Timeline for Multiple roots in the classical Laver tables
Current License: CC BY-SA 4.0
2 events
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| Mar 8, 2019 at 17:52 | comment | added | Joseph Van Name | Under large cardinal hypotheses, the answer is affirmative if one removes condition 3. If $r$ is a term, then define a term $r^{\sharp}(x)=r(x)*x$. Then $$s^{\sharp}\circ t^{\sharp}=(t\circ s)^{\sharp}$$, and therefore $$s^{\sharp}\circ t^{\sharp}=(t\circ s)^{\sharp}=((t*s)\circ t)^{\sharp}=t^{\sharp}\circ(t*s)^{\sharp}.$$ So, if $x=s^{\sharp}(a),y=(s*r)^{\sharp}(a)$, then $r^{\sharp}(x)=s^{\sharp}(y)$ and with large cardinals, one can make $u(x,y)\neq 2^{n}$. The problem with this example is that $$\gcd(2^{n},y)=\gcd(2^{n},a)<\gcd(2^{n},x).$$ | |
| Mar 8, 2019 at 14:07 | history | asked | Joseph Van Name | CC BY-SA 4.0 |