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TRS Standard pair #487069259
details
property
value
status
complete
benchmark
revlist.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n179.star.cs.uiowa.edu
space
Rubio_04
run statistics
property
value
solver
muterm 6.0.3
configuration
default
runtime (wallclock)
0.0599821 seconds
cpu usage
0.049732
user time
0.027075
system time
0.022657
max virtual memory
113188.0
max residence set size
5676.0
stage attributes
key
value
starexec-result
YES
output
YES Problem 1: (VAR v_NonEmpty:S L:S X:S Y:S) (RULES rev(cons(X:S,L:S)) -> cons(rev1(X:S,L:S),rev2(X:S,L:S)) rev(nil) -> nil rev1(0,nil) -> 0 rev1(s(X:S),nil) -> s(X:S) rev1(X:S,cons(Y:S,L:S)) -> rev1(Y:S,L:S) rev2(X:S,cons(Y:S,L:S)) -> rev(cons(X:S,rev(rev2(Y:S,L:S)))) rev2(X:S,nil) -> nil ) Problem 1: Innermost Equivalent Processor: -> Rules: rev(cons(X:S,L:S)) -> cons(rev1(X:S,L:S),rev2(X:S,L:S)) rev(nil) -> nil rev1(0,nil) -> 0 rev1(s(X:S),nil) -> s(X:S) rev1(X:S,cons(Y:S,L:S)) -> rev1(Y:S,L:S) rev2(X:S,cons(Y:S,L:S)) -> rev(cons(X:S,rev(rev2(Y:S,L:S)))) rev2(X:S,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:S,L:S)) -> REV1(X:S,L:S) REV(cons(X:S,L:S)) -> REV2(X:S,L:S) REV1(X:S,cons(Y:S,L:S)) -> REV1(Y:S,L:S) REV2(X:S,cons(Y:S,L:S)) -> REV(rev2(Y:S,L:S)) REV2(X:S,cons(Y:S,L:S)) -> REV(cons(X:S,rev(rev2(Y:S,L:S)))) REV2(X:S,cons(Y:S,L:S)) -> REV2(Y:S,L:S) -> Rules: rev(cons(X:S,L:S)) -> cons(rev1(X:S,L:S),rev2(X:S,L:S)) rev(nil) -> nil rev1(0,nil) -> 0 rev1(s(X:S),nil) -> s(X:S) rev1(X:S,cons(Y:S,L:S)) -> rev1(Y:S,L:S) rev2(X:S,cons(Y:S,L:S)) -> rev(cons(X:S,rev(rev2(Y:S,L:S)))) rev2(X:S,nil) -> nil Problem 1: SCC Processor: -> Pairs: REV(cons(X:S,L:S)) -> REV1(X:S,L:S) REV(cons(X:S,L:S)) -> REV2(X:S,L:S) REV1(X:S,cons(Y:S,L:S)) -> REV1(Y:S,L:S) REV2(X:S,cons(Y:S,L:S)) -> REV(rev2(Y:S,L:S)) REV2(X:S,cons(Y:S,L:S)) -> REV(cons(X:S,rev(rev2(Y:S,L:S)))) REV2(X:S,cons(Y:S,L:S)) -> REV2(Y:S,L:S) -> Rules: rev(cons(X:S,L:S)) -> cons(rev1(X:S,L:S),rev2(X:S,L:S)) rev(nil) -> nil rev1(0,nil) -> 0 rev1(s(X:S),nil) -> s(X:S) rev1(X:S,cons(Y:S,L:S)) -> rev1(Y:S,L:S) rev2(X:S,cons(Y:S,L:S)) -> rev(cons(X:S,rev(rev2(Y:S,L:S)))) rev2(X:S,nil) -> nil ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: REV1(X:S,cons(Y:S,L:S)) -> REV1(Y:S,L:S) ->->-> Rules: rev(cons(X:S,L:S)) -> cons(rev1(X:S,L:S),rev2(X:S,L:S)) rev(nil) -> nil rev1(0,nil) -> 0 rev1(s(X:S),nil) -> s(X:S) rev1(X:S,cons(Y:S,L:S)) -> rev1(Y:S,L:S) rev2(X:S,cons(Y:S,L:S)) -> rev(cons(X:S,rev(rev2(Y:S,L:S)))) rev2(X:S,nil) -> nil ->->Cycle: ->->-> Pairs: REV(cons(X:S,L:S)) -> REV2(X:S,L:S) REV2(X:S,cons(Y:S,L:S)) -> REV(rev2(Y:S,L:S)) REV2(X:S,cons(Y:S,L:S)) -> REV(cons(X:S,rev(rev2(Y:S,L:S)))) REV2(X:S,cons(Y:S,L:S)) -> REV2(Y:S,L:S) ->->-> Rules: rev(cons(X:S,L:S)) -> cons(rev1(X:S,L:S),rev2(X:S,L:S)) rev(nil) -> nil rev1(0,nil) -> 0 rev1(s(X:S),nil) -> s(X:S) rev1(X:S,cons(Y:S,L:S)) -> rev1(Y:S,L:S) rev2(X:S,cons(Y:S,L:S)) -> rev(cons(X:S,rev(rev2(Y:S,L:S)))) rev2(X:S,nil) -> nil The problem is decomposed in 2 subproblems. Problem 1.1: Subterm Processor: -> Pairs:
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