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===Step 5===
 
===Step 5===
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<pre>
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'''Existential Graph Format : Application Triples with Structure Sharing'''
| NB.  I am working from rough notes that
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| I wrote out in the Fall of 1996, and it
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| is not always easy to reconstruct what
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| I had in mind at the time.  I misread
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| this passage in my last posting of it,
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| causing me to leave out a few steps.
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Existential Graph Format, Application Triples with Structure Sharing.
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Redo the same development in Existential Graph notation. In the work below, the term development is carried out in reverse, that is, in application order.
 
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Same development in Existential Graph notation.
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Here I am carrying out the term development in
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reverse, that is, in application order.
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<pre>
 
o-----------------------------------------------------------o
 
o-----------------------------------------------------------o
 
|                                                          |
 
|                                                          |
Line 178: Line 169:  
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|                                                          |
 
o-----------------------------------------------------------o
 
o-----------------------------------------------------------o
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</pre>
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| I am still not sure what order I intended for the
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'''NB.'''  Looking at my notes from Fall Term 1996, I'm still not sure what order I intended for the application triples, but the above is one likely guess:
| application triples, but this is one likely guess:
      
For example:
 
For example:
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The nodes that are right-labeled <K, KS, K(KS)>,
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* The nodes that are right-labeled <math>(\operatorname{K}, \operatorname{K}\operatorname{S}, \operatorname{K}(\operatorname{K}\operatorname{S})),</math> in that order, constitute an application triple.
in that order, constitute an application triple.
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The type of the applicand K is A=>(B=>A).
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* The type of the applicand <math>\operatorname{K}</math> is <math>A \Rightarrow (B \Rightarrow A).</math>
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The type of the applicator KS is (A=>(B=>A))=>(A=>A).
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* The type of the applicator <math>\operatorname{K}\operatorname{S}</math> is <math>(A \Rightarrow (B \Rightarrow A)) \Rightarrow (A \Rightarrow A).</math>
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Therefore, the type of the application K(KS) is A=>A.
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* Therefore, the type of the application <math>\operatorname{K}(\operatorname{K}\operatorname{S})</math> is <math>A \Rightarrow A.</math>
</pre>
      
==Composition, or the Composer==
 
==Composition, or the Composer==
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