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Finalmente
S[dcl]/S[dcl]
 
tenho
(S[dcl]\NP)/NP
 
tempo
N/PP
 
para
N\N
 
N/PP
< 1×
responder
(S[ng]\NP)/NP
 
às
N
 
NP
*
S[ng]\NP
> 0
N\N
*
N/PP
< 1×
correspondências
PP/NP
 
que
N/N
 
eu
N
 
N
> 0
NP
*
PP
> 0
N
> 0
NP
*
S[dcl]\NP
> 0
recebi
N
 
NP
*
S[dcl]/(S[dcl]\NP)
T >
durante
(S[dcl]\NP)\(S[dcl]\NP)
 
S[dcl]\(S[dcl]\NP)
> 1×
S[dcl]
< 0
S[dcl]
> 0
estas
(S[ng]\NP)/NP
 
três
NP/N
 
semanas
N
 
NP
> 0
S[ng]\NP
> 0
S[dcl]/S[dcl]
*
.
S[dcl]\S[dcl]
 
S[dcl]\S[dcl]
> 1×
S[dcl]
< 0

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