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Integer_Transition_Systems 2019-03-29 01.54 pair #432274554
details
property
value
status
complete
benchmark
ex3.t2_fixed.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n174.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
1.8056 seconds
cpu usage
1.8398
user time
0.937959
system time
0.901836
max virtual memory
684948.0
max residence set size
8672.0
stage attributes
key
value
starexec-result
YES
output
1.75/1.80 YES 1.75/1.80 1.75/1.80 DP problem for innermost termination. 1.75/1.80 P = 1.75/1.80 f8#(x1, x2, x3) -> f7#(x1, x2, x3) 1.75/1.80 f7#(I0, I1, I2) -> f4#(I0, 0, 10) 1.75/1.80 f2#(I3, I4, I5) -> f6#(I3, I4, I5) [2 * I5 <= I4] 1.75/1.80 f2#(I6, I7, I8) -> f6#(I6, I7, I8) [1 + I7 <= 2 * I8] 1.75/1.80 f3#(I12, I13, I14) -> f1#(I12, I13, I14) 1.75/1.80 f4#(I18, I19, I20) -> f3#(0, I19, I20) [2 <= I20] 1.75/1.80 f1#(I21, I22, I23) -> f3#(1 + I21, 2 + I22, I23) [1 + I21 <= I23] 1.75/1.80 f1#(I24, I25, I26) -> f2#(I24, I25, I26) [I26 <= I24] 1.75/1.80 R = 1.75/1.80 f8(x1, x2, x3) -> f7(x1, x2, x3) 1.75/1.80 f7(I0, I1, I2) -> f4(I0, 0, 10) 1.75/1.80 f2(I3, I4, I5) -> f6(I3, I4, I5) [2 * I5 <= I4] 1.75/1.80 f2(I6, I7, I8) -> f6(I6, I7, I8) [1 + I7 <= 2 * I8] 1.75/1.80 f6(I9, I10, I11) -> f5(I9, I10, I11) 1.75/1.80 f3(I12, I13, I14) -> f1(I12, I13, I14) 1.75/1.80 f4(I15, I16, I17) -> f5(I15, I16, I17) [I17 <= 1] 1.75/1.80 f4(I18, I19, I20) -> f3(0, I19, I20) [2 <= I20] 1.75/1.80 f1(I21, I22, I23) -> f3(1 + I21, 2 + I22, I23) [1 + I21 <= I23] 1.75/1.80 f1(I24, I25, I26) -> f2(I24, I25, I26) [I26 <= I24] 1.75/1.80 1.75/1.80 The dependency graph for this problem is: 1.75/1.80 0 -> 1 1.75/1.80 1 -> 5 1.75/1.80 2 -> 1.75/1.80 3 -> 1.75/1.80 4 -> 6, 7 1.75/1.80 5 -> 4 1.75/1.80 6 -> 4 1.75/1.80 7 -> 2, 3 1.75/1.80 Where: 1.75/1.80 0) f8#(x1, x2, x3) -> f7#(x1, x2, x3) 1.75/1.80 1) f7#(I0, I1, I2) -> f4#(I0, 0, 10) 1.75/1.80 2) f2#(I3, I4, I5) -> f6#(I3, I4, I5) [2 * I5 <= I4] 1.75/1.80 3) f2#(I6, I7, I8) -> f6#(I6, I7, I8) [1 + I7 <= 2 * I8] 1.75/1.80 4) f3#(I12, I13, I14) -> f1#(I12, I13, I14) 1.75/1.80 5) f4#(I18, I19, I20) -> f3#(0, I19, I20) [2 <= I20] 1.75/1.80 6) f1#(I21, I22, I23) -> f3#(1 + I21, 2 + I22, I23) [1 + I21 <= I23] 1.75/1.80 7) f1#(I24, I25, I26) -> f2#(I24, I25, I26) [I26 <= I24] 1.75/1.80 1.75/1.80 We have the following SCCs. 1.75/1.80 { 4, 6 } 1.75/1.80 1.75/1.80 DP problem for innermost termination. 