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| ===Step 5=== | | ===Step 5=== |
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− | <pre>
| + | '''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.
| + | 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. |
− | | |
− | 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 |
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| o-----------------------------------------------------------o | | o-----------------------------------------------------------o |
| + | </pre> |
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− | | I am still not sure what order I intended for the
| + | '''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:
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| | | |
| For example: | | For example: |
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− | The nodes that are right-labeled <K, KS, K(KS)>, | + | * 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). | + | * 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). | + | * 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. | + | * Therefore, the type of the application <math>\operatorname{K}(\operatorname{K}\operatorname{S})</math> is <math>A \Rightarrow A.</math> |
− | </pre> | |
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| ==Composition, or the Composer== | | ==Composition, or the Composer== |