We are assuming that a formula 𝜙(x,y) with 2 free variables defines a function means that for every x, there is a unique y such that 𝜙(x,y).
We assume that the axiom schema of Successor cardinals and the axiom schema of Ordinal inaccessibility are the result of replacing "f is a definable function" by "𝜙(x,y)
defines a function", replacing "𝑓(𝜆)=𝛾" by 𝜙(𝜆,𝛾), and replacing 𝛼≤𝑓(𝛽) by "there is a b such that 𝜙(𝛽,b) and 𝛼≤b".
We note that the axiom schema of Successor cardinals, is an immediate consequence of the axiom schema of Ordinal inaccessibility.(Suppose that 𝜙(x,y) defines a function and c ia an ordinal.
By Ordinal inaccessibility, there is an ordinal 𝛼 with the property that ∀𝛽<c:𝜙(𝛽,b)-->b<𝛼. Then ¬∀𝛾<(𝛼+1)∃𝜆<c:𝜙(𝜆,𝛾), since ¬𝜙(𝜆,𝛼) for all 𝜆<c.)
We now show that every instance of replacement is provable in Z + Ranks + Ordinal inaccessibility. It suffices to prove that reflection(for every finite set of formulas and ordinal 𝛼 there is an
ordinal 𝛽>𝛼 such that for every formual F(x1,x2,...,xn) in our finite set of formulas "for all x1,x2,...xn in V𝛽("F holds relativized to V𝛽"<-->F") holds. Suppose S is a finite set of formulas for which
reflection does not hold. Let 𝛼 be the least ordinal such that S does not reflect to a V𝛽 where 𝛽>𝛼. Define a formula 𝜙(x,y) by "x is an ordinal and y is the least ordinal such that if ∃tG is a
a subformula of a formula in S and ∃tG holds with parameters from Vx then there is a t∈Vy so that G(t,..) holds with these parameters, or x is not an ordinal and y=0.
𝜙(x,y) defines a function.
Proof: Suppose not and let c be the least ordinal x such that for all y, ¬𝜙(x,y). Let x1,...xn be the variables occurring in the formulas in S. Define a formula 𝜓(x,y) by x=(H,p1,...pn) for H a subformula of a formula of S of the form ∃tG and p1,...,pn are elements of V and ∃tG' holds where G' is obtained from G by substituting pi for xi, and y is the least ordinal such that there is a t∈Vy so that G' holds, or x is not such an (n+1)-tuple and y=0. By Ordinal inaccessibility there is an ordinal b greater than any y for which 𝜓(x,y) for some x.
Then 𝜙(c,d) where d is the least such b.
Let θ(x,y) be the formula "(x is a natural number) and (there is a function f with domain x+1, f(0)=𝛼 and 𝜙(f(n),f(n+1)) for (n+1) in the domain of f) and fx=y; or x is not a natural number and
y=0. By Ordinal inaccessibility there is an ordinal b greater than any y for which θ(x,y) for some x. Let d be the least such b. Then Vd reflects S.