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Integer TRS Innermost pair #487100511
details
property
value
status
complete
benchmark
complete4.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n139.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
0.329005 seconds
cpu usage
0.328563
user time
0.181077
system time
0.147486
max virtual memory
113188.0
max residence set size
8528.0
stage attributes
key
value
starexec-result
YES
output
YES DP problem for innermost termination. P = eval#(x, y) -> eval#(x, y - 1) [y >= 0] eval#(I0, I1) -> eval#(I0 - 1, z) [I0 >= 0] R = eval(x, y) -> eval(x, y - 1) [y >= 0] eval(I0, I1) -> eval(I0 - 1, z) [I0 >= 0] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2)] = z1 This gives the following inequalities: y >= 0 ==> x >= x I0 >= 0 ==> I0 > I0 - 1 with I0 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = eval#(x, y) -> eval#(x, y - 1) [y >= 0] R = eval(x, y) -> eval(x, y - 1) [y >= 0] eval(I0, I1) -> eval(I0 - 1, z) [I0 >= 0] We use the reverse value criterion with the projection function NU: NU[eval#(z1,z2)] = z2 + -1 * 0 This gives the following inequalities: y >= 0 ==> y + -1 * 0 > y - 1 + -1 * 0 with y + -1 * 0 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
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