1.75/1.80 P = 1.75/1.80 f3#(I12, I13, I14) -> f1#(I12, I13, I14) 1.75/1.80 f1#(I21, I22, I23) -> f3#(1 + I21, 2 + I22, I23) [1 + I21 <= I23] 1.75/1.80 R = 1.75/1.80 f8(x1, x2, x3) -> f7(x1, x2, x3) 1.75/1.80 f7(I0, I1, I2) -> f4(I0, 0, 10) 1.75/1.80 f2(I3, I4, I5) -> f6(I3, I4, I5) [2 * I5 <= I4] 1.75/1.80 f2(I6, I7, I8) -> f6(I6, I7, I8) [1 + I7 <= 2 * I8] 1.75/1.80 f6(I9, I10, I11) -> f5(I9, I10, I11) 1.75/1.80 f3(I12, I13, I14) -> f1(I12, I13, I14) 1.75/1.80 f4(I15, I16, I17) -> f5(I15, I16, I17) [I17 <= 1] 1.75/1.80 f4(I18, I19, I20) -> f3(0, I19, I20) [2 <= I20] 1.75/1.80 f1(I21, I22, I23) -> f3(1 + I21, 2 + I22, I23) [1 + I21 <= I23] 1.75/1.80 f1(I24, I25, I26) -> f2(I24, I25, I26) [I26 <= I24] 1.75/1.80 1.75/1.80 We use the reverse value criterion with the projection function NU: 1.75/1.80 NU[f1#(z1,z2,z3)] = z3 + -1 * (1 + z1) 1.75/1.80 NU[f3#(z1,z2,z3)] = z3 + -1 * (1 + z1) 1.75/1.80 1.75/1.80 This gives the following inequalities: 1.75/1.80 ==> I14 + -1 * (1 + I12) >= I14 + -1 * (1 + I12) 1.75/1.80 1 + I21 <= I23 ==> I23 + -1 * (1 + I21) > I23 + -1 * (1 + (1 + I21)) with I23 + -1 * (1 + I21) >= 0 1.75/1.80 1.75/1.80 We remove all the strictly oriented dependency pairs. 1.75/1.80 1.75/1.80 DP problem for innermost termination. 1.75/1.80 P = 1.75/1.80 f3#(I12, I13, I14) -> f1#(I12, I13, I14) 1.75/1.80 R = 1.75/1.80 f8(x1, x2, x3) -> f7(x1, x2, x3) 1.75/1.80 f7(I0, I1, I2) -> f4(I0, 0, 10) 1.75/1.80 f2(I3, I4, I5) -> f6(I3, I4, I5) [2 * I5 <= I4] 1.75/1.80 f2(I6, I7, I8) -> f6(I6, I7, I8) [1 + I7 <= 2 * I8] 1.75/1.80 f6(I9, I10, I11) -> f5(I9, I10, I11) 1.75/1.80 f3(I12, I13, I14) -> f1(I12, I13, I14) 1.75/1.80 f4(I15, I16, I17) -> f5(I15, I16, I17) [I17 <= 1] 1.75/1.80 f4(I18, I19, I20) -> f3(0, I19, I20) [2 <= I20] 1.75/1.80 f1(I21, I22, I23) -> f3(1 + I21, 2 + I22, I23) [1 + I21 <= I23] 1.75/1.80 f1(I24, I25, I26) -> f2(I24, I25, I26) [I26 <= I24] 1.75/1.80 1.75/1.80 The dependency graph for this problem is: 1.75/1.80 4 -> 1.75/1.80 Where: 1.75/1.80 4) f3#(I12, I13, I14) -> f1#(I12, I13, I14) 1.75/1.80 1.75/1.80 We have the following SCCs. 1.75/1.80 1.75/4.78 EOF
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