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Integer_TRS_Innermost 2019-03-21 04.53 pair #429995121
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
n040.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
0.10283 seconds
cpu usage
0.104433
user time
0.05507
system time
0.049363
max virtual memory
113176.0
max residence set size
8296.0
stage attributes
key
value
starexec-result
MAYBE
output
0.00/0.10 MAYBE 0.00/0.10 0.00/0.10 DP problem for innermost termination. 0.00/0.10 P = 0.00/0.10 if_3#(pair(m, zs), x, ys) -> msort#(zs) 0.00/0.10 msort#(ins(I0, B0)) -> if_3#(min(I0, B0), I0, B0) 0.00/0.10 msort#(ins(I0, B0)) -> min#(I0, B0) 0.00/0.10 min#(I3, ins(I4, B2)) -> if_2#(min(I4, B2), I3, I4, B2) 0.00/0.10 min#(I3, ins(I4, B2)) -> min#(I4, B2) 0.00/0.10 min#(I8, ins(I9, B5)) -> if_1#(min(I9, B5), I8, I9, B5) 0.00/0.10 min#(I8, ins(I9, B5)) -> min#(I9, B5) 0.00/0.10 R = 0.00/0.10 if_3(pair(m, zs), x, ys) -> cons(m, msort(zs)) 0.00/0.10 msort(ins(I0, B0)) -> if_3(min(I0, B0), I0, B0) 0.00/0.10 msort(e) -> nil 0.00/0.10 if_2(pair(I1, zh), I2, y, B1) -> pair(I1, ins(I2, zh)) [I2 >= I1] 0.00/0.10 min(I3, ins(I4, B2)) -> if_2(min(I4, B2), I3, I4, B2) 0.00/0.10 if_1(pair(I5, B3), I6, I7, B4) -> pair(I6, ins(I5, B3)) [I5 > I6] 0.00/0.10 min(I8, ins(I9, B5)) -> if_1(min(I9, B5), I8, I9, B5) 0.00/0.10 min(I10, e) -> pair(I10, e) 0.00/0.10 0.00/0.10 The dependency graph for this problem is: 0.00/0.10 0 -> 1, 2 0.00/0.10 1 -> 0 0.00/0.10 2 -> 3, 4, 5, 6 0.00/0.10 3 -> 0.00/0.10 4 -> 3, 4, 5, 6 0.00/0.10 5 -> 0.00/0.10 6 -> 3, 4, 5, 6 0.00/0.10 Where: 0.00/0.10 0) if_3#(pair(m, zs), x, ys) -> msort#(zs) 0.00/0.10 1) msort#(ins(I0, B0)) -> if_3#(min(I0, B0), I0, B0) 0.00/0.10 2) msort#(ins(I0, B0)) -> min#(I0, B0) 0.00/0.10 3) min#(I3, ins(I4, B2)) -> if_2#(min(I4, B2), I3, I4, B2) 0.00/0.10 4) min#(I3, ins(I4, B2)) -> min#(I4, B2) 0.00/0.10 5) min#(I8, ins(I9, B5)) -> if_1#(min(I9, B5), I8, I9, B5) 0.00/0.10 6) min#(I8, ins(I9, B5)) -> min#(I9, B5) 0.00/0.10 0.00/0.10 We have the following SCCs. 0.00/0.10 { 0, 1 } 0.00/0.10 { 4, 6 } 0.00/0.10 0.00/0.10 DP problem for innermost termination. 0.00/0.10 P = 0.00/0.10 min#(I3, ins(I4, B2)) -> min#(I4, B2) 0.00/0.10 min#(I8, ins(I9, B5)) -> min#(I9, B5) 0.00/0.10 R = 0.00/0.10 if_3(pair(m, zs), x, ys) -> cons(m, msort(zs)) 0.00/0.10 msort(ins(I0, B0)) -> if_3(min(I0, B0), I0, B0) 0.00/0.10 msort(e) -> nil 0.00/0.10 if_2(pair(I1, zh), I2, y, B1) -> pair(I1, ins(I2, zh)) [I2 >= I1] 0.00/0.10 min(I3, ins(I4, B2)) -> if_2(min(I4, B2), I3, I4, B2) 0.00/0.10 if_1(pair(I5, B3), I6, I7, B4) -> pair(I6, ins(I5, B3)) [I5 > I6] 0.00/0.10 min(I8, ins(I9, B5)) -> if_1(min(I9, B5), I8, I9, B5) 0.00/0.10 min(I10, e) -> pair(I10, e) 0.00/0.10 0.00/0.10 We use the subterm criterion with the projection function NU: 0.00/0.10 NU[min#(x0,x1)] = x1 0.00/0.10 0.00/0.10 This gives the following inequalities: 0.00/0.10 ins(I4, B2) |> B2 0.00/0.10 ins(I9, B5) |> B5 0.00/0.10 0.00/0.10 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/0.10 0.00/0.10 DP problem for innermost termination. 0.00/0.10 P = 0.00/0.10 if_3#(pair(m, zs), x, ys) -> msort#(zs) 0.00/0.10 msort#(ins(I0, B0)) -> if_3#(min(I0, B0), I0, B0) 0.00/0.10 R = 0.00/0.10 if_3(pair(m, zs), x, ys) -> cons(m, msort(zs)) 0.00/0.10 msort(ins(I0, B0)) -> if_3(min(I0, B0), I0, B0) 0.00/0.10 msort(e) -> nil 0.00/0.10 if_2(pair(I1, zh), I2, y, B1) -> pair(I1, ins(I2, zh)) [I2 >= I1] 0.00/0.10 min(I3, ins(I4, B2)) -> if_2(min(I4, B2), I3, I4, B2) 0.00/0.10 if_1(pair(I5, B3), I6, I7, B4) -> pair(I6, ins(I5, B3)) [I5 > I6] 0.00/0.10 min(I8, ins(I9, B5)) -> if_1(min(I9, B5), I8, I9, B5) 0.00/0.10 min(I10, e) -> pair(I10, e) 0.00/0.10 0.00/3.08 EOF
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