details

property	value
status	complete
benchmark	A02.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n085.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.143174171448 seconds
cpu usage	0.146840072
max memory	8818688.0

stage attributes

key	value
output-size	1322
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_Itrs /export/starexec/sandbox/benchmark/theBenchmark.itrs /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES DP problem for innermost termination. P = f#(true, x, y) -> f#(x > y, x, y + 1) R = f(true, x, y) -> f(x > y, x, y + 1) 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, x, y + 1) f_1#(x, y) -> f_1#(x, y + 1) [x > y] R = f(true, x, y) -> f(x > y, x, y + 1) 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, x, y + 1) 2) f_1#(x, y) -> f_1#(x, y + 1) [x > y] We have the following SCCs. { 2 } DP problem for innermost termination. P = f_1#(x, y) -> f_1#(x, y + 1) [x > y] R = f(true, x, y) -> f(x > y, x, y + 1) 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 ==> x + -1 * y > x + -1 * (y + 1) with x + -1 * y >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. popout

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job log

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actions

all output return to Integ TRS Inner 43557