Prove: $\{α_1,...,α_n\} ⊨ α$ iff $\{α_1,...,α_{n−1}\} ⊨ (α_n→α)$. Recently began my second logic course and have been surprised at how very, very different it is from the first one.
My main struggle is that we have to prove things all the time, and I've never learnt how to properly set it up (I am a mere philosophy student who did the "easiest" maths in high school)
Thus, I would like to ask you guys for help, hoping to learn by example.
Our task is to prove

$\{α_1,...,α_n\} ⊨ α$ if and only if $\{α_1,...,α_{n−1}\} ⊨ (α_n→α)$

This seems intuitive enough to me, but I don't really know how to formally prove this (I don't think we are required to give a strictly formal proof only using symbols and such, but a proof that doesn't just explain in long, long sentences why it is so and so would be ideal). 
Our "hint" for this one is that we have to prove "in two directions". From left to right we are to assume $\{α_1,...,α_n\} ⊨ α$ to hold and "let truth assignment $ν$ satisfy $\{α_1,...,α_{n−1}\}$. Then show that, under the assumption, $ν$ satisfies $(α_n →α)$."
So here is what I am kind of thinking about:

1) Suppose $$\{α_1 , . . . , α_n \} \vDash α.$$
Let a truth assignment $v$ satisfy $\{α_1 , . . . , α_{n-1} \}$. Then either:

(i) $v$ assigns T to $\{α_1 , . . . , α_{n-1} \}$, $α_n$, and $α$, in which case it assigns T to $(α_n →α)$ (by the truth definition of implication); Or
(ii) $v$ assigns T to $\{α_1 , . . . , α_{n-1} \}$ but F to $α_n$, in which case whether it assigns T to $α$ or F to $α$ $(αn →α)$ will never the less be T; Or
(iii) $\{α_1 , . . . , α_n \}$ is unsatisfiable, in which case either $\{α_1 , . . . , α_{n-1} \}$ is also unsatisfiable and anything follows; or $\{α_1 , . . . , α_{n-1} \}$ is not unsatisfiable but αn contradicts it, in which case αn will be F whenever $\{α_1 , . . . , α_{n-1} \}$ is satisfied, and then whether or not $α$ is T, $(α_n →α)$ follows.

Thus: {α1,...,αn−1} |= (αn→α).
2) Suppose $$\{α_1,...,α_{n−1}\} ⊨ (α_n→α).$$
Let truth assignment $v$ satisfy $\{α_1 , . . . , α_n\}$. 
Then either
(1) v assigns T to $\{α_1,...,α_{n−1}\}$, $α_n$, and $α$ in which case it assigns T to both $\{α_1,...,α_n\}$ and $α$; Or
(2) v assigns T to $\{α_1,...,α_{n−1}\}$ and $α$, but not  $α_n$, in which case $\{α_1,...,α_{n}\}$ is unsatisfiable and anything follows; Or
(3) v assigns T to $\{α_1,...,α_{n−1}\}$, but not to $α$ and $α_n$, in which case $\{α_1,...,α_{n}\}$ is unsatisfiable and anything follows; Or
(4) $\{α_1,...,α_{n−1}\}$ is unsatisfiable in which case $\{α_1,...,α_{n}\}$ is also unsatisfiable and anything follows.
Thus: $\{α_1,...,α_n\} ⊨ α$
We have thus proven that if $\{α_1,...,α_n\} ⊨ α$ then $\{α_1,...,α_{n−1}\} ⊨ (α_n→α)$

Two question:
Is this even correct, or almost correct?
If so, how could it have been written better?
I also wonder about the "Let truth assignment v satisfy ..." phrases. Why are we told to use these? Doesn't that assume the sets are satisfiable? But they might not be.
 A: First, I would recommend you to have a look at Velleman's How to Prove it (2006). In my opinion, this book sheds light proof methods and strategies in an instructive and illuminating way. It is easy to cope with and I'm sure it is exactly what you need until get the feeling of how informal proofs work.
Your proof strategy is correct. (1) Whenever we want to prove that a statement $\varphi \leftrightarrow \psi$ in a form of a biconditional holds, we have to prove that both that a conditional holds in both directions, namely, $\varphi \rightarrow \psi$ and $\psi \rightarrow \varphi$. (2) In order to prove a conditional statement $\varphi \rightarrow \psi$, we have to assume that the antecedent $\varphi$ holds, and then obtain the consequent $\psi$. This is exactly what you are doing above.

Theorem:

$\{α_1, . . . , α_n\} \vDash α$ iff $\{α_1, . . . , α_{n-1}\} \vDash (α_n \to α)$

Proof ($\Rightarrow$): Our goal has a form of a  conditional statement. Hence we need to assume that the antecedent holds, derive the consequent.


*

*Suppose $$\{α_1, . . . , α_n\} \vDash α \tag{1}$$
Hence, by definition of semantic entailment it means that 

For all truth assignment $v$, if $[[α_i]]_v=1$, then $[[α]]_v=1$, where $i \leq n$.


*(Deriving the consequent) Now let $v'$ be a truth assignment such that $[[α_j]]_{v'}=1$, where $j \leq n-1$. We need to show that $[[α_n \to α]]_{v'}=1$. Since $α_n$ can only be true or false in $v'$ we can divide our proof in two cases:

*Case 1: $[[α_n]]_{v'}=1$. Then by our assumption (1) $[[α]]_{v'}=0$ cannot be the case. This shows that $[[α_n \to α]]_{v'}=1$.

*Case 2: $[[α_n]]_{v'}=0$. This shows that $[[α_n \to α]]_{v'}=1$ (by vacuity).
This shows that if $\{α_1, . . . , α_n\} \vDash α$ then $\{α_1, . . . , α_{n-1}\} \vDash (α_n \to α)$, as required.
Proof ($\Leftarrow$): Same strategy. Supose:
$$\{α_1, . . . , α_{n-1}\} \vDash (α_n \to α) \tag{2}$$
Derive $\{α_1, . . . , α_n\} \vDash α$.
Can your continue from here?
