details

property	value
status	complete
benchmark	#3.2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n032.star.cs.uiowa.edu
space	AG01

run statistics

property	value
solver	muterm 6.0.3
configuration	default
runtime (wallclock)	0.0585639476776 seconds
cpu usage	0.039666553
max memory	3387392.0

stage attributes

key	value
output-size	3706
starexec-result	"YES"

output

/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES Problem 1: (VAR v_NonEmpty:S x:S y:S) (RULES minus(x:S,0) -> x:S minus(x:S,s(y:S)) -> pred(minus(x:S,y:S)) pred(s(x:S)) -> x:S quot(0,s(y:S)) -> 0 quot(s(x:S),s(y:S)) -> s(quot(minus(x:S,y:S),s(y:S))) ) Problem 1: Innermost Equivalent Processor: -> Rules: minus(x:S,0) -> x:S minus(x:S,s(y:S)) -> pred(minus(x:S,y:S)) pred(s(x:S)) -> x:S quot(0,s(y:S)) -> 0 quot(s(x:S),s(y:S)) -> s(quot(minus(x:S,y:S),s(y:S))) -> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination. Problem 1: Dependency Pairs Processor: -> Pairs: MINUS(x:S,s(y:S)) -> MINUS(x:S,y:S) MINUS(x:S,s(y:S)) -> PRED(minus(x:S,y:S)) QUOT(s(x:S),s(y:S)) -> MINUS(x:S,y:S) QUOT(s(x:S),s(y:S)) -> QUOT(minus(x:S,y:S),s(y:S)) -> Rules: minus(x:S,0) -> x:S minus(x:S,s(y:S)) -> pred(minus(x:S,y:S)) pred(s(x:S)) -> x:S quot(0,s(y:S)) -> 0 quot(s(x:S),s(y:S)) -> s(quot(minus(x:S,y:S),s(y:S))) Problem 1: SCC Processor: -> Pairs: MINUS(x:S,s(y:S)) -> MINUS(x:S,y:S) MINUS(x:S,s(y:S)) -> PRED(minus(x:S,y:S)) QUOT(s(x:S),s(y:S)) -> MINUS(x:S,y:S) QUOT(s(x:S),s(y:S)) -> QUOT(minus(x:S,y:S),s(y:S)) -> Rules: minus(x:S,0) -> x:S minus(x:S,s(y:S)) -> pred(minus(x:S,y:S)) pred(s(x:S)) -> x:S quot(0,s(y:S)) -> 0 quot(s(x:S),s(y:S)) -> s(quot(minus(x:S,y:S),s(y:S))) ->Strongly Connected Components: ->->Cycle: ->->-> Pairs: MINUS(x:S,s(y:S)) -> MINUS(x:S,y:S) ->->-> Rules: minus(x:S,0) -> x:S minus(x:S,s(y:S)) -> pred(minus(x:S,y:S)) pred(s(x:S)) -> x:S quot(0,s(y:S)) -> 0 quot(s(x:S),s(y:S)) -> s(quot(minus(x:S,y:S),s(y:S))) ->->Cycle: ->->-> Pairs: QUOT(s(x:S),s(y:S)) -> QUOT(minus(x:S,y:S),s(y:S)) ->->-> Rules: minus(x:S,0) -> x:S minus(x:S,s(y:S)) -> pred(minus(x:S,y:S)) pred(s(x:S)) -> x:S quot(0,s(y:S)) -> 0 quot(s(x:S),s(y:S)) -> s(quot(minus(x:S,y:S),s(y:S))) The problem is decomposed in 2 subproblems. Problem 1.1: Subterm Processor: -> Pairs: MINUS(x:S,s(y:S)) -> MINUS(x:S,y:S) -> Rules: minus(x:S,0) -> x:S minus(x:S,s(y:S)) -> pred(minus(x:S,y:S)) pred(s(x:S)) -> x:S quot(0,s(y:S)) -> 0 quot(s(x:S),s(y:S)) -> s(quot(minus(x:S,y:S),s(y:S))) ->Projection: pi(MINUS) = 2 Problem 1.1: SCC Processor: popout

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

job log

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actions

all output return to TRS Standard