Sentence


Parse

No
NP/N
 
one
N
 
NP
> 0
knows
(S[dcl]\NP)/NP
 
when
(S/S)/S[dcl]
 
such
NP/NP
 
a
NP/N
 
custom
N
 
NP
> 0
NP
> 0
(S[X]\NP)\((S[X]\NP)/NP)
T <
((S/S)\NP)\((S[dcl]\NP)/NP)
> n
(S/S)\NP
< 0
S/S
< 0
first
N
 
NP
*
came
(S[dcl]\NP)/PP
 
into
PP/NP
 
existence
N
 
NP
*
.
.
 
NP
.
PP
> 0
S[dcl]\NP
> 0
S[dcl]
< 0
S[dcl]
> 0

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