Sentence


Parse

Um
S[dcl]/S[dcl]
 
alles
NP
 
zu
(S[to]\NP)/(S[b]\NP)
 
erklären
S[b]\NP
 
S[to]\NP
> 0
S[dcl]/S[dcl]
*
würde
(S[dcl]/NP)/(S[b]\NP)
 
ich
NP
 
S[dcl]\(S[dcl]/NP)
T <
S[dcl]/(S[b]\NP)
< 1×
ewig
(S[b]\NP)/(S[b]\NP)
 
brauchen
(S[b]\NP)\NP
 
(S[b]\NP)\NP
> 1×
S[dcl]\NP
> 1×
S[dcl]\NP
> 1×
S[dcl]
< 0
.
S[dcl]\S[dcl]
 
S[dcl]
< 0
S[dcl]
> 0

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