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Integer TRS Innermost pair #487100581
details
property
value
status
complete
benchmark
A12.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)
0.194647 seconds
cpu usage
0.198264
user time
0.098507
system time
0.099757
max virtual memory
113188.0
max residence set size
8668.0
stage attributes
key
value
starexec-result
YES
output
YES DP problem for innermost termination. P = f#(true, x, y) -> f#(x > y && y > 2, 1 + x, 2 * y) R = f(true, x, y) -> f(x > y && y > 2, 1 + x, 2 * y) This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = f#(true, x, y) -> f_1#(x, y) f_1#(x, y) -> f#(x > y && y > 2, 1 + x, 2 * y) f_1#(x, y) -> f_1#(1 + x, 2 * y) [x > y && y > 2] R = f(true, x, y) -> f(x > y && y > 2, 1 + x, 2 * y) The dependency graph for this problem is: 0 -> 2, 1 1 -> 2 -> 2, 1 Where: 0) f#(true, x, y) -> f_1#(x, y) 1) f_1#(x, y) -> f#(x > y && y > 2, 1 + x, 2 * y) 2) f_1#(x, y) -> f_1#(1 + x, 2 * y) [x > y && y > 2] We have the following SCCs. { 2 } DP problem for innermost termination. P = f_1#(x, y) -> f_1#(1 + x, 2 * y) [x > y && y > 2] R = f(true, x, y) -> f(x > y && y > 2, 1 + x, 2 * y) We use the reverse value criterion with the projection function NU: NU[f_1#(z1,z2)] = z1 + -1 * z2 This gives the following inequalities: x > y && y > 2 ==> x + -1 * y > 1 + x + -1 * (2 * y) with x + -1 * y >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
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