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ITS pair #487099021
details
property
value
status
complete
benchmark
241.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n144.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
3.01104 seconds
cpu usage
2.99424
user time
1.56353
system time
1.4307
max virtual memory
579452.0
max residence set size
8572.0
stage attributes
key
value
starexec-result
YES
output
YES DP problem for innermost termination. P = f5#(x1, x2) -> f4#(x1, x2) f2#(I0, I1) -> f1#(I0, I0) [1 <= I0] f4#(I2, I3) -> f2#(I2, I3) f3#(I4, I5) -> f1#(I4, I5) f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7] f1#(I8, I9) -> f2#(-1 + I8, 1 + I9) [I9 <= 0] R = f5(x1, x2) -> f4(x1, x2) f2(I0, I1) -> f1(I0, I0) [1 <= I0] f4(I2, I3) -> f2(I2, I3) f3(I4, I5) -> f1(I4, I5) f1(I6, I7) -> f3(I6, -1 + I7) [1 <= I7] f1(I8, I9) -> f2(-1 + I8, 1 + I9) [I9 <= 0] The dependency graph for this problem is: 0 -> 2 1 -> 4 2 -> 1 3 -> 4, 5 4 -> 3 5 -> 1 Where: 0) f5#(x1, x2) -> f4#(x1, x2) 1) f2#(I0, I1) -> f1#(I0, I0) [1 <= I0] 2) f4#(I2, I3) -> f2#(I2, I3) 3) f3#(I4, I5) -> f1#(I4, I5) 4) f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7] 5) f1#(I8, I9) -> f2#(-1 + I8, 1 + I9) [I9 <= 0] We have the following SCCs. { 1, 3, 4, 5 } DP problem for innermost termination. P = f2#(I0, I1) -> f1#(I0, I0) [1 <= I0] f3#(I4, I5) -> f1#(I4, I5) f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7] f1#(I8, I9) -> f2#(-1 + I8, 1 + I9) [I9 <= 0] R = f5(x1, x2) -> f4(x1, x2) f2(I0, I1) -> f1(I0, I0) [1 <= I0] f4(I2, I3) -> f2(I2, I3) f3(I4, I5) -> f1(I4, I5) f1(I6, I7) -> f3(I6, -1 + I7) [1 <= I7] f1(I8, I9) -> f2(-1 + I8, 1 + I9) [I9 <= 0] We use the extended value criterion with the projection function NU: NU[f3#(x0,x1)] = x0 - 2 NU[f1#(x0,x1)] = x0 - 2 NU[f2#(x0,x1)] = x0 - 1 This gives the following inequalities: 1 <= I0 ==> I0 - 1 > I0 - 2 with I0 - 1 >= 0 ==> I4 - 2 >= I4 - 2 1 <= I7 ==> I6 - 2 >= I6 - 2 I9 <= 0 ==> I8 - 2 >= (-1 + I8) - 1 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f3#(I4, I5) -> f1#(I4, I5) f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7] f1#(I8, I9) -> f2#(-1 + I8, 1 + I9) [I9 <= 0] R = f5(x1, x2) -> f4(x1, x2) f2(I0, I1) -> f1(I0, I0) [1 <= I0] f4(I2, I3) -> f2(I2, I3) f3(I4, I5) -> f1(I4, I5) f1(I6, I7) -> f3(I6, -1 + I7) [1 <= I7] f1(I8, I9) -> f2(-1 + I8, 1 + I9) [I9 <= 0] The dependency graph for this problem is: 3 -> 4, 5 4 -> 3 5 -> Where: 3) f3#(I4, I5) -> f1#(I4, I5) 4) f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7] 5) f1#(I8, I9) -> f2#(-1 + I8, 1 + I9) [I9 <= 0] We have the following SCCs. { 3, 4 } DP problem for innermost termination. P = f3#(I4, I5) -> f1#(I4, I5) f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7] R = f5(x1, x2) -> f4(x1, x2) f2(I0, I1) -> f1(I0, I0) [1 <= I0] f4(I2, I3) -> f2(I2, I3) f3(I4, I5) -> f1(I4, I5) f1(I6, I7) -> f3(I6, -1 + I7) [1 <= I7] f1(I8, I9) -> f2(-1 + I8, 1 + I9) [I9 <= 0]
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