Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
Integer TRS Innermost pair #487528938
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
n147.star.cs.uiowa.edu
space
Mixed_ITRS_2014
run statistics
property
value
solver
Ctrl
configuration
Itrs
runtime (wallclock)
0.627974033356 seconds
cpu usage
0.651878073
max memory
1.1251712E7
stage attributes
key
value
output-size
5964
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 = 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]
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to Integer TRS Innermost