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Integer TRS Innermost pair #487100517
details
property
value
status
complete
benchmark
c.05.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n146.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
0.456929 seconds
cpu usage
0.465136
user time
0.25227
system time
0.212866
max virtual memory
113188.0
max residence set size
8660.0
stage attributes
key
value
starexec-result
YES
output
YES DP problem for innermost termination. P = eval#(x, y) -> eval#(x, y - x) [y > x && x > 0 && y > 0 && y >= x] eval#(I0, I1) -> eval#(I0, I1 - I0) [I0 > I1 && I0 > 0 && I1 > 0 && I1 >= I0] eval#(I2, I3) -> eval#(I2 - I3, I3) [I3 > I2 && I2 > 0 && I3 > 0 && I2 > I3] eval#(I4, I5) -> eval#(I4 - I5, I5) [I4 > I5 && I4 > 0 && I5 > 0] R = eval(x, y) -> eval(x, y - x) [y > x && x > 0 && y > 0 && y >= x] eval(I0, I1) -> eval(I0, I1 - I0) [I0 > I1 && I0 > 0 && I1 > 0 && I1 >= I0] eval(I2, I3) -> eval(I2 - I3, I3) [I3 > I2 && I2 > 0 && I3 > 0 && I2 > I3] eval(I4, I5) -> eval(I4 - I5, I5) [I4 > I5 && I4 > 0 && I5 > 0] The dependency graph for this problem is: 0 -> 0, 3 1 -> 2 -> 3 -> 0, 3 Where: 0) eval#(x, y) -> eval#(x, y - x) [y > x && x > 0 && y > 0 && y >= x] 1) eval#(I0, I1) -> eval#(I0, I1 - I0) [I0 > I1 && I0 > 0 && I1 > 0 && I1 >= I0] 2) eval#(I2, I3) -> eval#(I2 - I3, I3) [I3 > I2 && I2 > 0 && I3 > 0 && I2 > I3] 3) eval#(I4, I5) -> eval#(I4 - I5, I5) [I4 > I5 && I4 > 0 && I5 > 0] We have the following SCCs. { 0, 3 } DP problem for innermost termination. P = eval#(x, y) -> eval#(x, y - x) [y > x && x > 0 && y > 0 && y >= x] eval#(I4, I5) -> eval#(I4 - I5, I5) [I4 > I5 && I4 > 0 && I5 > 0] R = eval(x, y) -> eval(x, y - x) [y > x && x > 0 && y > 0 && y >= x] eval(I0, I1) -> eval(I0, I1 - I0) [I0 > I1 && I0 > 0 && I1 > 0 && I1 >= I0] eval(I2, I3) -> eval(I2 - I3, I3) [I3 > I2 && I2 > 0 && I3 > 0 && I2 > I3] eval(I4, I5) -> eval(I4 - I5, I5) [I4 > I5 && I4 > 0 && I5 > 0] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2)] = z1 This gives the following inequalities: y > x && x > 0 && y > 0 && y >= x ==> x >= x I4 > I5 && I4 > 0 && I5 > 0 ==> I4 > I4 - I5 with I4 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = eval#(x, y) -> eval#(x, y - x) [y > x && x > 0 && y > 0 && y >= x] R = eval(x, y) -> eval(x, y - x) [y > x && x > 0 && y > 0 && y >= x] eval(I0, I1) -> eval(I0, I1 - I0) [I0 > I1 && I0 > 0 && I1 > 0 && I1 >= I0] eval(I2, I3) -> eval(I2 - I3, I3) [I3 > I2 && I2 > 0 && I3 > 0 && I2 > I3] eval(I4, I5) -> eval(I4 - I5, I5) [I4 > I5 && I4 > 0 && I5 > 0] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2)] = z2 + -1 * z1 This gives the following inequalities: y > x && x > 0 && y > 0 && y >= x ==> y + -1 * x > y - x + -1 * x with y + -1 * x >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
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