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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.)

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.)

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.)

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.)

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).

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.

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).

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.)