details

property	value
status	complete
benchmark	22.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n184.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	3.05879592896 seconds
cpu usage	3.268627505
max memory	1.6080896E7

stage attributes

key	value
output-size	2088
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 && 0 >= y] eval_2#(I0, I1) -> eval_2#(I0, I1 - 1) [I0 > 0 && I1 > 0] eval_1#(I2, I3) -> eval_2#(I2, I3) [I2 > 0] R = eval_2(x, y) -> eval_1(x - 1, y) [x > 0 && 0 >= y] eval_2(I0, I1) -> eval_2(I0, I1 - 1) [I0 > 0 && I1 > 0] eval_1(I2, I3) -> eval_2(I2, I3) [I2 > 0] We use the reverse value criterion with the projection function NU: NU[eval_1#(z1,z2)] = z1 NU[eval_2#(z1,z2)] = z1 - 1 + -1 * 0 This gives the following inequalities: x > 0 && 0 >= y ==> x - 1 + -1 * 0 >= x - 1 I0 > 0 && I1 > 0 ==> I0 - 1 + -1 * 0 >= I0 - 1 + -1 * 0 I2 > 0 ==> I2 > I2 - 1 + -1 * 0 with I2 >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = eval_2#(x, y) -> eval_1#(x - 1, y) [x > 0 && 0 >= y] eval_2#(I0, I1) -> eval_2#(I0, I1 - 1) [I0 > 0 && I1 > 0] R = eval_2(x, y) -> eval_1(x - 1, y) [x > 0 && 0 >= y] eval_2(I0, I1) -> eval_2(I0, I1 - 1) [I0 > 0 && I1 > 0] eval_1(I2, I3) -> eval_2(I2, I3) [I2 > 0] The dependency graph for this problem is: 0 -> 1 -> 0, 1 Where: 0) eval_2#(x, y) -> eval_1#(x - 1, y) [x > 0 && 0 >= y] 1) eval_2#(I0, I1) -> eval_2#(I0, I1 - 1) [I0 > 0 && I1 > 0] We have the following SCCs. { 1 } DP problem for innermost termination. P = eval_2#(I0, I1) -> eval_2#(I0, I1 - 1) [I0 > 0 && I1 > 0] R = eval_2(x, y) -> eval_1(x - 1, y) [x > 0 && 0 >= y] eval_2(I0, I1) -> eval_2(I0, I1 - 1) [I0 > 0 && I1 > 0] eval_1(I2, I3) -> eval_2(I2, I3) [I2 > 0] We use the reverse value criterion with the projection function NU: NU[eval_2#(z1,z2)] = z2 This gives the following inequalities: I0 > 0 && I1 > 0 ==> I1 > I1 - 1 with 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