Sentence


Parse

I
NP
 
should
(S[dcl]\NP)/(S[b]\NP)
 
n't
(S\NP)\(S\NP)
 
(S[dcl]\NP)/(S[b]\NP)
< 1×
have
(S[b]\NP)/(S[pt]\NP)
 
logged
(S[pt]\NP)/PR
 
off
PR
 
.
.
 
PR
.
S[pt]\NP
> 0
S[b]\NP
> 0
S[dcl]\NP
> 0
S[dcl]
< 0

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