Sentence


Parse

With
(S/S)/NP
 
pure
N/N
 
logic
N
 
N
> 0
NP
*
S/S
> 0
you
NP
 
can
(S[dcl]\NP)/(S[b]\NP)
 
prove
(S[b]\NP)/NP
 
the
NP/N
 
biggest
N/N
 
nonsense
N
 
N
> 0
NP
> 0
.
.
 
NP
.
S[b]\NP
> 0
S[dcl]\NP
> 0
S[dcl]
< 0
S[dcl]
> 0

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