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TRS Standard pair #487072654
details
property
value
status
complete
benchmark
6.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n185.star.cs.uiowa.edu
space
Beerendonk_07
run statistics
property
value
solver
muterm 6.0.3
configuration
default
runtime (wallclock)
0.0804571 seconds
cpu usage
0.060136
user time
0.02504
system time
0.035096
max virtual memory
113188.0
max residence set size
5528.0
stage attributes
key
value
starexec-result
YES
output
YES Problem 1: (VAR v_NonEmpty:S x:S) (RULES cond(ttrue,x:S) -> cond(odd(x:S),p(x:S)) odd(0) -> ffalse odd(s(0)) -> ttrue odd(s(s(x:S))) -> odd(x:S) p(0) -> 0 p(s(x:S)) -> x:S ) Problem 1: Innermost Equivalent Processor: -> Rules: cond(ttrue,x:S) -> cond(odd(x:S),p(x:S)) odd(0) -> ffalse odd(s(0)) -> ttrue odd(s(s(x:S))) -> odd(x:S) p(0) -> 0 p(s(x:S)) -> x:S -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination. Problem 1: Dependency Pairs Processor: -> Pairs: COND(ttrue,x:S) -> COND(odd(x:S),p(x:S)) COND(ttrue,x:S) -> ODD(x:S) COND(ttrue,x:S) -> P(x:S) ODD(s(s(x:S))) -> ODD(x:S) -> Rules: cond(ttrue,x:S) -> cond(odd(x:S),p(x:S)) odd(0) -> ffalse odd(s(0)) -> ttrue odd(s(s(x:S))) -> odd(x:S) p(0) -> 0 p(s(x:S)) -> x:S Problem 1: SCC Processor: -> Pairs: COND(ttrue,x:S) -> COND(odd(x:S),p(x:S)) COND(ttrue,x:S) -> ODD(x:S) COND(ttrue,x:S) -> P(x:S) ODD(s(s(x:S))) -> ODD(x:S) -> Rules: cond(ttrue,x:S) -> cond(odd(x:S),p(x:S)) odd(0) -> ffalse odd(s(0)) -> ttrue odd(s(s(x:S))) -> odd(x:S) p(0) -> 0 p(s(x:S)) -> x:S ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: ODD(s(s(x:S))) -> ODD(x:S) ->->-> Rules: cond(ttrue,x:S) -> cond(odd(x:S),p(x:S)) odd(0) -> ffalse odd(s(0)) -> ttrue odd(s(s(x:S))) -> odd(x:S) p(0) -> 0 p(s(x:S)) -> x:S ->->Cycle: ->->-> Pairs: COND(ttrue,x:S) -> COND(odd(x:S),p(x:S)) ->->-> Rules: cond(ttrue,x:S) -> cond(odd(x:S),p(x:S)) odd(0) -> ffalse odd(s(0)) -> ttrue odd(s(s(x:S))) -> odd(x:S) p(0) -> 0 p(s(x:S)) -> x:S The problem is decomposed in 2 subproblems. Problem 1.1: Subterm Processor: -> Pairs: ODD(s(s(x:S))) -> ODD(x:S) -> Rules: cond(ttrue,x:S) -> cond(odd(x:S),p(x:S)) odd(0) -> ffalse odd(s(0)) -> ttrue odd(s(s(x:S))) -> odd(x:S) p(0) -> 0 p(s(x:S)) -> x:S ->Projection: pi(ODD) = 1 Problem 1.1:
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