Sentence


Parse

I
NP
 
'm
(S[dcl]\NP)/(S[ng]\NP)
 
leaving
((S[ng]\NP)/(S[to]\NP))/NP
 
it
NP
 
(S[ng]\NP)/(S[to]\NP)
> 0
to
((S\NP)\(S\NP))/NP
 
you
NP
 
(S\NP)\(S\NP)
> 0
(S[ng]\NP)/(S[to]\NP)
< 1×
to
(S[to]\NP)/(S[b]\NP)
 
decide
(S[b]\NP)/NP
 
that
NP
 
.
.
 
NP
.
S[b]\NP
> 0
S[to]\NP
> 0
S[ng]\NP
> 0
S[dcl]\NP
> 0
S[dcl]
< 0

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