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Integ TRS Inner 43557 pair #381740632
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
n052.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
0.445364952087 seconds
cpu usage
0.468534574
max memory
1.0477568E7
stage attributes
key
value
output-size
4996
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 = 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))
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