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	n022.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.101500988007 seconds
cpu usage	0.093470842
max memory	8794112.0

stage attributes

key	value
output-size	2917
starexec-result	MAYBE

output

/export/starexec/sandbox/solver/bin/starexec_run_Itrs /export/starexec/sandbox/benchmark/theBenchmark.itrs /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- 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

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actions

all output return to Integ TRS Inner 43557