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TRS Conditional pair #487562961
details
property
value
status
complete
benchmark
quotrem.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n023.star.cs.uiowa.edu
space
Mixed_CTRS
run statistics
property
value
solver
muterm 6.0.3
configuration
default
runtime (wallclock)
0.123543977737 seconds
cpu usage
0.043079661
max memory
3784704.0
stage attributes
key
value
output-size
9257
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 q:S r:S x:S y:S) (RULES less(0,s(x:S)) -> ttrue less(s(x:S),s(y:S)) -> less(x:S,y:S) less(x:S,0) -> ffalse minus(0,s(y:S)) -> 0 minus(s(x:S),s(y:S)) -> minus(x:S,y:S) minus(x:S,0) -> x:S quotrem(0,s(y:S)) -> pair(0,0) quotrem(s(x:S),s(y:S)) -> pair(0,s(x:S)) | less(x:S,y:S) ->* ttrue quotrem(s(x:S),s(y:S)) -> pair(s(q:S),r:S) | less(x:S,y:S) ->* ffalse, quotrem(minus(x:S,y:S),s(y:S)) ->* pair(q:S,r:S) ) Problem 1: Valid CTRS Processor: -> Rules: less(0,s(x:S)) -> ttrue less(s(x:S),s(y:S)) -> less(x:S,y:S) less(x:S,0) -> ffalse minus(0,s(y:S)) -> 0 minus(s(x:S),s(y:S)) -> minus(x:S,y:S) minus(x:S,0) -> x:S quotrem(0,s(y:S)) -> pair(0,0) quotrem(s(x:S),s(y:S)) -> pair(0,s(x:S)) | less(x:S,y:S) ->* ttrue quotrem(s(x:S),s(y:S)) -> pair(s(q:S),r:S) | less(x:S,y:S) ->* ffalse, quotrem(minus(x:S,y:S),s(y:S)) ->* pair(q:S,r:S) -> The system is a deterministic 3-CTRS. Problem 1: Dependency Pairs Processor: Conditional Termination Problem 1: -> Pairs: LESS(s(x:S),s(y:S)) -> LESS(x:S,y:S) MINUS(s(x:S),s(y:S)) -> MINUS(x:S,y:S) -> QPairs: Empty -> Rules: less(0,s(x:S)) -> ttrue less(s(x:S),s(y:S)) -> less(x:S,y:S) less(x:S,0) -> ffalse minus(0,s(y:S)) -> 0 minus(s(x:S),s(y:S)) -> minus(x:S,y:S) minus(x:S,0) -> x:S quotrem(0,s(y:S)) -> pair(0,0) quotrem(s(x:S),s(y:S)) -> pair(0,s(x:S)) | less(x:S,y:S) ->* ttrue quotrem(s(x:S),s(y:S)) -> pair(s(q:S),r:S) | less(x:S,y:S) ->* ffalse, quotrem(minus(x:S,y:S),s(y:S)) ->* pair(q:S,r:S) Conditional Termination Problem 2: -> Pairs: QUOTREM(s(x:S),s(y:S)) -> LESS(x:S,y:S) QUOTREM(s(x:S),s(y:S)) -> MINUS(x:S,y:S) | less(x:S,y:S) ->* ffalse QUOTREM(s(x:S),s(y:S)) -> QUOTREM(minus(x:S,y:S),s(y:S)) | less(x:S,y:S) ->* ffalse -> QPairs: LESS(s(x:S),s(y:S)) -> LESS(x:S,y:S) MINUS(s(x:S),s(y:S)) -> MINUS(x:S,y:S) -> Rules: less(0,s(x:S)) -> ttrue less(s(x:S),s(y:S)) -> less(x:S,y:S) less(x:S,0) -> ffalse minus(0,s(y:S)) -> 0 minus(s(x:S),s(y:S)) -> minus(x:S,y:S) minus(x:S,0) -> x:S quotrem(0,s(y:S)) -> pair(0,0) quotrem(s(x:S),s(y:S)) -> pair(0,s(x:S)) | less(x:S,y:S) ->* ttrue quotrem(s(x:S),s(y:S)) -> pair(s(q:S),r:S) | less(x:S,y:S) ->* ffalse, quotrem(minus(x:S,y:S),s(y:S)) ->* pair(q:S,r:S) The problem is decomposed in 2 subproblems. Problem 1.1: SCC Processor: -> Pairs: LESS(s(x:S),s(y:S)) -> LESS(x:S,y:S) MINUS(s(x:S),s(y:S)) -> MINUS(x:S,y:S) -> QPairs: Empty -> Rules: less(0,s(x:S)) -> ttrue less(s(x:S),s(y:S)) -> less(x:S,y:S) less(x:S,0) -> ffalse minus(0,s(y:S)) -> 0 minus(s(x:S),s(y:S)) -> minus(x:S,y:S) minus(x:S,0) -> x:S quotrem(0,s(y:S)) -> pair(0,0) quotrem(s(x:S),s(y:S)) -> pair(0,s(x:S)) | less(x:S,y:S) ->* ttrue quotrem(s(x:S),s(y:S)) -> pair(s(q:S),r:S) | less(x:S,y:S) ->* ffalse, quotrem(minus(x:S,y:S),s(y:S)) ->* pair(q:S,r:S) ->Strongly Connected Components: ->->Cycle:
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