details

property	value
status	complete
benchmark	A16.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n022.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.0749938488007 seconds
cpu usage	0.072796082
max memory	8626176.0

stage attributes

key	value
output-size	2099
starexec-result	MAYBE

output

/export/starexec/sandbox2/solver/bin/starexec_run_Itrs /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE DP problem for innermost termination. P = length#(cons(x, l)) -> length#(l) cond#(false, n, B2) -> tail#(nthtail(n + 1, B2)) cond#(false, n, B2) -> nthtail#(n + 1, B2) nthtail#(I1, B4) -> cond#(I1 >= length(B4), I1, B4) nthtail#(I1, B4) -> length#(B4) R = length(cons(x, l)) -> 1 + length(l) length(nil) -> 0 tail(cons(B0, B1)) -> B1 tail(nil) -> nil cond(false, n, B2) -> tail(nthtail(n + 1, B2)) cond(true, I0, B3) -> B3 nthtail(I1, B4) -> cond(I1 >= length(B4), I1, B4) The dependency graph for this problem is: 0 -> 0 1 -> 2 -> 3, 4 3 -> 1, 2 4 -> 0 Where: 0) length#(cons(x, l)) -> length#(l) 1) cond#(false, n, B2) -> tail#(nthtail(n + 1, B2)) 2) cond#(false, n, B2) -> nthtail#(n + 1, B2) 3) nthtail#(I1, B4) -> cond#(I1 >= length(B4), I1, B4) 4) nthtail#(I1, B4) -> length#(B4) We have the following SCCs. { 2, 3 } { 0 } DP problem for innermost termination. P = length#(cons(x, l)) -> length#(l) R = length(cons(x, l)) -> 1 + length(l) length(nil) -> 0 tail(cons(B0, B1)) -> B1 tail(nil) -> nil cond(false, n, B2) -> tail(nthtail(n + 1, B2)) cond(true, I0, B3) -> B3 nthtail(I1, B4) -> cond(I1 >= length(B4), I1, B4) We use the subterm criterion with the projection function NU: NU[length#(x0)] = x0 This gives the following inequalities: cons(x, l) |> l All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = cond#(false, n, B2) -> nthtail#(n + 1, B2) nthtail#(I1, B4) -> cond#(I1 >= length(B4), I1, B4) R = length(cons(x, l)) -> 1 + length(l) length(nil) -> 0 tail(cons(B0, B1)) -> B1 tail(nil) -> nil cond(false, n, B2) -> tail(nthtail(n + 1, B2)) cond(true, I0, B3) -> B3 nthtail(I1, B4) -> cond(I1 >= length(B4), I1, B4) popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to Integer TRS Innermost