details

property	value
status	complete
benchmark	268.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n135.star.cs.uiowa.edu
space	COPS

run statistics

property	value
solver	muterm 6.0.3
configuration	default
runtime (wallclock)	40.1668 seconds
cpu usage	40.1103
user time	39.4968
system time	0.613493
max virtual memory	1122240.0
max residence set size	289344.0

stage attributes

key	value
starexec-result	NO

output

40.07/40.16 NO 40.07/40.16 40.07/40.16 Problem 1: 40.07/40.16 40.07/40.16 (VAR v_NonEmpty:S x:S) 40.07/40.16 (RULES 40.07/40.16 a(x:S) -> b | p(x:S) ->* 0 40.07/40.16 c -> c 40.07/40.16 p(q(x:S)) -> 0 40.07/40.16 ) 40.07/40.16 40.07/40.16 Problem 1: 40.07/40.16 Valid CTRS Processor: 40.07/40.16 -> Rules: 40.07/40.16 a(x:S) -> b | p(x:S) ->* 0 40.07/40.16 c -> c 40.07/40.16 p(q(x:S)) -> 0 40.07/40.16 -> The system is a deterministic 3-CTRS. 40.07/40.16 40.07/40.16 Problem 1: 40.07/40.16 40.07/40.16 Dependency Pairs Processor: 40.07/40.16 40.07/40.16 Conditional Termination Problem 1: 40.07/40.16 -> Pairs: 40.07/40.16 C -> C 40.07/40.16 -> QPairs: 40.07/40.16 Empty 40.07/40.16 -> Rules: 40.07/40.16 a(x:S) -> b | p(x:S) ->* 0 40.07/40.16 c -> c 40.07/40.16 p(q(x:S)) -> 0 40.07/40.16 40.07/40.16 Conditional Termination Problem 2: 40.07/40.16 -> Pairs: 40.07/40.16 A(x:S) -> P(x:S) 40.07/40.16 -> QPairs: 40.07/40.16 Empty 40.07/40.16 -> Rules: 40.07/40.16 a(x:S) -> b | p(x:S) ->* 0 40.07/40.16 c -> c 40.07/40.16 p(q(x:S)) -> 0 40.07/40.16 40.07/40.16 Problem 1: 40.07/40.16 40.07/40.16 SCC Processor: 40.07/40.16 -> Pairs: 40.07/40.16 C -> C 40.07/40.16 -> QPairs: 40.07/40.16 Empty 40.07/40.16 -> Rules: 40.07/40.16 a(x:S) -> b | p(x:S) ->* 0 40.07/40.16 c -> c 40.07/40.16 p(q(x:S)) -> 0 40.07/40.16 ->Strongly Connected Components: 40.07/40.16 ->->Cycle: 40.07/40.16 ->->-> Pairs: 40.07/40.16 C -> C 40.07/40.16 -> QPairs: 40.07/40.16 Empty 40.07/40.16 ->->-> Rules: 40.07/40.16 a(x:S) -> b | p(x:S) ->* 0 40.07/40.16 c -> c 40.07/40.16 p(q(x:S)) -> 0 40.07/40.16 40.07/40.16 Problem 1: 40.07/40.16 40.07/40.16 Infinite Processor: 40.07/40.16 -> Pairs: 40.07/40.16 C -> C 40.07/40.16 -> QPairs: 40.07/40.16 Empty 40.07/40.16 -> Rules: 40.07/40.16 a(x:S) -> b | p(x:S) ->* 0 40.07/40.16 c -> c 40.07/40.16 p(q(x:S)) -> 0 40.07/40.16 -> Pairs in cycle: 40.07/40.16 C -> C 40.07/40.16 40.07/40.16 The problem is infinite. 40.07/40.16 EOF popout

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job log

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actions

all output return to TRS_Conditional 2019-03-28 22.59