Changes

→‎Step 4: markup
Line 1,059: Line 1,059:  
For example, we might begin as follows:
 
For example, we might begin as follows:
   −
<pre>
+
{| align="center" cellpadding="8" width="90%"
o-----------------------------------------------------------o
+
|
|                                                          |
+
<math>\begin{array}{l}
|            B=>C  C                                        |
+
(y \overset{ }{\underset{B}{\Downarrow}} ~
(y: (x: z:    ): ):                                    |
+
(x \overset{ }{\underset{A}{\Downarrow}} ~
|    B  A A     B C                                     |
+
  z \overset{A}{\underset{B \Rightarrow C}{\Downarrow}}
|                                                          |
+
  ) \overset{B}{\underset{C}{\Downarrow}}
o===========================================================o
+
  ) \overset{ }{\underset{C}{\Downarrow}}
|                                                          |
+
\\ \\
|            A=>B B         B=>C  C                       |
+
=
((x: (y: K:    ): ): (x: z:    ): ):                    |
+
\\ \\
|      A   B  B    A A  A    B C                   |
+
((x \overset{ }{\underset{A}{\Downarrow}} ~
|                                                          |
+
(y \overset{ }{\underset{B}{\Downarrow}} ~
o===========================================================o
+
  K \overset{B}{\underset{A \Rightarrow B}{\Downarrow}}
|                                                          |
+
  ) \overset{A}{\underset{B}{\Downarrow}}
|            A=>B  B  B=>C  (A=>B)=>(A=>C) A=>C C      |
+
  ) \overset{ }{\underset{B}{\Downarrow}} ~
|   (x: ((y: K:    ): (z:    S:              ):    ): ):    |
+
(x \overset{ }{\underset{A}{\Downarrow}} ~
|    A   B B    A  A     A=>(B=>C)       A=>B  A  C   |
+
   z \overset{A}{\underset{B \Rightarrow C}{\Downarrow}}
|                                                          |
+
   ) \overset{B}{\underset{C}{\Downarrow}}
o===========================================================o
+
  ) \overset{ }{\underset{C}{\Downarrow}}
|                                                          |
+
\\ \\
|  ...                                                    |
+
=
|                                                          |
+
\\ \\
o-----------------------------------------------------------o
+
(x \overset{ }{\underset{A}{\Downarrow}} ~
</pre>
+
((y \overset{ }{\underset{B}{\Downarrow}} ~
 +
   K \overset{B}{\underset{A \Rightarrow B}{\Downarrow}}
 +
  ) \overset{A}{\underset{B}{\Downarrow}} ~
 +
  (z \overset{A}{\underset{B \Rightarrow C}{\Downarrow}} ~
 +
   S \overset{A \Rightarrow (B \Rightarrow C)}{\underset{(A \Rightarrow B) \Rightarrow (A \Rightarrow C)}{\Downarrow}}
 +
  ) \overset{A \Rightarrow B}{\underset{A \Rightarrow C}{\Downarrow}}
 +
   ) \overset{A}{\underset{C}{\Downarrow}}
 +
  ) \overset{ }{\underset{C}{\Downarrow}}
 +
\\ \\
 +
=
 +
\\ \\
 +
\ldots
 +
\end{array}</math>
 +
|}
    
If this strategy is successful it suggests that the proof tree can be grown in a stepwise equational fashion from a seed term of the appropriate species, in other words, from a contextual, embedded, or paraphrastic specification of the desired term.
 
If this strategy is successful it suggests that the proof tree can be grown in a stepwise equational fashion from a seed term of the appropriate species, in other words, from a contextual, embedded, or paraphrastic specification of the desired term.
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