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EDIT: About your question below, on the version of the ABC conjecture claimed in Mochizuki's work, it is clearly stated in Theorem A of the 4th paper. Anyways, for the benefit of the people that might read this question, I will state in very elementary terms a corollary of Theorem A in the following context: X is the projective line with the usual projective coordinates [x:y], and D is the divisor $[0:1] + [1:0] + [1:1]$ which makes the curve U=X\D hyperbolic (the degree of the canonical divisor $\omega$ of X in this case is -2 and the degree of D is 3, hence the degree of $\omega(D)$ is 1>0). Ok, here is the corollary (the notation is explained below):

Statement: Let $d$ be a positive integer and let $\epsilon>0$. There is a constant $C>0$ depending only on $d$ and $\epsilon$ such that the following is true: If $A,B$ are non-zero algebraic numbers with $A+B=1$, and if the degree over Q of the number field $K=Q(A)$ is at most d, then we have $H(A,B,1) < C(\Delta_K N_K(A,B,1))^{1+\epsilon}.$

Notation: Here I am using the same definition of $\Delta_K$, H(a,b,c) and $N_K(a,b,c)$ as in the paper on Siegel zeros of the question (this notation is explained in the first page of the paper). Well, if you check the reference you'll see that actually there is one difference: the paper uses N(a,b,c), not $N_K(a,b,c)$. However, in the above statement it is crucial that we must compute N(A,B,1) using the number filed K=Q(A), that's why I added this subscript.

I hope that the readers can see the difference between this version and the uniform ABC conjecture for the paper on Siegel zeros: the fact that here the constant C also depends on d, not only $\epsilon$.

A last trivial remark. To get the classical ABC conjecture with coprime integers a+b=c you take A=a/c, B=b/c and hence K=Q which makes $\Delta_K=1$, and N(A,B,1)=rad(abc).

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I don't think so. Mochizuki claims to have proved a diophantine result for points of bounded degree, while you need a uniform form of the ABC conjecture for the application that you mention.