Sentence


Parse

I
NP
 
'll
(S[dcl]\NP)/(S[b]\NP)
 
be
(S[b]\NP)/(S[ng]\NP)
 
going
S[ng]\NP
 
with
((S\NP)\(S\NP))/NP
 
you
NP
 
(S\NP)\(S\NP)
> 0
.
.
 
(S\NP)\(S\NP)
.
S[ng]\NP
< 0
S[b]\NP
> 0
S[dcl]\NP
> 0
S[dcl]
< 0

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