details

property	value
status	complete
benchmark	quicksort.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n147.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	0.397904 seconds
cpu usage	0.405553
user time	0.208935
system time	0.196618
max virtual memory	113176.0
max residence set size	8704.0

stage attributes

key	value
starexec-result	MAYBE

output

0.00/0.39 MAYBE 0.00/0.39 0.00/0.39 DP problem for innermost termination. 0.00/0.39 P = 0.00/0.39 if_3#(pair(yl, yh), x, ys) -> app#(qsort(yl), cons(x, qsort(yh))) 0.00/0.39 if_3#(pair(yl, yh), x, ys) -> qsort#(yl) 0.00/0.39 if_3#(pair(yl, yh), x, ys) -> qsort#(yh) 0.00/0.39 qsort#(ins(I0, B0)) -> if_3#(split(I0, B0), I0, B0) 0.00/0.39 qsort#(ins(I0, B0)) -> split#(I0, B0) 0.00/0.39 split#(I2, ins(I3, B1)) -> if_2#(split(I2, B1), I2, I3, B1) [I3 >= I2] 0.00/0.39 split#(I2, ins(I3, B1)) -> split#(I2, B1) [I3 >= I2] 0.00/0.39 split#(I6, ins(I7, B5)) -> if_1#(split(I6, B5), I6, I7, B5) [I6 > I7] 0.00/0.39 split#(I6, ins(I7, B5)) -> split#(I6, B5) [I6 > I7] 0.00/0.39 app#(cons(I9, B6), B7) -> app#(B6, B7) 0.00/0.39 R = 0.00/0.39 if_3(pair(yl, yh), x, ys) -> app(qsort(yl), cons(x, qsort(yh))) 0.00/0.39 qsort(ins(I0, B0)) -> if_3(split(I0, B0), I0, B0) 0.00/0.39 qsort(e) -> nil 0.00/0.39 if_2(pair(zl, zh), I1, y, zs) -> pair(zl, ins(y, zh)) [y >= I1] 0.00/0.39 split(I2, ins(I3, B1)) -> if_2(split(I2, B1), I2, I3, B1) [I3 >= I2] 0.00/0.39 if_1(pair(B2, B3), I4, I5, B4) -> pair(ins(I5, B2), B3) [I4 > I5] 0.00/0.39 split(I6, ins(I7, B5)) -> if_1(split(I6, B5), I6, I7, B5) [I6 > I7] 0.00/0.39 split(I8, e) -> pair(e, e) 0.00/0.39 app(cons(I9, B6), B7) -> cons(I9, app(B6, B7)) 0.00/0.39 app(nil, B8) -> B8 0.00/0.39 0.00/0.39 The dependency graph for this problem is: 0.00/0.39 0 -> 9 0.00/0.39 1 -> 3, 4 0.00/0.39 2 -> 3, 4 0.00/0.39 3 -> 0, 1, 2 0.00/0.39 4 -> 5, 6, 7, 8 0.00/0.39 5 -> 0.00/0.39 6 -> 5, 6, 7, 8 0.00/0.39 7 -> 0.00/0.39 8 -> 5, 6, 7, 8 0.00/0.39 9 -> 9 0.00/0.39 Where: 0.00/0.39 0) if_3#(pair(yl, yh), x, ys) -> app#(qsort(yl), cons(x, qsort(yh))) 0.00/0.39 1) if_3#(pair(yl, yh), x, ys) -> qsort#(yl) 0.00/0.39 2) if_3#(pair(yl, yh), x, ys) -> qsort#(yh) 0.00/0.39 3) qsort#(ins(I0, B0)) -> if_3#(split(I0, B0), I0, B0) 0.00/0.39 4) qsort#(ins(I0, B0)) -> split#(I0, B0) 0.00/0.39 5) split#(I2, ins(I3, B1)) -> if_2#(split(I2, B1), I2, I3, B1) [I3 >= I2] 0.00/0.39 6) split#(I2, ins(I3, B1)) -> split#(I2, B1) [I3 >= I2] 0.00/0.39 7) split#(I6, ins(I7, B5)) -> if_1#(split(I6, B5), I6, I7, B5) [I6 > I7] 0.00/0.39 8) split#(I6, ins(I7, B5)) -> split#(I6, B5) [I6 > I7] 0.00/0.39 9) app#(cons(I9, B6), B7) -> app#(B6, B7) 0.00/0.39 0.00/0.39 We have the following SCCs. 0.00/0.39 { 1, 2, 3 } 0.00/0.39 { 6, 8 } 0.00/0.39 { 9 } 0.00/0.39 0.00/0.39 DP problem for innermost termination. 