0
$\begingroup$

Here :

http://googology.wikia.com/wiki/Chained_arrow_notation

the definition of the chained arrow notation is given.

How can I prove $$X\rightarrow 1\rightarrow Y=X$$ for every chains $X,Y$ using this definition ?

I thought about using this property :

If $X$ is an arbitary chain and we define $$f(n):=X\rightarrow n$$ $$g(n):=X\rightarrow n\rightarrow 2$$

Then, the rules give us $f(1)=g(1)=(X)$ (the value of the chain $X$) and we have $$X\rightarrow (n+1)\rightarrow 2=X\rightarrow (X\rightarrow n\rightarrow 2)\rightarrow 1=X\rightarrow (X\rightarrow n\rightarrow 2)$$ hence $$g(n+1)=X\rightarrow g(n)=f(g(n))$$ With induction this leads to $g(n)=f^n(1)$. Similar , defining $h(n):=X\rightarrow n\rightarrow 3$ , we get $h(n)=g^n(1)$ and so on. But I am not sure whether this approach can prove the claim for arbitary chains $Y$.

$\endgroup$
  • $\begingroup$ See if definition 4 helps you, sir. $\endgroup$ – Rohan Dec 26 '17 at 8:36
  • 1
    $\begingroup$ @Rohan I'm not quite sure what you're trying to say. If you mean the 4th rule in the definition of Conway's chains, then yes, it will probably be needed, but I'm not sure if you mean anything else. $\endgroup$ – Simply Beautiful Art Dec 26 '17 at 22:58
  • 1
    $\begingroup$ Also note that Peter is intending to use the definitions provided on googology.wikia $\endgroup$ – Simply Beautiful Art Dec 26 '17 at 22:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.