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Accepting the convention that it is a logical axiom that the universe is nonempty, the answer is yes. We will formalize the transitive closure axiom schema (TC) as follows: for any definable (with parameters) binary relation $R,$ if for all $x,$ $\{y: y R x\}$ is a set, then for all $x,$ there is a set $T$ such that $x \in T$ and $T$ is closed downwards under $R.$ (*) Of course, this can only be weaker than asserting the existence of a minimum such $T.$

For efficiency, we will prove Pairing, Union, Infinity, and Replacement from Extensionality, Separation, and TC.

Pairing: We first note that $\emptyset$ exists by applying separation to an arbitrary set. Next, for all $x,$ $\{x\}$ exists by applying TC to $x$ and the empty relation. Finally, for all $x, y,$ we get $\{x,y\}$ by applying TC to $x$ and the relation defined by $a R b$ iff $a = x$ and $b=y.$

Union: Fix a set $S.$ By Separation and Russell's paradox, there is $x \not \in S.$ Define $R$ by $a R b$ iff $a = x$ and $b \in S$ or $a \in S$ and $b \in a.$ Then we get $\bigcup S$ by applying TC to $x$ and $R.$

Infinity: Define a relation $R$ by $a R b$ iff $a$ and $b$ are natural numbers and $a=b+1.$ Then $\omega$ exists by applying TC to $\emptyset$ and $R.$

Replacement: Fix a set $S$ and a definable function $F.$ Fix $x \not \in S.$ Define $R$ by $a R b$ iff $a=x$ and $b \in S$ or $a \in S$ and $b = F(a).$ Then we get $F"S$ by applying TC to $x$ and $R.$

(*) Note that my formulation of TC only makes sense under the convention that the transitive closure of a relation is reflexive. Without this convention, then it's not clear we can prove the existence of $\{x\}$ from the axioms I specified. Of course, we can prove it exists from Separation and Power Set, which is included in the axioms listed in the question, but that feels overpowered for our purposes.

Edit: The question was updated with the intended formalization of the transitive closure schema. My TC here follows from Vladimir's version plus existence of $\{x\}$ for all $x,$ and the latter follows from Separation and Power Set.

Accepting the convention that it is a logical axiom that the universe is nonempty, the answer is yes. We will formalize the transitive closure axiom schema (TC) as follows: for any definable (with parameters) binary relation $R,$ if for all $x,$ $\{y: y R x\}$ is a set, then for all $x,$ there is a set $T$ such that $x \in T$ and $T$ is closed downwards under $R.$ (*) Of course, this can only be weaker than asserting the existence of a minimum such $T.$

For efficiency, we will prove Pairing, Union, Infinity, and Replacement from Extensionality, Separation, and TC.

Pairing: We first note that $\emptyset$ exists by applying separation to an arbitrary set. Next, for all $x,$ $\{x\}$ exists by applying TC to $x$ and the empty relation. Finally, for all $x, y,$ we get $\{x,y\}$ by applying TC to $x$ and the relation defined by $a R b$ iff $b = x$ and $a=y.$

Union: Fix a set $S.$ By Separation and Russell's paradox, there is $x \not \in S.$ Define $R$ by $a R b$ iff $b = x$ and $a \in S$ or $b \in S$ and $a \in b.$ Then we get $\bigcup S$ by applying TC to $x$ and $R.$

Infinity: Define a relation $R$ by $a R b$ iff $a$ and $b$ are natural numbers and $a=b+1.$ Then $\omega$ exists by applying TC to $\emptyset$ and $R.$

Replacement: Fix a set $S$ and a definable function $F.$ Fix $x \not \in S.$ Define $R$ by $a R b$ iff $b=x$ and $a \in S$ or $b \in S$ and $a = F(b).$ Then we get $F"S$ by applying TC to $x$ and $R.$

(*) Note that my formulation of TC only makes sense under the convention that the transitive closure of a relation is reflexive. Without this convention, then it's not clear we can prove the existence of $\{x\}$ from the axioms I specified. Of course, we can prove it exists from Separation and Power Set, which is included in the axioms listed in the question, but that feels overpowered for our purposes.

