User blog comment:B1mb0w/Strong D Function/@comment-1605058-20150624061803

I don't know how strong the Stromg D function is, but I can say with certainty that it's (a lot) weaker than \(f_{\varepsilon_0}\). The reason - PA can prove this function to be total, argument for which I'm going to sketch now:

PA can show that ordinal \(\omega^\omega\) is well-founded. To every instance of D function, \(D(x_1,...,x_n)\) we connect an ordinal \(F(x_1,...,x_n)=\omega^n+\omega^{n-1}x_1+\omega^{n-2}x_2+...+\omega^0x_n\). Now, by transfinite recursion of \(F(x_1,...,x_n)\) up to \(\omega_\omega\) we verify that \(D(x_1,...,x_n)\) is well-defined.

Let's say it's clear for two-argument arrays, so for ordinal \(<\omega^3\). Now assume function is well-defined whenever it has ordinal less than \(F(x_1,...,x_n)\). Now, \(D(x_1,...,x_n)=D(x_1-1,D(x_1,...,x_n-1),...,D(x_1,...,x_n-1)\) (or some special case, it is similar to check other cases). \(F(x_1,...,x_n-1)<F(x_1,...,x_n)\), so \(D(x_1,...,x_n-1\) is well-defined and equal to, say, \(D\). Then \(D(x_1,...,x_n)=D(x_1-1,D,...,D)\). Now again \(F(x_1-1,D,...,D)<F(x_1,...,x_n)\), so it is well-defined completing induction step.

Now every function provably total in PA is slower than \(f_{\varepsilon_0}\), and the proof suggests the function isn't even stronger than \(f_{\omega^\omega}\).