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		338 bytes added
	
		,  19:26, 24 March 2009
	
 
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|   | ===Step 4===  |   | ===Step 4===  | 
|   | + |  | 
|   | + | Repeat the development in Step 1, but this time articulating the type information as we go.  | 
|   |  |   |  | 
|   | <pre>  |   | <pre>  | 
| − | Repeat the development in Step 1,
  |   | 
| − | but this time articulating the
  |   | 
| − | type information as we go.
  |   | 
| − | 
  |   | 
|   | o---------------------------------------------------------------------o  |   | o---------------------------------------------------------------------o  | 
|   | |                                                                     |  |   | |                                                                     |  | 
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|   | |                                                                     |  |   | |                                                                     |  | 
|   | o---------------------------------------------------------------------o  |   | o---------------------------------------------------------------------o  | 
|   | + | </pre>  | 
|   |  |   |  | 
| − | The foregoing development has taken us from the  | + | The foregoing development has taken us from the typed parse tree for the definiens <math>((xy)z)\!</math> to the typed parse tree for the explicated definiendum <math>(x(y(z(\operatorname{K}((\operatorname{S}\operatorname{K})\operatorname{S}))~))),</math> which gives us both the construction of the composition combinator <math>\operatorname{P}</math> in terms of primitive combinators:  | 
| − | typed parse tree for the definiens ((xy)z) to the  |   | 
| − | typed parse tree for the explicated definiendum  |   | 
| − | (x(y(z(K((SK)S)) ))), which gives us both the  |   | 
| − | construction of the composition combinator P  |   | 
| − | in terms of primitive combinators:  |   | 
|   |  |   |  | 
| − |    P  =  (K((SK)S))
  | + | {| align="center" cellpadding="8" width="90%"  | 
|   | + | | <math>\begin{matrix}\operatorname{P} & = & (\operatorname{K}((\operatorname{S}\operatorname{K})\operatorname{S}))\end{matrix}</math>  | 
|   | + | |}  | 
|   |  |   |  | 
| − | and also the proof tree for the proposition type of P:  | + | and also the proof tree for the type proposition of <math>\operatorname{P},</math> as follows:  | 
|   |  |   |  | 
|   | + | <pre>  | 
|   |          S   K  |   |          S   K  | 
|   |          o   o  |   |          o   o  |