Sentence


Parse

Doing
S[ng]\NP
 
S/S
*
that
NP
 
will
(S[dcl]\NP)/(S[b]\NP)
 
be
(S[b]\NP)/(S[adj]\NP)
 
as
(S[adj]\NP)/(S[adj]\NP)
 
easy
S[adj]\NP
 
S[adj]\NP
> 0
S[b]\NP
> 0
S[dcl]\NP
> 0
as
((S\NP)\(S\NP))/NP
 
pie
N
 
NP
*
(S\NP)\(S\NP)
> 0
S[dcl]\NP
< 0
S[dcl]
< 0
S[dcl]
> 0
.
.
 
S[dcl]
.

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