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==Cleanup==
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== Simpler solution ==
I came up with a simpler lambda calculus solution for
<
(lambda f: lambda x: f(f)(x)) \
(lambda f: lambda x: 1 if x == 0 else x * f(f)(x-1))
</syntaxhighlight>
But, this is original research I guess, so I won't put it on the page. Also the existing explanation might help people trying to understand the Y combinator. <small class="autosigned">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/15.203.233.79|15.203.233.79]] ([[User talk:15.203.233.79|talk]]) 22:19, 27 April 2015 (UTC)</small><!-- Template:Unsigned IP --> <!--Autosigned by SineBot-->
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