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Integer TRS Innermost pair #487100593
details
property
value
status
complete
benchmark
maxsort.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n149.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
0.573679 seconds
cpu usage
0.583654
user time
0.284582
system time
0.299072
max virtual memory
237428.0
max residence set size
8664.0
stage attributes
key
value
starexec-result
MAYBE
output
MAYBE DP problem for innermost termination. P = sort#(cons(x, xs)) -> max#(cons(x, xs)) sort#(cons(x, xs)) -> sort#(del(max(cons(x, xs)), cons(x, xs))) sort#(cons(x, xs)) -> del#(max(cons(x, xs)), cons(x, xs)) sort#(cons(x, xs)) -> max#(cons(x, xs)) if2#(false, I0, y, B0) -> del#(I0, B0) del#(I3, cons(I4, B2)) -> if2#(I3 = I4, I3, I4, B2) if1#(false, I6, I7, B3) -> max#(cons(I7, B3)) if1#(true, I8, I9, B4) -> max#(cons(I8, B4)) max#(cons(I10, cons(I11, B5))) -> if1#(I10 >= I11, I10, I11, B5) R = sort(cons(x, xs)) -> cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs)))) sort(nil) -> nil if2(false, I0, y, B0) -> cons(y, del(I0, B0)) if2(true, I1, I2, B1) -> B1 del(I3, cons(I4, B2)) -> if2(I3 = I4, I3, I4, B2) del(I5, nil) -> nil if1(false, I6, I7, B3) -> max(cons(I7, B3)) if1(true, I8, I9, B4) -> max(cons(I8, B4)) max(cons(I10, cons(I11, B5))) -> if1(I10 >= I11, I10, I11, B5) max(cons(I12, nil)) -> I12 max(nil) -> 0 This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = sort#(cons(x, xs)) -> max#(cons(x, xs)) sort#(cons(x, xs)) -> sort#(del(max(cons(x, xs)), cons(x, xs))) sort#(cons(x, xs)) -> del#(max(cons(x, xs)), cons(x, xs)) sort#(cons(x, xs)) -> max#(cons(x, xs)) if2#(false, I0, y, B0) -> del#(I0, B0) del#(I3, cons(I4, B2)) -> if2#(I3 = I4, I3, I4, B2) if1#(false, I6, I7, B3) -> max#(cons(I7, B3)) if1#(true, I8, I9, B4) -> max#(cons(I8, B4)) max#(cons(I10, cons(I11, B5))) -> if1#(I10 >= I11, I10, I11, B5) max#(cons(I10, cons(I11, B5))) -> max#(cons(I11, B5)) [not(I10 >= I11)] max#(cons(I10, cons(I11, B5))) -> max#(cons(I10, B5)) [I10 >= I11] del#(I3, cons(I4, B2)) -> del#(I3, B2) [not(I3 = I4)] R = sort(cons(x, xs)) -> cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs)))) sort(nil) -> nil if2(false, I0, y, B0) -> cons(y, del(I0, B0)) if2(true, I1, I2, B1) -> B1 del(I3, cons(I4, B2)) -> if2(I3 = I4, I3, I4, B2) del(I5, nil) -> nil if1(false, I6, I7, B3) -> max(cons(I7, B3)) if1(true, I8, I9, B4) -> max(cons(I8, B4)) max(cons(I10, cons(I11, B5))) -> if1(I10 >= I11, I10, I11, B5) max(cons(I12, nil)) -> I12 max(nil) -> 0 The dependency graph for this problem is: 0 -> 10, 9, 8 1 -> 0, 1, 2, 3 2 -> 11, 5 3 -> 10, 9, 8 4 -> 11, 5 5 -> 6 -> 10, 9, 8 7 -> 10, 9, 8 8 -> 9 -> 10, 9, 8 10 -> 10, 8, 9 11 -> 11, 5 Where: 0) sort#(cons(x, xs)) -> max#(cons(x, xs)) 1) sort#(cons(x, xs)) -> sort#(del(max(cons(x, xs)), cons(x, xs))) 2) sort#(cons(x, xs)) -> del#(max(cons(x, xs)), cons(x, xs)) 3) sort#(cons(x, xs)) -> max#(cons(x, xs)) 4) if2#(false, I0, y, B0) -> del#(I0, B0) 5) del#(I3, cons(I4, B2)) -> if2#(I3 = I4, I3, I4, B2) 6) if1#(false, I6, I7, B3) -> max#(cons(I7, B3)) 7) if1#(true, I8, I9, B4) -> max#(cons(I8, B4)) 8) max#(cons(I10, cons(I11, B5))) -> if1#(I10 >= I11, I10, I11, B5) 9) max#(cons(I10, cons(I11, B5))) -> max#(cons(I11, B5)) [not(I10 >= I11)] 10) max#(cons(I10, cons(I11, B5))) -> max#(cons(I10, B5)) [I10 >= I11] 11) del#(I3, cons(I4, B2)) -> del#(I3, B2) [not(I3 = I4)] We have the following SCCs. { 1 } { 9, 10 } { 11 } DP problem for innermost termination. P = del#(I3, cons(I4, B2)) -> del#(I3, B2) [not(I3 = I4)] R = sort(cons(x, xs)) -> cons(max(cons(x, xs)), sort(del(max(cons(x, xs)), cons(x, xs)))) sort(nil) -> nil if2(false, I0, y, B0) -> cons(y, del(I0, B0)) if2(true, I1, I2, B1) -> B1 del(I3, cons(I4, B2)) -> if2(I3 = I4, I3, I4, B2) del(I5, nil) -> nil if1(false, I6, I7, B3) -> max(cons(I7, B3)) if1(true, I8, I9, B4) -> max(cons(I8, B4)) max(cons(I10, cons(I11, B5))) -> if1(I10 >= I11, I10, I11, B5)
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