details

property	value
status	complete
benchmark	conv.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n145.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.217830896378 seconds
cpu usage	0.173987702
max memory	8691712.0

stage attributes

key	value
output-size	1839
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 = conv#(x) -> conv#(x / 2) [x > 0] conv#(x) -> lastbit#(x) [x > 0] lastbit#(I0) -> lastbit#(I0 - 2) [I0 > 1] R = conv(x) -> cons(conv(x / 2), lastbit(x)) [x > 0] conv(0) -> cons(nil, 0) lastbit(I0) -> lastbit(I0 - 2) [I0 > 1] lastbit(1) -> 1 lastbit(0) -> 0 The dependency graph for this problem is: 0 -> 0, 1 1 -> 2 2 -> 2 Where: 0) conv#(x) -> conv#(x / 2) [x > 0] 1) conv#(x) -> lastbit#(x) [x > 0] 2) lastbit#(I0) -> lastbit#(I0 - 2) [I0 > 1] We have the following SCCs. { 0 } { 2 } DP problem for innermost termination. P = lastbit#(I0) -> lastbit#(I0 - 2) [I0 > 1] R = conv(x) -> cons(conv(x / 2), lastbit(x)) [x > 0] conv(0) -> cons(nil, 0) lastbit(I0) -> lastbit(I0 - 2) [I0 > 1] lastbit(1) -> 1 lastbit(0) -> 0 We use the reverse value criterion with the projection function NU: NU[lastbit#(z1)] = z1 This gives the following inequalities: I0 > 1 ==> I0 > I0 - 2 with I0 >= 0 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = conv#(x) -> conv#(x / 2) [x > 0] R = conv(x) -> cons(conv(x / 2), lastbit(x)) [x > 0] conv(0) -> cons(nil, 0) lastbit(I0) -> lastbit(I0 - 2) [I0 > 1] lastbit(1) -> 1 lastbit(0) -> 0 We use the reverse value criterion with the projection function NU: NU[conv#(z1)] = z1 This gives the following inequalities: x > 0 ==> x > x / 2 with x >= 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