1#(I0, I1, I2) 13.16/13.08 2) f6#(I3, I4, I5) -> f1#(I3, I4, I5) 13.16/13.08 3) f1#(I6, I7, I8) -> f6#(I6, -99 + I7, 1 + I8) [-1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7] 13.16/13.08 4) f5#(I9, I10, I11) -> f1#(I9, I10, I11) 13.16/13.08 5) f4#(I12, I13, I14) -> f5#(I12, 1 + I13, 1 + I14) 13.16/13.08 6) f3#(I15, I16, I17) -> f4#(I15, I16, I17) [0 <= I17] 13.16/13.08 7) f3#(I18, I19, I20) -> f4#(I18, I19, I20) [1 + I20 <= -1] 13.16/13.08 8) f1#(I21, I22, I23) -> f3#(I21, I22, I23) [0 <= -1 - I22] 13.16/13.08 13.16/13.08 We have the following SCCs. 13.16/13.08 { 2, 3, 4, 5, 6, 7, 8 } 13.16/13.08 13.16/13.08 DP problem for innermost termination. 13.16/13.08 P = 13.16/13.08 f6#(I3, I4, I5) -> f1#(I3, I4, I5) 13.16/13.08 f1#(I6, I7, I8) -> f6#(I6, -99 + I7, 1 + I8) [-1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7] 13.16/13.08 f5#(I9, I10, I11) -> f1#(I9, I10, I11) 13.16/13.08 f4#(I12, I13, I14) -> f5#(I12, 1 + I13, 1 + I14) 13.16/13.08 f3#(I15, I16, I17) -> f4#(I15, I16, I17) [0 <= I17] 13.16/13.08 f3#(I18, I19, I20) -> f4#(I18, I19, I20) [1 + I20 <= -1] 13.16/13.08 f1#(I21, I22, I23) -> f3#(I21, I22, I23) [0 <= -1 - I22] 13.16/13.08 R = 13.16/13.08 f8(x1, x2, x3) -> f7(x1, x2, x3) 13.16/13.08 f7(I0, I1, I2) -> f1(I0, I1, I2) 13.16/13.08 f6(I3, I4, I5) -> f1(I3, I4, I5) 13.16/13.08 f1(I6, I7, I8) -> f6(I6, -99 + I7, 1 + I8) [-1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7] 13.16/13.08 f5(I9, I10, I11) -> f1(I9, I10, I11) 13.16/13.08 f4(I12, I13, I14) -> f5(I12, 1 + I13, 1 + I14) 13.16/13.08 f3(I15, I16, I17) -> f4(I15, I16, I17) [0 <= I17] 13.16/13.08 f3(I18, I19, I20) -> f4(I18, I19, I20) [1 + I20 <= -1] 13.16/13.08 f1(I21, I22, I23) -> f3(I21, I22, I23) [0 <= -1 - I22] 13.16/13.08 f1(I24, I25, I26) -> f2(rnd1, I25, I26) [rnd1 = rnd1 /\ -1 * I25 <= 0] 13.16/13.08 13.16/13.08 We use the extended value criterion with the projection function NU: 13.16/13.08 NU[f3#(x0,x1,x2)] = -x2 - 2 13.16/13.08 NU[f4#(x0,x1,x2)] = -x2 - 2 13.16/13.08 NU[f5#(x0,x1,x2)] = -x2 - 1 13.16/13.08 NU[f1#(x0,x1,x2)] = -x2 - 1 13.16/13.08 NU[f6#(x0,x1,x2)] = -x2 - 1 13.16/13.08 13.16/13.08 This gives the following inequalities: 13.16/13.08 ==> -I5 - 1 >= -I5 - 1 13.16/13.08 -1 <= I8 /\ I8 <= -1 /\ 0 <= -1 - I7 ==> -I8 - 1 > -(1 + I8) - 1 with -I8 - 1 >= 0 13.16/13.08 ==> -I11 - 1 >= -I11 - 1 13.16/13.08 ==> -I14 - 2 >= -(1 + I14) - 1 13.16/13.08 0 <= I17 ==> -I17 - 2 >= -I17 - 2 13.16/13.08 1 + I20 <= -1 ==> -I20 - 2 >= -I20 - 2 13.16/13.08 0 <= -1 - I22 ==> -I23 - 1 >= -I23 - 2 13.16/13.08 13.16/13.08 We remove all the strictly oriented dependency pairs. 13.16/13.08 13.16/13.08 DP problem for innermost termination. 13.16/13.08 P = 13.16/13.08 f6#(I3, I4, I5) -> f1#(I3, I4, I5) 13.16/13.08 f5#(I9, I10, I11) -> f1#(I9, I10, I11) 13.16/13.08 f4#(I12, I13, I14) -> f5#(I12, 1 + I13, 1 + I14) 13.16/13.08 f3#(I15, I16, I17) -> f4#(I15, I16, I17) [0 <= I17] 13.16/13.08 f3#(I18, I19, I20) -> f4#(I18, I19, I20) [1 + I20 <= -1] 13.16/13.08 f1#(I21, I22, I23) -> f3#(I21, I22, I23) [0 <= -1 - I22] 13.16/13.08 R = 13.16/13.08 f8(x1, x2, x3) -> f7(x1, x2, x3) 13.16/13.08 f7(I0, I1, I2) -> f1(I0, I1, I2) 13.16/13.08 f6(I3, I4, I5) -> f1(I3, I4, I5)
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