Existence of an axiom question in relation to $\mathsf{Infinity}$

Question

Original Post
This may be a stupid question, but does there axist an axiom $\phi$ that is independent of $\mathsf{ZFC}$, and not equivalent to the axiom of $\mathsf{Infinity}$, such that $\left(\mathsf{ZFC} -\mathsf{Infinity}+\phi\right) \vdash \mathsf{Infinity}$? Can such a $\phi$ exist? Or will all $\phi$ be equivalent to $\mathsf{Infinity}$?

Edit1: Danul's answer certainly answers the question, but I am wondering if there exists a weaker axiom $\phi$.
Edit2 I realize I had no idea what I meant by weaker axiom

Edit3, Conclusions
I guess I want to know whether or not every axiom $\phi$ such that $\mathsf{ZFC}-\mathsf{Infinity} + \phi$ proves that "There exists an inductive set" is such that $\phi$ is either stronger than $\mathsf{Infinity}$ or equivalent to $\mathsf{Infinity}$ and I guess this is the case as $\left(\mathsf{ZFC} -\mathsf{Infinity} + \phi\right) \vdash \mathsf{Infinity} \Longrightarrow \left(\phi \Longrightarrow \mathsf{Infinity}\right) \lor \left(\phi \longleftrightarrow \mathsf{Infinity}\right)$

In this beginning, I really just wanted to know if we could prove that a countably infinite set existed without simply postulating that one such exists. Can we? What is the most minimal assumption that we need to make? Is it really just $\mathsf{Infinity}$?
Edit4
AH, correct, Thanks Danul. I guess what I meant was: $$ \left(\mathsf{ZFC} -\mathsf{Infinity} + \phi\right) \vdash \mathsf{Infinity} \Longrightarrow \left(\phi \vdash \mathsf{Infinity}\right) $$ As the axiom of infinity cannot be derived from the rest of the axioms of $\mathsf{ZFC}$, so it must be the case that $\phi \vdash \mathsf{Infinity}$. Steven Stadniski answered my question in the comments. Thanks everyone.

Edit5
I am making all sorts of reasoning errors! We have that $$(\mathsf{ZFC} - \mathsf{Infinity} + \phi) \vdash \mathsf{Infinity} \not{\rightarrow} \phi \vdash \mathsf{Infinity}$$

Because $\phi = \mathsf{Extensionality} \longrightarrow \mathsf{Infinity} $ is a counterexample.

False logic will be the end of me.

Answer 1

Make $\varphi$ be Axiom of infinity plus $GCH$. Then you have that $ZFC-\text{infinity}+\varphi\vdash\text{infinity}$, but $\varphi$ is definitely much stronger than $\text{infinity}$.

Edit: I don't know what you mean by weaker axioms. But remember that $ZFC$ can produce a model (in-fact a countable transitive model) for $ZFC-\text{infinity}+\neg{\text{infinty}}$, namely the hereditarily finite sets (or $V_{\omega}$ if you prefer). So in a sense you can't really be any weaker.

Edit 2: At your current edit (You should keep track of the edits btw or else the entire conversation will become useless, so maybe not delete the original post) the question is trivial. Let $\varphi=\neg{\text{infinity}}$. Now the right hand side is trivial because of the way $\implies$ is defined.

Edit 3: Edit 2 is meant to show that your reasoning is faulty. $\neg{\text{infinity}}$ and $\text{infinity}$ are contradictory, but if you set $\varphi=\neg{\text{infinity}}$, you still get $ZFC-\text{infinity}+\varphi\vdash \text{Infinity}\implies(\varphi\implies\text{Infinity})\lor(\varphi\iff{\text{infinity}})$.

For an answer to your informal question, see the first edit for this solution.

Answer 2

If you are looking for a simpler, more intuitive alternative to the ZF axiom of infinity -- reading between the lines here -- you might consider simply assuming that there exists a set $S$ on which there is defined an injective, but not surjective function $f$, i.e. $S$ is Dedekind-infinite. Could it get more minimal? From such a set, you can extract at least one subset $N$ that satisfies Peano's axioms where $f$ serves as the successor function.