Sentence


Parse

J'
N/N
 
en
S[dcl]/S[dcl]
 
ai
NP/N
 
assez
N
 
N\(N/N)
T <
NP\(N/N)
> 1×
supporté
S[dcl]\NP
 
S[dcl]\(N/N)
< 1
S[dcl]\(N/N)
> 1×
S[dcl]
< 0
.
S[dcl]\S[dcl]
 
S[dcl]
< 0

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