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TRS Stand 20472 pair #381713890
details
property
value
status
complete
benchmark
#3.38.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n079.star.cs.uiowa.edu
space
AG01
run statistics
property
value
solver
muterm 5.18
configuration
default
runtime (wallclock)
0.0586590766907 seconds
cpu usage
0.038827221
max memory
4096000.0
stage attributes
key
value
output-size
5248
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES Problem 1: (VAR l x y) (RULES rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l)) rev(nil) -> nil rev1(0,nil) -> 0 rev1(s(x),nil) -> s(x) rev1(x,cons(y,l)) -> rev1(y,l) rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l))) rev2(x,nil) -> nil ) Problem 1: Innermost Equivalent Processor: -> Rules: rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l)) rev(nil) -> nil rev1(0,nil) -> 0 rev1(s(x),nil) -> s(x) rev1(x,cons(y,l)) -> rev1(y,l) rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l))) rev2(x,nil) -> nil -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination. Problem 1: Dependency Pairs Processor: -> Pairs: REV(cons(x,l)) -> REV1(x,l) REV(cons(x,l)) -> REV2(x,l) REV1(x,cons(y,l)) -> REV1(y,l) REV2(x,cons(y,l)) -> REV(cons(x,rev2(y,l))) REV2(x,cons(y,l)) -> REV2(y,l) -> Rules: rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l)) rev(nil) -> nil rev1(0,nil) -> 0 rev1(s(x),nil) -> s(x) rev1(x,cons(y,l)) -> rev1(y,l) rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l))) rev2(x,nil) -> nil Problem 1: SCC Processor: -> Pairs: REV(cons(x,l)) -> REV1(x,l) REV(cons(x,l)) -> REV2(x,l) REV1(x,cons(y,l)) -> REV1(y,l) REV2(x,cons(y,l)) -> REV(cons(x,rev2(y,l))) REV2(x,cons(y,l)) -> REV2(y,l) -> Rules: rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l)) rev(nil) -> nil rev1(0,nil) -> 0 rev1(s(x),nil) -> s(x) rev1(x,cons(y,l)) -> rev1(y,l) rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l))) rev2(x,nil) -> nil ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: REV1(x,cons(y,l)) -> REV1(y,l) ->->-> Rules: rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l)) rev(nil) -> nil rev1(0,nil) -> 0 rev1(s(x),nil) -> s(x) rev1(x,cons(y,l)) -> rev1(y,l) rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l))) rev2(x,nil) -> nil ->->Cycle: ->->-> Pairs: REV(cons(x,l)) -> REV2(x,l) REV2(x,cons(y,l)) -> REV(cons(x,rev2(y,l))) REV2(x,cons(y,l)) -> REV2(y,l) ->->-> Rules: rev(cons(x,l)) -> cons(rev1(x,l),rev2(x,l)) rev(nil) -> nil rev1(0,nil) -> 0 rev1(s(x),nil) -> s(x) rev1(x,cons(y,l)) -> rev1(y,l) rev2(x,cons(y,l)) -> rev(cons(x,rev2(y,l))) rev2(x,nil) -> nil The problem is decomposed in 2 subproblems. Problem 1.1:
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