0.00/0.39 P = 0.00/0.39 app#(cons(I9, B6), B7) -> app#(B6, B7) 0.00/0.39 R = 0.00/0.39 if_3(pair(yl, yh), x, ys) -> app(qsort(yl), cons(x, qsort(yh))) 0.00/0.39 qsort(ins(I0, B0)) -> if_3(split(I0, B0), I0, B0) 0.00/0.39 qsort(e) -> nil 0.00/0.39 if_2(pair(zl, zh), I1, y, zs) -> pair(zl, ins(y, zh)) [y >= I1] 0.00/0.39 split(I2, ins(I3, B1)) -> if_2(split(I2, B1), I2, I3, B1) [I3 >= I2] 0.00/0.39 if_1(pair(B2, B3), I4, I5, B4) -> pair(ins(I5, B2), B3) [I4 > I5] 0.00/0.39 split(I6, ins(I7, B5)) -> if_1(split(I6, B5), I6, I7, B5) [I6 > I7] 0.00/0.39 split(I8, e) -> pair(e, e) 0.00/0.39 app(cons(I9, B6), B7) -> cons(I9, app(B6, B7)) 0.00/0.39 app(nil, B8) -> B8 0.00/0.39 0.00/0.39 We use the subterm criterion with the projection function NU: 0.00/0.39 NU[app#(x0,x1)] = x0 0.00/0.39 0.00/0.39 This gives the following inequalities: 0.00/0.39 cons(I9, B6) |> B6 0.00/0.39 0.00/0.39 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/0.39 0.00/0.39 DP problem for innermost termination. 0.00/0.39 P = 0.00/0.39 split#(I2, ins(I3, B1)) -> split#(I2, B1) [I3 >= I2] 0.00/0.39 split#(I6, ins(I7, B5)) -> split#(I6, B5) [I6 > I7] 0.00/0.39 R = 0.00/0.39 if_3(pair(yl, yh), x, ys) -> app(qsort(yl), cons(x, qsort(yh))) 0.00/0.39 qsort(ins(I0, B0)) -> if_3(split(I0, B0), I0, B0) 0.00/0.39 qsort(e) -> nil 0.00/0.39 if_2(pair(zl, zh), I1, y, zs) -> pair(zl, ins(y, zh)) [y >= I1] 0.00/0.39 split(I2, ins(I3, B1)) -> if_2(split(I2, B1), I2, I3, B1) [I3 >= I2] 0.00/0.39 if_1(pair(B2, B3), I4, I5, B4) -> pair(ins(I5, B2), B3) [I4 > I5] 0.00/0.39 split(I6, ins(I7, B5)) -> if_1(split(I6, B5), I6, I7, B5) [I6 > I7] 0.00/0.39 split(I8, e) -> pair(e, e) 0.00/0.39 app(cons(I9, B6), B7) -> cons(I9, app(B6, B7)) 0.00/0.39 app(nil, B8) -> B8 0.00/0.39 0.00/0.39 We use the subterm criterion with the projection function NU: 0.00/0.39 NU[split#(x0,x1)] = x1 0.00/0.39 0.00/0.39 This gives the following inequalities: 0.00/0.39 ins(I3, B1) |> B1 0.00/0.39 ins(I7, B5) |> B5 0.00/0.39 popout

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job log

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actions

all output return to Integer_TRS_Innermost 2019-03-21 04.53