TPID 4

[sheldonison]

Indeed it applies to all real-analytic superfunctions !

How so? Here \( \theta(z) \) is an entire 1-cyclic function.
\( \text{tet}(z+1)=\exp(\text{tet}(z)) \)
\( \text{tet}(z+1)\times \theta(z) \;<>\; \exp(\theta(z) \times \text{tet}(z))\;\; \) unless theta(z)=1 everywhere

But if you replace tet(z) with b^z, then it works, so that's how I interpreted the Op's proof, given that the proof never mentioned superfunctions or anything like that.

\( b^{z+1}\times \theta(z) \;=\; b^{\theta(z) \times b^z}\;\; \) for any entire theta(z) function