details

property	value
status	complete
benchmark	minsort.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n148.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.106631 seconds
cpu usage	0.107252
user time	0.060065
system time	0.047187
max virtual memory	113188.0
max residence set size	8304.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE DP problem for innermost termination. P = if_3#(pair(m, zs), x, ys) -> msort#(zs) msort#(ins(I0, B0)) -> if_3#(min(I0, B0), I0, B0) msort#(ins(I0, B0)) -> min#(I0, B0) min#(I3, ins(I4, B2)) -> if_2#(min(I4, B2), I3, I4, B2) min#(I3, ins(I4, B2)) -> min#(I4, B2) min#(I8, ins(I9, B5)) -> if_1#(min(I9, B5), I8, I9, B5) min#(I8, ins(I9, B5)) -> min#(I9, B5) R = if_3(pair(m, zs), x, ys) -> cons(m, msort(zs)) msort(ins(I0, B0)) -> if_3(min(I0, B0), I0, B0) msort(e) -> nil if_2(pair(I1, zh), I2, y, B1) -> pair(I1, ins(I2, zh)) [I2 >= I1] min(I3, ins(I4, B2)) -> if_2(min(I4, B2), I3, I4, B2) if_1(pair(I5, B3), I6, I7, B4) -> pair(I6, ins(I5, B3)) [I5 > I6] min(I8, ins(I9, B5)) -> if_1(min(I9, B5), I8, I9, B5) min(I10, e) -> pair(I10, e) The dependency graph for this problem is: 0 -> 1, 2 1 -> 0 2 -> 3, 4, 5, 6 3 -> 4 -> 3, 4, 5, 6 5 -> 6 -> 3, 4, 5, 6 Where: 0) if_3#(pair(m, zs), x, ys) -> msort#(zs) 1) msort#(ins(I0, B0)) -> if_3#(min(I0, B0), I0, B0) 2) msort#(ins(I0, B0)) -> min#(I0, B0) 3) min#(I3, ins(I4, B2)) -> if_2#(min(I4, B2), I3, I4, B2) 4) min#(I3, ins(I4, B2)) -> min#(I4, B2) 5) min#(I8, ins(I9, B5)) -> if_1#(min(I9, B5), I8, I9, B5) 6) min#(I8, ins(I9, B5)) -> min#(I9, B5) We have the following SCCs. { 0, 1 } { 4, 6 } DP problem for innermost termination. P = min#(I3, ins(I4, B2)) -> min#(I4, B2) min#(I8, ins(I9, B5)) -> min#(I9, B5) R = if_3(pair(m, zs), x, ys) -> cons(m, msort(zs)) msort(ins(I0, B0)) -> if_3(min(I0, B0), I0, B0) msort(e) -> nil if_2(pair(I1, zh), I2, y, B1) -> pair(I1, ins(I2, zh)) [I2 >= I1] min(I3, ins(I4, B2)) -> if_2(min(I4, B2), I3, I4, B2) if_1(pair(I5, B3), I6, I7, B4) -> pair(I6, ins(I5, B3)) [I5 > I6] min(I8, ins(I9, B5)) -> if_1(min(I9, B5), I8, I9, B5) min(I10, e) -> pair(I10, e) We use the subterm criterion with the projection function NU: NU[min#(x0,x1)] = x1 This gives the following inequalities: ins(I4, B2) |> B2 ins(I9, B5) |> B5 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. DP problem for innermost termination. P = if_3#(pair(m, zs), x, ys) -> msort#(zs) msort#(ins(I0, B0)) -> if_3#(min(I0, B0), I0, B0) R = if_3(pair(m, zs), x, ys) -> cons(m, msort(zs)) msort(ins(I0, B0)) -> if_3(min(I0, B0), I0, B0) msort(e) -> nil if_2(pair(I1, zh), I2, y, B1) -> pair(I1, ins(I2, zh)) [I2 >= I1] min(I3, ins(I4, B2)) -> if_2(min(I4, B2), I3, I4, B2) if_1(pair(I5, B3), I6, I7, B4) -> pair(I6, ins(I5, B3)) [I5 > I6] min(I8, ins(I9, B5)) -> if_1(min(I9, B5), I8, I9, B5) min(I10, e) -> pair(I10, e) popout

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

job log

popout

actions

all output return to Integer TRS Innermost