Sentence


Parse

Endlich
S[dcl]/S[dcl]
 
habe
(S[dcl]/NP)/(S[b]\NP)
 
ich
NP
 
S[dcl]\(S[dcl]/NP)
T <
S[dcl]/(S[b]\NP)
< 1×
Zeit
N
 
NP
*
NP/(NP\NP)
T >
,
(NP\NP)/NP
 
die
NP/N
 
Post
N
 
N/(N\N)
T >
NP/(N\N)
> 1
(NP\NP)/(N\N)
> 1
NP/(N\N)
> 1
zu
(S[to]\NP)/(S[b]\NP)
 
beantworten
(S[b]\NP)/NP
 
,
NP/NP
 
die
S[dcl]/NP
 
N\N
*
NP\(NP/(N\N))
T <
NP\(NP/(N\N))
> 1×
(S[b]\NP)\(NP/(N\N))
> 1×
(S[to]\NP)\(NP/(N\N))
> 1×
S[to]\NP
< 0
S[dcl]/S[dcl]
*
ich
N
 
in
(N\N)/NP
 
den
NP/N
 
letzten
N/N
 
drei
N/N
 
Wochen
N
 
N
> 0
N
> 0
NP
> 0
N\N
> 0
N
< 0
NP
*
erhalten
(S[pt]\NP)\NP
 
S[pt]\NP
< 0
habe
(S[b]\NP)\(S[pt]\NP)
 
S[b]\NP
< 0
S[dcl]\(S[dcl]/(S[b]\NP))
T <
S[dcl]\(S[dcl]/(S[b]\NP))
> 1×
S[dcl]
< 0
.
S[dcl]\S[dcl]
 
S[dcl]
< 0
S[dcl]
> 0

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