Edit: The question was updated with the intended formalization of the transitive closure schema. My TC here follows from Vladimir's version plus existence of $\{x\}$ for all $x,$ and the latter follows from Separation and Power Set.

Accepting the convention that it is a logical axiom that the universe is nonempty, the answer is yes. We will formalize the transitive closure axiom schema (TC) as follows: for any definable (with parameters) binary relation $R,$ if for all $x,$ $\{y: y R x\}$ is a set, then for all $x,$ there is a set $T$ such that $x \in T$ and $T$ is closed downwards under $R.$ (*) Of course, this can only be weaker than asserting the existence of a minimum such $T.$

For efficiency, we will prove Pairing, Union, Infinity, and Replacement from Extensionality, Separation, and TC.

Pairing: We first note that $\emptyset$ exists by applying separation to an arbitrary set. Next, for all $x,$ $\{x\}$ exists by applying TC to $x$ and the empty relation. Finally, for all $x, y,$ we get $\{x,y\}$ by applying TC to $x$ and the relation defined by $a R b$ iff $a = x$ and $b=y.$

Union: Fix a set $S.$ By Separation and Russell's paradox, there is $x \not \in S.$ Define $R$ by $a R b$ iff $a = x$ and $b \in S$ or $a \in S$ and $b \in a.$ Then we get $\bigcup S$ by applying TC to $x$ and $R.$

Infinity: Define a relation $R$ by $a R b$ iff $a$ and $b$ are natural numbers and $a=b+1.$ Then $\omega$ exists by applying TC to $\emptyset$ and $R.$

Replacement: Fix a set $S$ and a definable function $F.$ Fix $x \not \in S.$ Define $R$ by $a R b$ iff $a=x$ and $b \in S$ or $a \in S$ and $b = F(a).$ Then we get $F"S$ by applying TC to $x$ and $R.$

(*) Note that my formulation of TC only makes sense under the convention that the transitive closure of a relation is reflexive. Without this convention, then it's not clear we can prove the existence of $\{x\}$ from the axioms I specified. Of course, we can prove it exists from Separation and Power Set, which is included in the axioms listed in the question, but that feels overpowered for our purposes.

Accepting the convention that it is a logical axiom that the universe is nonempty, the answer is yes. We will formalize the transitive closure axiom schema (TC) as follows: for any definable (with parameters) binary relation $R,$ if for all $x,$ $\{y: y R x\}$ is a set, then for all $x,$ there is a set $T$ such that $x \in T$ and $T$ is closed downwards under $R.$ (*) Of course, this can only be weaker than asserting the existence of a minimum such $T.$

For efficiency, we will prove Pairing, Union, Infinity, and Replacement from Extensionality, Separation, and TC.

Pairing: We first note that $\emptyset$ exists by applying separation to an arbitrary set. Next, for all $x,$ $\{x\}$ exists by applying TC to $x$ and the empty relation. Finally, for all $x, y,$ we get $\{x,y\}$ by applying TC to $x$ and the relation defined by $a R b$ iff $a = x$ and $b=y.$

Union: Fix a set $S.$ By Separation and Russell's paradox, there is $x \not \in S.$ Define $R$ by $a R b$ iff $a = x$ and $b \in S$ or $a \in S$ and $b \in a.$ Then we get $\bigcup S$ by applying TC to $x$ and $R.$

Infinity: Define a relation $R$ by $a R b$ iff $a$ and $b$ are natural numbers and $a=b+1.$ Then $\omega$ exists by applying TC to $\emptyset$ and $R.$

Replacement: Fix a set $S$ and a definable function $F.$ Fix $x \not \in S.$ Define $R$ by $a R b$ iff $a=x$ and $b \in S$ or $a \in S$ and $b = F(a).$ Then we get $F"S$ by applying TC to $x$ and $R.$

(*) Note that my formulation of TC only makes sense under the convention that the transitive closure of a relation is reflexive. Without this convention, then it's not clear we can prove the existence of $\{x\}$ from the axioms I specified. Of course, we can prove it exists from Separation and Power Set, which is included in the axioms listed in the question, but that feels overpowered for our purposes.

