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Synthesis of (lambda(f) ...) and `(lambda(x0) ...)' in [28]
Here we have another conflict between a free variable and one which is lambda-bound.
The conflict exists between the `x' in `(lambda(x)'
and the `x' in the expression `(x(x(x x0)))'. To dissolve the
conflict the lambda-bound `x' has been renamed `x1'.
![\begin{lstlisting}[language=Scm]{}
(lambda(x)
((lambda(f)
(lambda(x1)
(f(f x1)))))
(lambda(x0)
(x(x(x x0))))) [29]
\end{lstlisting}](img217.gif)
Now the synthesis can be performed:
![\begin{lstlisting}[language=Scm]{}
(lambda(x)
(lambda(x1)
((lambda(x0)
(x(x(x x0))))
((lambda(x0)
(x(x(x x0))))
x1)))) [30]
\end{lstlisting}](img218.gif)
Georg P. Loczewski
2004-03-05