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Parse

I
NP
 
do
(S[dcl]\NP)/(S[b]\NP)
 
n't
(S\NP)\(S\NP)
 
(S[dcl]\NP)/(S[b]\NP)
< 1×
think
(S[b]\NP)/S[dcl]
 
I
NP
 
will
(S[dcl]\NP)/(S[b]\NP)
 
get
(S[b]\NP)/PP
 
through
PP/NP
 
all
NP/NP
 
this
NP/N
 
work
N
 
NP
> 0
NP
> 0
PP
> 0
S[b]\NP
> 0
this
((S\NP)\(S\NP))/N
 
afternoon
N
 
(S\NP)\(S\NP)
> 0
.
.
 
(S\NP)\(S\NP)
.
S[b]\NP
< 0
S[dcl]\NP
> 0
S[dcl]
< 0
S[b]\NP
> 0
S[dcl]\NP
> 0
S[dcl]
< 0

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