Edit: The question was updated with the intended formalization of the transitive closure schema. My TC here follows from Vladimir's version plus existence of $\{x\}$ for all $x,$ and the latter follows from Separation and Power Set.

Accepting the convention that it is a logical axiom that the universe is nonempty, the answer is yes. We will formalize the transitive closure axiom schema (TC) as follows: for any definable (with parameters) binary relation $R,$ if for all $x,$ $\{y: y R x\}$ is a set, then for all $x,$ there is a set $T$ such that $x \in T$ and $T$ is closed downwards under $R.$ (*) Of course, this can only be weaker than asserting the existence of a minimum such $T.$

For efficiency, we will prove Pairing, Union, Infinity, and Replacement from Extensionality, Separation, and TC.

Pairing: We first note that $\emptyset$ exists by applying separation to an arbitrary set. Next, for all $x,$ $\{x\}$ exists by applying TC to $x$ and the empty relation. Finally, for all $x, y,$ we get $\{x,y\}$ by applying TC to $x$ and the relation defined by $a R b$ iff $a = x$ and $b=y.$

Union: Fix a set $S.$ By Separation and Russell's paradox, there is $x \not \in S.$ Define $R$ by $a R b$ iff $a = x$ and $b = S$ or $a \in S$ and $b \in a.$ Then we get $\bigcup S$ by applying TC to $x$ and $R.$

Infinity: Define a relation $R$ by $a R b$ iff $a$ and $b$ are natural numbers and $a=b+1.$ Then $\omega$ exists by applying TC to $\emptyset$ and $R.$

Replacement: Fix a set $S$ and a definable function $F.$ Fix $x \not \in S.$ Define $R$ by $a R b$ iff $a=x$ and $b=S$ or $a \in S$ and $b = F(a).$ Then we get $F"S$ by applying TC to $x$ and $R.$

(*) Note that my formulation of TC only makes sense under the convention that the transitive closure of a relation is reflexive. Without this convention, then it's not clear we can prove the existence of $\{x\}$ from the axioms I specified. Of course, we can prove it exists from Separation and Power Set, which is included in the axioms listed in the question, but that feels overpowered for our purposes.

Accepting the convention that it is a logical axiom that the universe is nonempty, the answer is yes. We will formalize the transitive closure axiom schema (TC) as follows: for any definable (with parameters) binary relation $R,$ if for all $x,$ $\{y: y R x\}$ is a set, then for all $x,$ there is a set $T$ such that $x \in T$ and $T$ is closed downwards under $R.$ (*) Of course, this can only be weaker than asserting the existence of a minimum such $T.$

For efficiency, we will prove Pairing, Union, Infinity, and Replacement from Extensionality, Separation, and TC.

Pairing: We first note that $\emptyset$ exists by applying separation to an arbitrary set. Next, for all $x,$ $\{x\}$ exists by applying TC to $x$ and the empty relation. Finally, for all $x, y,$ we get $\{x,y\}$ by applying TC to $x$ and the relation defined by $a R b$ iff $a = x$ and $b=y.$

Union: Fix a set $S.$ By Separation and Russell's paradox, there is $x \not \in S.$ Define $R$ by $a R b$ iff $a = x$ and $b \in S$ or $a \in S$ and $b \in a.$ Then we get $\bigcup S$ by applying TC to $x$ and $R.$

Infinity: Define a relation $R$ by $a R b$ iff $a$ and $b$ are natural numbers and $a=b+1.$ Then $\omega$ exists by applying TC to $\emptyset$ and $R.$

Replacement: Fix a set $S$ and a definable function $F.$ Fix $x \not \in S.$ Define $R$ by $a R b$ iff $a=x$ and $b \in S$ or $a \in S$ and $b = F(a).$ Then we get $F"S$ by applying TC to $x$ and $R.$

(*) Note that my formulation of TC only makes sense under the convention that the transitive closure of a relation is reflexive. Without this convention, then it's not clear we can prove the existence of $\{x\}$ from the axioms I specified. Of course, we can prove it exists from Separation and Power Set, which is included in the axioms listed in the question, but that feels overpowered for our purposes.