Sentence


Parse

Did
(S[q]/(S[b]\NP))/NP
 
n't
S\S
 
(S[q]/(S[b]\NP))/NP
< n×
you
NP
 
S[q]/(S[b]\NP)
> 0
know
(S[b]\NP)/S[dcl]
 
Tom
N
 
NP
*
and
conj
 
I
NP
 
NP\NP
NP
< 0
were
(S[dcl]\NP)/(S[pss]\NP)
 
married
S[pss]\NP
 
?
.
 
S[pss]\NP
.
S[dcl]\NP
> 0
S[dcl]
< 0
S[b]\NP
> 0
S[q]
> 0

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