details

property	value
status	complete
benchmark	2.39.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n072.star.cs.uiowa.edu
space	SK90

run statistics

property	value
solver	muterm 6.0.3
configuration	default
runtime (wallclock)	0.0200089 seconds
cpu usage	0.020417
user time	0.011137
system time	0.00928
max virtual memory	113188.0
max residence set size	5316.0

stage attributes

key	value
starexec-result	YES

output

YES Problem 1: (VAR v_NonEmpty:S x:S y:S z:S) (RULES ++(.(x:S,y:S),z:S) -> .(x:S,++(y:S,z:S)) ++(nil,y:S) -> y:S car(.(x:S,y:S)) -> x:S cdr(.(x:S,y:S)) -> y:S null(.(x:S,y:S)) -> ffalse null(nil) -> ttrue rev(.(x:S,y:S)) -> ++(rev(y:S),.(x:S,nil)) rev(nil) -> nil ) Problem 1: Innermost Equivalent Processor: -> Rules: ++(.(x:S,y:S),z:S) -> .(x:S,++(y:S,z:S)) ++(nil,y:S) -> y:S car(.(x:S,y:S)) -> x:S cdr(.(x:S,y:S)) -> y:S null(.(x:S,y:S)) -> ffalse null(nil) -> ttrue rev(.(x:S,y:S)) -> ++(rev(y:S),.(x:S,nil)) rev(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: ++#(.(x:S,y:S),z:S) -> ++#(y:S,z:S) REV(.(x:S,y:S)) -> ++#(rev(y:S),.(x:S,nil)) REV(.(x:S,y:S)) -> REV(y:S) -> Rules: ++(.(x:S,y:S),z:S) -> .(x:S,++(y:S,z:S)) ++(nil,y:S) -> y:S car(.(x:S,y:S)) -> x:S cdr(.(x:S,y:S)) -> y:S null(.(x:S,y:S)) -> ffalse null(nil) -> ttrue rev(.(x:S,y:S)) -> ++(rev(y:S),.(x:S,nil)) rev(nil) -> nil Problem 1: SCC Processor: -> Pairs: ++#(.(x:S,y:S),z:S) -> ++#(y:S,z:S) REV(.(x:S,y:S)) -> ++#(rev(y:S),.(x:S,nil)) REV(.(x:S,y:S)) -> REV(y:S) -> Rules: ++(.(x:S,y:S),z:S) -> .(x:S,++(y:S,z:S)) ++(nil,y:S) -> y:S car(.(x:S,y:S)) -> x:S cdr(.(x:S,y:S)) -> y:S null(.(x:S,y:S)) -> ffalse null(nil) -> ttrue rev(.(x:S,y:S)) -> ++(rev(y:S),.(x:S,nil)) rev(nil) -> nil ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: ++#(.(x:S,y:S),z:S) -> ++#(y:S,z:S) ->->-> Rules: ++(.(x:S,y:S),z:S) -> .(x:S,++(y:S,z:S)) ++(nil,y:S) -> y:S car(.(x:S,y:S)) -> x:S cdr(.(x:S,y:S)) -> y:S null(.(x:S,y:S)) -> ffalse null(nil) -> ttrue rev(.(x:S,y:S)) -> ++(rev(y:S),.(x:S,nil)) rev(nil) -> nil ->->Cycle: ->->-> Pairs: REV(.(x:S,y:S)) -> REV(y:S) ->->-> Rules: ++(.(x:S,y:S),z:S) -> .(x:S,++(y:S,z:S)) ++(nil,y:S) -> y:S car(.(x:S,y:S)) -> x:S cdr(.(x:S,y:S)) -> y:S null(.(x:S,y:S)) -> ffalse null(nil) -> ttrue rev(.(x:S,y:S)) -> ++(rev(y:S),.(x:S,nil)) rev(nil) -> nil The problem is decomposed in 2 subproblems. Problem 1.1: Subterm Processor: -> Pairs: ++#(.(x:S,y:S),z:S) -> ++#(y:S,z:S) -> Rules: ++(.(x:S,y:S),z:S) -> .(x:S,++(y:S,z:S)) popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to TRS Standard