Proof of a tautological entailment in Peter Smith's Introduction to Formal Logic I am currently working my way through Peter Smith's Introduction to Formal Logic and tried to prove the following theorem:
Theorem. (α ⟷ β), (β ⟷ γ) ⊨ (α ⟷ γ).
Proof. Suppose (α ⟷ β) and (β ⟷ γ) are true. We will consider two cases.

Case 1. α is true. Then since (α ⟷ β) is true, β is true. Then since (β ⟷ γ) is true, γ is true. Thus, α ⊨ γ.


Case 2. γ is true. Then since (β ⟷ γ) is true, β is true. Then since (α ⟷ β) is true, α is true. Thus, γ ⊨ α.

Since α ⊨ γ and γ ⊨ α, α ≈ γ and therefore, ⊨ (α ⟷ γ). Thus, we can conclude that (α ⟷ β), (β ⟷ γ) ⊨ (α ⟷ γ).    ◻
I am unsure whether my proof works (as this is one of my first proofs in logic). Would this be correct? Thank you in advance for your help!
EDIT: After some feedback, I tried to correct my proof. I hope that's better now:
Proof. Suppose (α ⟷ β) and (β ⟷ γ) are true under the interpretation being considered. We will prove that it follows under that valuation of the relevant atoms that (α ⟷ γ) is true. Suppose α is true under that interpretation. Then since (α ⟷ β) is true under that interpretation, β is true. Then since β and (β ⟷ γ) are true under that interpretation, γ is true. Thus, if α is true under the interpretation being considered, γ is true und therefore, (α → γ) is true. Suppose α is false under that interpretation. Then obviously, (α → γ) is true. 
Now suppose γ is true under that interpretation. Then since (β ⟷ γ) is true under that interpretation, β is true. Then since β and (α ⟷ β) are true under that interpretation, α is true. Thus, if γ is true under the interpretation being considered, α is true und therefore, (γ → α) is true. Suppose γ is false under that interpretation. Then obviously, (γ → α) is true.
Since (α → γ) and (γ → α) are true under the given valuation of atoms, (α ⟷ γ) is true under that interpretation. Thus, we can conclude that (α ⟷ β), (β ⟷ γ) ⊨ (α ⟷ γ).    ◻
 A: You have the right idea, but there are some small errors.  First of all, $(\alpha \leftrightarrow \beta), (\beta \leftrightarrow \gamma) \vDash \alpha \leftrightarrow \gamma$ means that under all interpretations in which both $\alpha \leftrightarrow \beta$ and $\beta \leftrightarrow \gamma$ are true, $\alpha \leftrightarrow \gamma$ is also true.  (In propositional logic, an interpretation is given by an assignment of truth values to all the propositional letters--what Smith calls a valuation.  In quantified logic, interpretations are more complicated.  I'm not sure where you are in Smith's book, so I'm not sure which kind of logic you're referring to.)
To prove something about all interpretations, you should start by saying that you are going to consider an arbitrary interpretation.  So in your first sentence, when you say that $\alpha \leftrightarrow \beta$ and $\beta \leftrightarrow \gamma$ are true, what you mean is that they are true under the interpretation being considered.  You now have to prove that $\alpha \leftrightarrow \gamma$ is true under that interpretation.
You then say that you are going to consider two cases, but what you present are not really cases.  Rather, what you should say is that you are going to prove two claims.  The first claim is that $\alpha \to \gamma$ is true under the interpretation you are considering.  You assume that $\alpha$ is true under the interpretation you are considering and show that $\gamma$ must also be true, so that establishes that $\alpha \to \gamma$ is true under that interpretation.  But you haven't shown at this point that for every interpretation under which $\alpha$ is true, $\gamma$ is true, so you shouldn't say $\alpha \vDash \gamma$.  What you should say is that $\alpha \to \gamma$ is true in the interpretation under consideration.  Similarly, the second argument establishes the claim that $\gamma \to \alpha$ is true under that interpretation, and the two claims together imply that $\alpha \leftrightarrow \gamma$ is true, as required.
Notice that you have not proven that $\alpha \leftrightarrow \gamma$ is true under all interpretations; you have only shown that it is true under all interpretations in which $\alpha \leftrightarrow \beta$ and $\beta \leftrightarrow \gamma$ are both true.  So you have not shown $\vDash \alpha \leftrightarrow \gamma$; you have only shown $(\alpha \leftrightarrow \beta), (\beta \leftrightarrow \gamma) \vDash \alpha \leftrightarrow \gamma$.
A: I've not read the book, but $\vDash$ indicates semantic entailment and calls for the use of a truth tree, truth table, or a metalogic proof. You seem to want a metalogic proof. To do that we need to show that there exists no interpretation where the premises are true and the conclusion is false. We tend to do this via reductio ad absurdum/proof by contradiction, and the valuation function. Different authors do approach things in different ways, though.
Proof
We'll prove that $\{\alpha\leftrightarrow\beta,~\beta\leftrightarrow\gamma\}\vDash\alpha\leftrightarrow\gamma$.
Assume for reductio that $v(\alpha\leftrightarrow\beta)=v(\beta\leftrightarrow\gamma)=1$ and $v(\alpha\leftrightarrow\gamma)=0$. As $v(\alpha\leftrightarrow\gamma)=0$ then $v(\alpha)\neq v(\gamma)$. Wlog, let $v(\alpha)=1$, then $v(\beta)=1$, which means $v(\gamma)=v(\alpha)=1$; a contradiction.
