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Hättest
(S[q]/(S[b]\NP))/NP
 
du
NP
 
S[q]/(S[b]\NP)
> 0
etwas
NP
 
(S[b]\NP)/((S[b]\NP)\NP)
T >
dagegen
S[adj]\NP
 
N/N
*
,
((S[b]\NP)\(S[b]\NP))/(S[b]\NP)
 
wenn
(S[b]\NP)/NP
 
ich
NP/NP
 
die
NP/N
 
Tür
N
 
N\(N/N)
T <
NP\(N/N)
> 1×
NP\(N/N)
> 1×
(S[b]\NP)\(N/N)
> 1×
((S[b]\NP)\(S[b]\NP))\(N/N)
> 1×
(S[b]\NP)\(S[b]\NP)
< 0
(S[b]\NP)/((S[b]\NP)\NP)
< 1×
schlösse
(S[b]\NP)\NP
 
S[b]\NP
> 0
S[q]
> 0
?
S[q]\S[q]
 
S[q]
< 0

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