details

property	value
status	complete
benchmark	c.01.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n149.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	2.41246199608 seconds
cpu usage	2.564062646
max memory	1.814528E7

stage attributes

key	value
output-size	2230
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 = eval_2#(x, y) -> eval_1#(x - 1, y) [x >= 0 && y > 0 && y >= x] eval_2#(I0, I1) -> eval_2#(I0, 2 * I1) [I0 >= 0 && I1 > 0 && I0 > I1] eval_1#(I2, I3) -> eval_2#(I2, 1) [I2 >= 0] R = eval_2(x, y) -> eval_1(x - 1, y) [x >= 0 && y > 0 && y >= x] eval_2(I0, I1) -> eval_2(I0, 2 * I1) [I0 >= 0 && I1 > 0 && I0 > I1] eval_1(I2, I3) -> eval_2(I2, 1) [I2 >= 0] We use the reverse value criterion with the projection function NU: NU[eval_1#(z1,z2)] = z1 + -1 * 0 NU[eval_2#(z1,z2)] = z1 This gives the following inequalities: x >= 0 && y > 0 && y >= x ==> x > x - 1 + -1 * 0 with x >= 0 I0 >= 0 && I1 > 0 && I0 > I1 ==> I0 >= I0 I2 >= 0 ==> I2 + -1 * 0 >= I2 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = eval_2#(I0, I1) -> eval_2#(I0, 2 * I1) [I0 >= 0 && I1 > 0 && I0 > I1] eval_1#(I2, I3) -> eval_2#(I2, 1) [I2 >= 0] R = eval_2(x, y) -> eval_1(x - 1, y) [x >= 0 && y > 0 && y >= x] eval_2(I0, I1) -> eval_2(I0, 2 * I1) [I0 >= 0 && I1 > 0 && I0 > I1] eval_1(I2, I3) -> eval_2(I2, 1) [I2 >= 0] The dependency graph for this problem is: 1 -> 1 2 -> 1 Where: 1) eval_2#(I0, I1) -> eval_2#(I0, 2 * I1) [I0 >= 0 && I1 > 0 && I0 > I1] 2) eval_1#(I2, I3) -> eval_2#(I2, 1) [I2 >= 0] We have the following SCCs. { 1 } DP problem for innermost termination. P = eval_2#(I0, I1) -> eval_2#(I0, 2 * I1) [I0 >= 0 && I1 > 0 && I0 > I1] R = eval_2(x, y) -> eval_1(x - 1, y) [x >= 0 && y > 0 && y >= x] eval_2(I0, I1) -> eval_2(I0, 2 * I1) [I0 >= 0 && I1 > 0 && I0 > I1] eval_1(I2, I3) -> eval_2(I2, 1) [I2 >= 0] We use the reverse value criterion with the projection function NU: NU[eval_2#(z1,z2)] = z1 + -1 * z2 This gives the following inequalities: I0 >= 0 && I1 > 0 && I0 > I1 ==> I0 + -1 * I1 > I0 + -1 * (2 * I1) with I0 + -1 * I1 >= 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 Integer TRS Innermost