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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 is not an ordinal" or b<𝛼. Then ¬∀𝛾<(𝛼+1)∃𝜆<c:𝜙(𝜆,𝛾), since ¬𝜙(𝜆,𝛼) for all 𝜆<c.)

Z + Ranks + Ordinal inaccessibility has the same consequences as Z + Ranks + Ordinal Replacement.

Proof: Suppose Ordinal Replacement holds, 𝜙(x,y) defines a function, 𝛾 is an ordinal, and for every ordinal 𝛼 (∃𝛽<𝛾∃b:𝛼≤b∧𝜙(𝛽,b)). Let 𝜓(x,y) be the formula ("x is an ordinal" and "y is an ordinal" and 𝜙(x,y)) or ("x is an ordinal" and (∀t(𝜙(x,y)-->"t is not an ordinal") and y=0) or ("x is not an ordinal" and y=x). By Ordinal Replacement ∃𝐵∀𝑦(𝑦∈𝐵↔∃𝑥∈𝛾𝜓(𝑥,𝑦)). But then every ordinal is in UB. Now suppose Ordinal inaccessibility holds, 𝜙(x,y) is a formula in two free variables x,y, ∀𝑥[𝑜𝑟𝑑𝑖𝑛𝑎𝑙(𝑥)→∃!𝑦(𝑜𝑟𝑑𝑖𝑛𝑎𝑙(𝑦)∧𝜙(𝑥,𝑦))], and A is a set of ordinals. Let 𝜓(x,y) be the formula ("x is an ordinal" and "y is an ordinal" and 𝜙(x,y)) or ("x is not an ordinal" and x=y). Then 𝜓(x,y) defines a function. By Ordinal inaccessibility, there is an ordinal 𝛼 with the property ∀𝛽<(UA)∀b(𝜓(𝛽,b)-->b<𝛼). Then there is a set B such that

   b∈B<-->b∈𝛼∧∃𝑥∈𝐴(𝜙(x,b)). Therefore Ordinal Replacement holds.

If ZF is consistent then Z + Ranks + Ordinal inaccessibility + Choice does not prove Replacement.

Proof: See the answer of Joel David Hamkins to "Is full Replacement provable in Z + Ordinal Replacement?".(His argument shows that in a model of ZFC, ⟨𝑉𝜔1,∈⟩ satisfies Z + Ranks + Ordinal Replacement + Choice but not all instancess of Replacement.)