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Integer TRS Innermost pair #487100461
details
property
value
status
complete
benchmark
A15.itrs
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n144.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
0.68526 seconds
cpu usage
0.697457
user time
0.371283
system time
0.326174
max virtual memory
685080.0
max residence set size
8220.0
stage attributes
key
value
starexec-result
MAYBE
output
MAYBE DP problem for innermost termination. P = max#(cons(I2, l)) -> if#(I2 > max(l), I2, max(l)) max#(cons(I2, l)) -> max#(l) max#(cons(I2, l)) -> max#(l) member#(n, cons(m, B0)) -> member#(n, B0) cond2#(false, I4, B1) -> st#(I4 + 1, B1) cond1#(false, I6, B3) -> cond2#(I6 > max(B3), I6, B3) cond1#(false, I6, B3) -> max#(B3) cond1#(true, I7, B4) -> st#(I7 + 1, B4) stNat#(true, I8, B5) -> cond1#(member(I8, B5), I8, B5) stNat#(true, I8, B5) -> member#(I8, B5) st#(I9, B6) -> stNat#(I9 >= 0, I9, B6) sort#(B7) -> st#(0, B7) R = if(false, u, v) -> v if(true, I0, I1) -> I0 max(cons(I2, l)) -> if(I2 > max(l), I2, max(l)) max(nil) -> 0 member(n, cons(m, B0)) -> n = m || member(n, B0) member(I3, nil) -> false cond2(false, I4, B1) -> st(I4 + 1, B1) cond2(true, I5, B2) -> nil cond1(false, I6, B3) -> cond2(I6 > max(B3), I6, B3) cond1(true, I7, B4) -> cons(I7, st(I7 + 1, B4)) stNat(true, I8, B5) -> cond1(member(I8, B5), I8, B5) st(I9, B6) -> stNat(I9 >= 0, I9, B6) sort(B7) -> st(0, B7) This problem is converted using chaining, where edges between chained DPs are removed. DP problem for innermost termination. P = max#(cons(I2, l)) -> if#(I2 > max(l), I2, max(l)) max#(cons(I2, l)) -> max#(l) max#(cons(I2, l)) -> max#(l) member#(n, cons(m, B0)) -> member#(n, B0) cond2#(false, I4, B1) -> st#(I4 + 1, B1) cond1#(false, I6, B3) -> cond2#(I6 > max(B3), I6, B3) cond1#(false, I6, B3) -> max#(B3) cond1#(true, I7, B4) -> st#(I7 + 1, B4) stNat#(true, I8, B5) -> cond1#(member(I8, B5), I8, B5) stNat#(true, I8, B5) -> member#(I8, B5) st#(I9, B6) -> stNat#(I9 >= 0, I9, B6) sort#(B7) -> st#(0, B7) st#(I9, B6) -> cond1#(member(I9, B6), I9, B6) [I9 >= 0] st#(I9, B6) -> member#(I9, B6) [I9 >= 0] R = if(false, u, v) -> v if(true, I0, I1) -> I0 max(cons(I2, l)) -> if(I2 > max(l), I2, max(l)) max(nil) -> 0 member(n, cons(m, B0)) -> n = m || member(n, B0) member(I3, nil) -> false cond2(false, I4, B1) -> st(I4 + 1, B1) cond2(true, I5, B2) -> nil cond1(false, I6, B3) -> cond2(I6 > max(B3), I6, B3) cond1(true, I7, B4) -> cons(I7, st(I7 + 1, B4)) stNat(true, I8, B5) -> cond1(member(I8, B5), I8, B5) st(I9, B6) -> stNat(I9 >= 0, I9, B6) sort(B7) -> st(0, B7) The dependency graph for this problem is: 0 -> 1 -> 0, 1, 2 2 -> 0, 1, 2 3 -> 3 4 -> 13, 12, 10 5 -> 4 6 -> 0, 1, 2 7 -> 13, 12, 10 8 -> 5, 6, 7 9 -> 3 10 -> 11 -> 13, 12, 10 12 -> 7, 6, 5 13 -> 3 Where: 0) max#(cons(I2, l)) -> if#(I2 > max(l), I2, max(l)) 1) max#(cons(I2, l)) -> max#(l) 2) max#(cons(I2, l)) -> max#(l) 3) member#(n, cons(m, B0)) -> member#(n, B0) 4) cond2#(false, I4, B1) -> st#(I4 + 1, B1) 5) cond1#(false, I6, B3) -> cond2#(I6 > max(B3), I6, B3) 6) cond1#(false, I6, B3) -> max#(B3) 7) cond1#(true, I7, B4) -> st#(I7 + 1, B4) 8) stNat#(true, I8, B5) -> cond1#(member(I8, B5), I8, B5) 9) stNat#(true, I8, B5) -> member#(I8, B5) 10) st#(I9, B6) -> stNat#(I9 >= 0, I9, B6) 11) sort#(B7) -> st#(0, B7) 12) st#(I9, B6) -> cond1#(member(I9, B6), I9, B6) [I9 >= 0] 13) st#(I9, B6) -> member#(I9, B6) [I9 >= 0] We have the following SCCs. { 4, 5, 7, 12 } { 1, 2 } { 3 }
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