/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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NO

Problem 1: 

(VAR v_NonEmpty:S x:S)
(RULES
eq(x:S,x:S) -> eq(T,T)
even(0) -> ttrue
even(s(x:S)) -> ffalse | eq(odd(x:S),ffalse) ->* eq(T,T)
even(s(x:S)) -> ttrue | eq(odd(x:S),ttrue) ->* eq(T,T)
odd(0) -> ffalse
odd(s(x:S)) -> ffalse | eq(even(x:S),ffalse) ->* eq(T,T)
odd(s(x:S)) -> ttrue | eq(even(x:S),ttrue) ->* eq(T,T)
)

Problem 1: 
Valid CTRS Processor:
-> Rules:
 eq(x:S,x:S) -> eq(T,T)
 even(0) -> ttrue
 even(s(x:S)) -> ffalse | eq(odd(x:S),ffalse) ->* eq(T,T)
 even(s(x:S)) -> ttrue | eq(odd(x:S),ttrue) ->* eq(T,T)
 odd(0) -> ffalse
 odd(s(x:S)) -> ffalse | eq(even(x:S),ffalse) ->* eq(T,T)
 odd(s(x:S)) -> ttrue | eq(even(x:S),ttrue) ->* eq(T,T)
-> The system is a deterministic 3-CTRS.

Problem 1: 

Dependency Pairs Processor:

Conditional Termination Problem 1:
-> Pairs:
 EQ(x:S,x:S) -> EQ(T,T)
-> QPairs:
 Empty
-> Rules:
 eq(x:S,x:S) -> eq(T,T)
 even(0) -> ttrue
 even(s(x:S)) -> ffalse | eq(odd(x:S),ffalse) ->* eq(T,T)
 even(s(x:S)) -> ttrue | eq(odd(x:S),ttrue) ->* eq(T,T)
 odd(0) -> ffalse
 odd(s(x:S)) -> ffalse | eq(even(x:S),ffalse) ->* eq(T,T)
 odd(s(x:S)) -> ttrue | eq(even(x:S),ttrue) ->* eq(T,T)

Conditional Termination Problem 2:
-> Pairs:
 EVEN(s(x:S)) -> EQ(odd(x:S),ffalse)
 EVEN(s(x:S)) -> EQ(odd(x:S),ttrue)
 EVEN(s(x:S)) -> ODD(x:S)
 ODD(s(x:S)) -> EQ(even(x:S),ffalse)
 ODD(s(x:S)) -> EQ(even(x:S),ttrue)
 ODD(s(x:S)) -> EVEN(x:S)
-> QPairs:
 EQ(x:S,x:S) -> EQ(T,T)
-> Rules:
 eq(x:S,x:S) -> eq(T,T)
 even(0) -> ttrue
 even(s(x:S)) -> ffalse | eq(odd(x:S),ffalse) ->* eq(T,T)
 even(s(x:S)) -> ttrue | eq(odd(x:S),ttrue) ->* eq(T,T)
 odd(0) -> ffalse
 odd(s(x:S)) -> ffalse | eq(even(x:S),ffalse) ->* eq(T,T)
 odd(s(x:S)) -> ttrue | eq(even(x:S),ttrue) ->* eq(T,T)

Problem 1: 

SCC Processor:
-> Pairs:
 EQ(x:S,x:S) -> EQ(T,T)
-> QPairs:
 Empty
-> Rules:
 eq(x:S,x:S) -> eq(T,T)
 even(0) -> ttrue
 even(s(x:S)) -> ffalse | eq(odd(x:S),ffalse) ->* eq(T,T)
 even(s(x:S)) -> ttrue | eq(odd(x:S),ttrue) ->* eq(T,T)
 odd(0) -> ffalse
 odd(s(x:S)) -> ffalse | eq(even(x:S),ffalse) ->* eq(T,T)
 odd(s(x:S)) -> ttrue | eq(even(x:S),ttrue) ->* eq(T,T)
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 EQ(x:S,x:S) -> EQ(T,T)
-> QPairs:
 Empty
->->-> Rules:
 eq(x:S,x:S) -> eq(T,T)
 even(0) -> ttrue
 even(s(x:S)) -> ffalse | eq(odd(x:S),ffalse) ->* eq(T,T)
 even(s(x:S)) -> ttrue | eq(odd(x:S),ttrue) ->* eq(T,T)
 odd(0) -> ffalse
 odd(s(x:S)) -> ffalse | eq(even(x:S),ffalse) ->* eq(T,T)
 odd(s(x:S)) -> ttrue | eq(even(x:S),ttrue) ->* eq(T,T)

Problem 1: 

Infinite Processor:
-> Pairs:
 EQ(x:S,x:S) -> EQ(T,T)
-> QPairs:
 Empty
-> Rules:
 eq(x:S,x:S) -> eq(T,T)
 even(0) -> ttrue
 even(s(x:S)) -> ffalse | eq(odd(x:S),ffalse) ->* eq(T,T)
 even(s(x:S)) -> ttrue | eq(odd(x:S),ttrue) ->* eq(T,T)
 odd(0) -> ffalse
 odd(s(x:S)) -> ffalse | eq(even(x:S),ffalse) ->* eq(T,T)
 odd(s(x:S)) -> ttrue | eq(even(x:S),ttrue) ->* eq(T,T)
-> Pairs in cycle:
 EQ(x:S,x:S) -> EQ(T,T)

The problem is infinite.