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TRS Conditional pair #516978576
details
property
value
status
complete
benchmark
361.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n078.star.cs.uiowa.edu
space
COPS
run statistics
property
value
solver
muterm 6.0.3
configuration
default
runtime (wallclock)
101.719713926 seconds
cpu usage
101.533273238
max memory
1.14450432E8
stage attributes
key
value
output-size
5109
starexec-result
NO
output
/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO Problem 1: (VAR v_NonEmpty:S q:S r:S x:S y:S) (RULES div(x:S,y:S) -> pair(0,y:S) | greater(y:S,x:S) ->* ttrue div(x:S,y:S) -> pair(s(q:S),r:S) | leq(y:S,x:S) ->* ttrue, div(m(x:S,y:S),y:S) ->* pair(q:S,r:S) greater(s(x:S),0) -> ttrue greater(s(x:S),s(y:S)) -> greater(x:S,y:S) leq(0,x:S) -> ttrue leq(s(x:S),s(y:S)) -> leq(x:S,y:S) m(0,y:S) -> 0 m(s(x:S),s(y:S)) -> m(x:S,y:S) m(x:S,0) -> x:S ) Problem 1: Valid CTRS Processor: -> Rules: div(x:S,y:S) -> pair(0,y:S) | greater(y:S,x:S) ->* ttrue div(x:S,y:S) -> pair(s(q:S),r:S) | leq(y:S,x:S) ->* ttrue, div(m(x:S,y:S),y:S) ->* pair(q:S,r:S) greater(s(x:S),0) -> ttrue greater(s(x:S),s(y:S)) -> greater(x:S,y:S) leq(0,x:S) -> ttrue leq(s(x:S),s(y:S)) -> leq(x:S,y:S) m(0,y:S) -> 0 m(s(x:S),s(y:S)) -> m(x:S,y:S) m(x:S,0) -> x:S -> The system is a deterministic 3-CTRS. Problem 1: Dependency Pairs Processor: Conditional Termination Problem 1: -> Pairs: GREATER(s(x:S),s(y:S)) -> GREATER(x:S,y:S) LEQ(s(x:S),s(y:S)) -> LEQ(x:S,y:S) M(s(x:S),s(y:S)) -> M(x:S,y:S) -> QPairs: Empty -> Rules: div(x:S,y:S) -> pair(0,y:S) | greater(y:S,x:S) ->* ttrue div(x:S,y:S) -> pair(s(q:S),r:S) | leq(y:S,x:S) ->* ttrue, div(m(x:S,y:S),y:S) ->* pair(q:S,r:S) greater(s(x:S),0) -> ttrue greater(s(x:S),s(y:S)) -> greater(x:S,y:S) leq(0,x:S) -> ttrue leq(s(x:S),s(y:S)) -> leq(x:S,y:S) m(0,y:S) -> 0 m(s(x:S),s(y:S)) -> m(x:S,y:S) m(x:S,0) -> x:S Conditional Termination Problem 2: -> Pairs: DIV(x:S,y:S) -> DIV(m(x:S,y:S),y:S) | leq(y:S,x:S) ->* ttrue DIV(x:S,y:S) -> GREATER(y:S,x:S) DIV(x:S,y:S) -> LEQ(y:S,x:S) DIV(x:S,y:S) -> M(x:S,y:S) | leq(y:S,x:S) ->* ttrue -> QPairs: GREATER(s(x:S),s(y:S)) -> GREATER(x:S,y:S) LEQ(s(x:S),s(y:S)) -> LEQ(x:S,y:S) M(s(x:S),s(y:S)) -> M(x:S,y:S) -> Rules: div(x:S,y:S) -> pair(0,y:S) | greater(y:S,x:S) ->* ttrue div(x:S,y:S) -> pair(s(q:S),r:S) | leq(y:S,x:S) ->* ttrue, div(m(x:S,y:S),y:S) ->* pair(q:S,r:S) greater(s(x:S),0) -> ttrue greater(s(x:S),s(y:S)) -> greater(x:S,y:S) leq(0,x:S) -> ttrue leq(s(x:S),s(y:S)) -> leq(x:S,y:S) m(0,y:S) -> 0 m(s(x:S),s(y:S)) -> m(x:S,y:S) m(x:S,0) -> x:S Problem 1: SCC Processor: -> Pairs: DIV(x:S,y:S) -> DIV(m(x:S,y:S),y:S) | leq(y:S,x:S) ->* ttrue DIV(x:S,y:S) -> GREATER(y:S,x:S) DIV(x:S,y:S) -> LEQ(y:S,x:S) DIV(x:S,y:S) -> M(x:S,y:S) | leq(y:S,x:S) ->* ttrue -> QPairs: GREATER(s(x:S),s(y:S)) -> GREATER(x:S,y:S) LEQ(s(x:S),s(y:S)) -> LEQ(x:S,y:S) M(s(x:S),s(y:S)) -> M(x:S,y:S) -> Rules: div(x:S,y:S) -> pair(0,y:S) | greater(y:S,x:S) ->* ttrue div(x:S,y:S) -> pair(s(q:S),r:S) | leq(y:S,x:S) ->* ttrue, div(m(x:S,y:S),y:S) ->* pair(q:S,r:S) greater(s(x:S),0) -> ttrue greater(s(x:S),s(y:S)) -> greater(x:S,y:S) leq(0,x:S) -> ttrue leq(s(x:S),s(y:S)) -> leq(x:S,y:S) m(0,y:S) -> 0
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