details

property	value
status	complete
benchmark	append.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n175.star.cs.uiowa.edu
space	CiME_04

run statistics

property	value
solver	muterm 6.0.3
configuration	default
runtime (wallclock)	0.0253889 seconds
cpu usage	0.023928
user time	0.006683
system time	0.017245
max virtual memory	0.0
max residence set size	5260.0

stage attributes

key	value
starexec-result	YES

output

YES Problem 1: (VAR v_NonEmpty:S l:S l1:S l2:S x:S) (RULES append(l1:S,l2:S) -> ifappend(l1:S,l2:S,l1:S) hd(cons(x:S,l:S)) -> x:S ifappend(l1:S,l2:S,cons(x:S,l:S)) -> cons(x:S,append(l:S,l2:S)) ifappend(l1:S,l2:S,nil) -> l2:S is_empty(cons(x:S,l:S)) -> ffalse is_empty(nil) -> ttrue tl(cons(x:S,l:S)) -> l:S ) Problem 1: Innermost Equivalent Processor: -> Rules: append(l1:S,l2:S) -> ifappend(l1:S,l2:S,l1:S) hd(cons(x:S,l:S)) -> x:S ifappend(l1:S,l2:S,cons(x:S,l:S)) -> cons(x:S,append(l:S,l2:S)) ifappend(l1:S,l2:S,nil) -> l2:S is_empty(cons(x:S,l:S)) -> ffalse is_empty(nil) -> ttrue tl(cons(x:S,l:S)) -> l:S -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination. Problem 1: Dependency Pairs Processor: -> Pairs: APPEND(l1:S,l2:S) -> IFAPPEND(l1:S,l2:S,l1:S) IFAPPEND(l1:S,l2:S,cons(x:S,l:S)) -> APPEND(l:S,l2:S) -> Rules: append(l1:S,l2:S) -> ifappend(l1:S,l2:S,l1:S) hd(cons(x:S,l:S)) -> x:S ifappend(l1:S,l2:S,cons(x:S,l:S)) -> cons(x:S,append(l:S,l2:S)) ifappend(l1:S,l2:S,nil) -> l2:S is_empty(cons(x:S,l:S)) -> ffalse is_empty(nil) -> ttrue tl(cons(x:S,l:S)) -> l:S Problem 1: SCC Processor: -> Pairs: APPEND(l1:S,l2:S) -> IFAPPEND(l1:S,l2:S,l1:S) IFAPPEND(l1:S,l2:S,cons(x:S,l:S)) -> APPEND(l:S,l2:S) -> Rules: append(l1:S,l2:S) -> ifappend(l1:S,l2:S,l1:S) hd(cons(x:S,l:S)) -> x:S ifappend(l1:S,l2:S,cons(x:S,l:S)) -> cons(x:S,append(l:S,l2:S)) ifappend(l1:S,l2:S,nil) -> l2:S is_empty(cons(x:S,l:S)) -> ffalse is_empty(nil) -> ttrue tl(cons(x:S,l:S)) -> l:S ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: APPEND(l1:S,l2:S) -> IFAPPEND(l1:S,l2:S,l1:S) IFAPPEND(l1:S,l2:S,cons(x:S,l:S)) -> APPEND(l:S,l2:S) ->->-> Rules: append(l1:S,l2:S) -> ifappend(l1:S,l2:S,l1:S) hd(cons(x:S,l:S)) -> x:S ifappend(l1:S,l2:S,cons(x:S,l:S)) -> cons(x:S,append(l:S,l2:S)) ifappend(l1:S,l2:S,nil) -> l2:S is_empty(cons(x:S,l:S)) -> ffalse is_empty(nil) -> ttrue tl(cons(x:S,l:S)) -> l:S Problem 1: Subterm Processor: -> Pairs: APPEND(l1:S,l2:S) -> IFAPPEND(l1:S,l2:S,l1:S) IFAPPEND(l1:S,l2:S,cons(x:S,l:S)) -> APPEND(l:S,l2:S) -> Rules: append(l1:S,l2:S) -> ifappend(l1:S,l2:S,l1:S) hd(cons(x:S,l:S)) -> x:S ifappend(l1:S,l2:S,cons(x:S,l:S)) -> cons(x:S,append(l:S,l2:S)) ifappend(l1:S,l2:S,nil) -> l2:S is_empty(cons(x:S,l:S)) -> ffalse is_empty(nil) -> ttrue tl(cons(x:S,l:S)) -> l:S ->Projection: pi(APPEND) = 1 pi(IFAPPEND) = 3 Problem 1: SCC Processor: -> Pairs: APPEND(l1:S,l2:S) -> IFAPPEND(l1:S,l2:S,l1:S) -> Rules: append(l1:S,l2:S) -> ifappend(l1:S,l2:S,l1:S) hd(cons(x:S,l:S)) -> x:S ifappend(l1:S,l2:S,cons(x:S,l:S)) -> cons(x:S,append(l:S,l2:S)) ifappend(l1:S,l2:S,nil) -> l2:S popout

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

job log

popout

actions

all output return to TRS Standard